Single Topic Note

Chapter 1

Example 1. Simplify the following: (a)$i^{49}$, (b)$i^{103}$.

Solution.$(a)$We divide 49 by 4 and we get$$ \begin{aligned} 49 & =4 \times 12+1 \\ i^{49} & =i^{4 \times 12+1}=\left(i^4 \right)^{12}\left(i^1 \right)=(1)^{12}(i)=i \end{aligned} $$(b) we divide 103 by 4, we get$$ \begin{aligned} & 103=4 \times 25+3 \\ & i^{103}=i^{4 \times 25+3}=\left(i^4 \right)^{25}\left(i^3 \right)=(1)^{25}(-i)=-i \qquad \textbf{Ans}. \end{aligned} $$

Example 2. Multiply 3 + 4i by 7 − 3i

$$ \begin{aligned} z_1 \cdot z_2 &=(3+4 i) \cdot(7-3 i) \\ &=21-9 i+28 i-12 i^2 \\ &=21-9 i+28 i-12(-1) \qquad{\left[because i^2=-1 \right]} \\ &=21-9 i+28 i+12 \\ &=33+19 i \qquad \text{\textbf{Ans}. } \end{aligned} $$

Example 3. Find out the conjugate of a complex number 7 + 6i.

Solution.$Le : z = 7 + 6i$To find the conjugate complex number of$7 + 6i$we change the sign of an imaginary number.
Conjugate of$z = \overline{z} = 7 - 6i$ Ans.

Example 4. Divide 1 + i by 3 + 4i.

Solution.$$ \begin{aligned} \frac{1+i}{3+4i} &= \frac{1+i}{3+4i}timesfrac{3-4i}{3-4i} \\ & =\frac{3-4i+3i-4i^2}{9-16i^2} \\ & =\frac{3-4i+3i-4i^2}{9-16i^2} \\ & =\frac{3-i+4}{9+16} = \frac{7}{25}-\frac{1}{25}i \end{aligned} $$

Example:5 If$\begin{aligned} a&=\cos \theta+i \sin \theta, \text{Prove that}: 1+a+a^2=(l+2 \cos \theta)(\cos \theta+i \sin \theta) \end{aligned}$.

Solution. Here we have$a=\cos \theta+i \sin \theta$$$ \begin{aligned} 1+a+a^2 & =1+(\cos \theta+i \sin \theta)+(\cos \theta+i \sin \theta)^2 \\ & =1+\cos \theta+i \sin \theta+\cos ^2 \theta+2 i \sin \theta \cos \theta-\sin ^2 \theta \\ & =(\cos \theta+i \sin \theta)+\left(1-\sin ^2 \theta \right))+\cos ^2 \theta+2 i \sin \theta \cos \theta \\ & =(\cos \theta+i \sin \theta)+\cos ^2 \theta+\cos ^2 \theta+2 i \sin \theta \cos \theta \\ & =(\cos \theta+i \sin \theta)+2 \cos ^2 \theta+2 i \sin \theta \cos \theta \\ & =(\cos \theta+i \sin \theta)+2 \cos \theta(\cos \theta+i \sin \theta) \\ & =(\cos \theta+i \sin \theta)(1+2 \cos \theta) \qquad \textbf{\textit{Proved}} \end{aligned} $$

Example 6. If$a^2+b^2+c^2=1$and$b+ic=(1+a)z$, prove that$\frac{a+ib}{1+c}=\frac{1+iz}{1-iz}$.

Solution. Here, we have$$ \begin{aligned} b+i c&=(1+a) z \Rightarrow z=\frac{b+i c}{1+a} \\ \frac{1+i z}{1-i z} & =\frac{1+i \frac}{b+i c}{1+a}{1-i \frac{b+i c}{1+a}}=\frac{1+a+i b-c}{1+a-i b+c} \\ & =\frac{[(1+a+i b)-c]}{(1+a+c-i b)} \times \frac{(1+a+i b+c)}{(1+a+c+i b)}=\frac{(1+a+i b)^2-c^2}{(1+a+c)^2+b^2} \\ & =\frac{1+a^2-b^2+2 a+2 i b+2 i a b-c^2}{1+a^2+c^2+2 a+2 c+2 a c+b^2}=\frac{1+a^2-b^2-c^2+2 a+2 i b+2 i a b}{1+\left(a^2+b^2+c^2 \right)+2 a+2 c+2 a c} \end{aligned} $$Putting the value of$a^2+b^2+c^2=1$in the above, we get$$ \begin{aligned} & =\frac{1+a^2-\left(1-a^2 \right)+2 a+2 i b+2 i a b}{1+1+2 a+2 c+2 a c}=\frac{2\left(a^2+a+i b+i a b \right)}{2(1+a+c+a c)}=\frac{2(1+a)(a+i b)}{2(1+a)(1+c)}=\frac{a+i b}{1+c} \end{aligned} $$ Proved.

Example 7. If$z=\cos \theta+i \sin \theta$, prove that$\frac{2}{1+z}=1-i \tan \frac{\theta}{2}$.

Solution. Here, we have$z=\cos \theta+i \sin \theta$.$$ \begin{aligned} \text{(a) } \frac{2}{1+z} &= \frac{2}{1+(\cos \theta+i \sin \theta)} \\ &= \frac{2}{(1+\cos \theta)+i \sin \theta} \times \frac{(1+\cos \theta)-i \sin \theta}{(1+\cos \theta)-i \sin \theta} \\ &= \frac{2[(1+\cos \theta)-i \sin \theta]}{(1+\cos \theta)^2+\sin^2 \theta} \\ &= \frac{2[(1+\cos \theta)-i \sin \theta]}{1+2\cos \theta+\cos^2 \theta+\sin^2 \theta} \\ &= \frac{2[(1+\cos \theta)-i \sin \theta]}{2(1+\cos \theta)} \\ &= 1 - \frac{i \sin \theta}{1+\cos \theta} \\ &= 1 - i \frac{2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)}{2 \cos^2\left(\frac{\theta}{2}\right)} \\ &= 1 - i \tan\left(\frac{\theta}{2}\right) \qquad \textbf{Proved.} \end{aligned} $$

Example 8. If$x=\cos \theta+i \sin \theta, y=\cos \phi+i \sin \phi$, prove that$\begin{aligned} \frac{x-y}{x+y}&=i \tan \left(\frac{\theta-\phi}{2}\right)\qquad(M.U. 2008) \end{aligned}$

Solution. We have,$$ \begin{aligned} \frac{x-y}{x+y} & =\frac{(\cos \theta+i \sin \theta)-(\cos \phi+i \sin \phi)}{(\cos \theta+i \sin \theta)+(\cos \phi+i \sin \phi)} \ \\ & = \frac{(\cos \theta-\cos \phi)+i(\sin \theta-\sin \phi)}{(\cos \theta+\cos \phi)+i(\sin \theta+\sin \phi)} \ \\ & = \frac{\left[-2 \sin \left(\frac{\theta+\phi}{2}\right) \sin \left(\frac{\theta-\phi}{2}\right)+2 i \cos \left(\frac{\theta+\phi}{2}\right) \sin \left(\frac{\theta-\phi}{2}\right)\right]}{\left[2 \cos \left(\frac{\theta+\phi}{2}\right) \cos \left(\frac{\theta-\phi}{2}\right)+2 i \sin \left(\frac{\theta+\phi}{2}\right) \cos \left(\frac{\theta-\phi}{2}\right)\right]} \ \\ & = \frac{2 i \sin \left(\frac{\theta-\phi}{2}\right)\left[\cos \left(\frac{\theta+\phi}{2}\right)+i \sin \left(\frac{\theta+\phi}{2}\right)\right]}{2 \cos \left(\frac{\theta-\phi}{2}\right)\left[\cos \left(\frac{\theta+\phi}{2}\right)+i \sin \left(\frac{\theta+\phi}{2}\right)\right]}=i \tan \left(\frac{\theta-\phi}{2}\right)\qquad \text{\textbf{Proved}.} \end{aligned} $$

Example 9. Find the modulus and principal argument of the complex number

$$ \begin{aligned} &\textbf{Solution}.\qquad \frac{1+2 i}{1-(1-i)^2}=\frac{1+2 i}{1-(1-1-2 i)}=\frac{1+2 i}{1+2 i}=1=1+0 i \\ &thereforehspace{0.7in}\left|\frac{1+2 i}{1-(1-i)^2}\right|=|1+0 i|=\sqrt{1^2}=1 \qquad \textbf{Ans.} \end{aligned} $$Principal argument of$$ \begin{aligned} \frac{1+2 i}{1-(1-i)^2}&= \end{aligned} $$Principal argument of$1+0 i$$=\tan ^{-1} \frac{0}{1}=\tan ^{-1} 0=0^{circ} \text {. }$$Hence modulus$=1$and principal argument$=0^{circ}. Ans.

Example 10. Find the modulus and principal argument of the complex number : $ 1+\cos \alpha+i \sin \alpha \left(0<\alpha<\frac{\pi}{2}\right) $

Solution. Let$(1+\cos \alpha)+i \sin \alpha=r(\cos \theta+i \sin \theta)$Equating real and imaginary parts, we get$$ \begin{aligned} 1+\cos \alpha & =r \cos \theta \qquad..(1) \end{aligned} $$And$$ \begin{aligned} \sin \alpha & =r \sin thetahspace{3.84in}..(2) \end{aligned} $$Squaring and adding (1) and (2), we get$$ \begin{aligned} &r^2\left(\cos ^2 \theta+\sin ^2 \theta \right))=(1+\cos \alpha)^2+(\sin \alpha)^2 \\ &\Rightarrow r^2 =1+\cos ^2 \alpha+2 \cos \alpha+\sin ^2 \alpha=1+2 \cos \alpha+1 \\ & =2(1+\cos \alpha)=2\left(1+2 \cos ^2 \frac{\alpha}{2}-1 \right)=4 \cos ^2 \frac{\alpha}{2} \\ &\Rightarrow r =2 \cos \frac{\alpha}{2} \end{aligned} $$From (1), we have,$$ \begin{aligned} \cos \theta&=\frac{1+\cos \alpha}{r}=\frac{1+2 \cos ^2 \frac{\alpha}{2}-1}{2 \cos \frac{\alpha}{2}}=\cos \frac{\alpha}{2}\qquad..(3) \end{aligned} $$

From (2), we have,$$ \begin{aligned} \sin \theta&=\frac{\sin \alpha}{\gamma}=\frac{\sin \alpha}{2 \cos \frac{\alpha}{2}}=\frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos \frac{\alpha}{2}}=\sin \frac{\alpha}{2}\qquad..(4) \end{aligned} $$

Argument$$ \begin{aligned} &=\tan ^{-1} \frac{\sin \alpha}{1+\cos \alpha}=\tan ^{-1} \frac{2 \sin \frac{\alpha^2}{2} \cos \frac{\alpha}{2}}{1+2 \cos ^2 \frac{\alpha}{2}-1}=\tan ^{-1} \tan \frac{\alpha}{2}=\frac{\alpha}{2} \end{aligned} $$

General value of argument$$ \begin{aligned} &=2 \pi k+\frac{\alpha}{2} \end{aligned} $$\newline$$ \begin{aligned} \theta&=\frac{\alpha}{2} \end{aligned} $$satisfied both equations, (1) and (2),

Arg$$ \begin{aligned} (1+\cos \alpha+i \sin \alpha)&=\frac{\alpha}{2} \end{aligned} $$and modulus of$$ \begin{aligned} (1+\cos \alpha+i \sin \alpha)&=r=2 \cos \frac{\alpha}{2} \end{aligned} $$ Ans.
v

Example 11. If$z_1$and$z_2$are any two complex numbers, prove that$\left|z_1+z_2\right|^2+\left|z_1-z_2\right|^2=2\left[\left|z_1\right|^2+\left|z_2\right|^2\right]$

Solution. Let$$ \begin{aligned} z_1&=x_1+i y_1 \\ z_2&=x_2+i y_2 \\ \left|z_1+z_2\right|^2&=\left|\left(x_1+i y_1\right)+\left(x_2+i y_2\right)\right|^2 \\ &=\left|\left(x_1+x_2\right)+i\left(y_1+y_2\right)\right|^2 \\ &=\left(x_1+x_2\right)^2+\left(y_1+y_2\right)^2 \qquad \dots (1) \\ \text { Similarly } \qquad\quad\left|z_1-z_2\right|^2&=\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2 \qquad \dots (2) \\ \text { and }\qquad\left|z_1\right|^2&=x_1^2+y_1^2 \qquad \dots (3) \\ \left|z_2\right|^2&=x_2^2+y_2^2 \qquad \dots (4) \\ \text { L.H.S. }&=\left|z_1+z_2\right|^2+\left|z_1-z_2\right|^2=\left(x_1+x_2\right)^2+\left(y_1+y_2\right)^2+\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2 \end{aligned}$\hspace{5.5in} [Using (1) and (2)] $$$$ \begin{aligned} & \qquad=x_1^2+2 x_1 x_2+x_2^2+y_1^2+2 y_1 y_2+y_2^2 \\ & \qquad+x_1^2-2 x_1 x_2+x_2^2+y_1^2-2 y_1 y_2+y_2^2 \\ & \qquad =2\left[x_1^2+x_2^2+y_1^2+y_2^2\right]=2\left[\left(x_1^2+y_1^2\right)+\left(x_2^2+y_2^2\right)\right] \qquad \dots (5) \\ & \qquad=2\left[\left|z_1\right|^2+\left|z_2\right|^2\right]=\text { R.H.S. } \end{aligned} $$Proved.

Example 12. If$z_1$and$z_2$are two complex numbers such that$\left|z_1+z_2\right|=\left|z_1-z_2\right| \text {, prove that }$$arg \cdot z_1-arg \cdot z_2=\frac{\pi}{2}$(M. 2002, 2007)

Solution. Let$$ \begin{aligned} z_1&=x_1+i y_1 \\ z_2&=x_2+i y_2 \end{aligned} $$Given that$\hspace{.3in}\quad \left||z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|$\begin{array}{llrl}
\Rightarrow \left|\left(x_1+iy_1 \right)+\left(x_2+iy_2 \right)\right| =\left|\left(x_1+iy_1 \right)-\left(x_2+i y_2 \right)\right|
Rightarrow \left|\left(x_1+x_2 \right)+i \left(y_1+y_2 \right)\right| =\left|\left(x_1-x_2 \right)+\left(y_1-y_2 \right) i\right|
\Rightarrow \left(x_1+x_2 \right)^2+\left(y_1+y_2 \right)^2 =\left(x_1-x_2 \right)^2+\left(y_1-y_2 \right)^2
\Rightarrow x_1^2+x_2^2+2 x_1 x_2+y_1^2+y_2^2+2 y_1 y_2 =x_1^2+x_2^2-2 x_1 x_2+y_1^2+y_2^2-2 y_1 y_2

\end{array}$$ \begin{aligned} & \Rightarrow 2 x_1 x_2+2 y_1 y_2=-2 x_1 x_2-2 y_1 y_2 \\ & \Rightarrow 4 x_1 x_2+4 y_1 y_2=0 \\ & \Rightarrow x_1 x_2+y_1 y_2=0 \qquad \dots(1) \end{aligned} $$\text { Now, }$\quad arg \cdot z_1-arg \cdot z_2=\tan ^{-1}\left(\frac{y_1}{x_1}\right)-\tan ^{-1}\left(\frac{y_2}{x_2}\right)$$$ \begin{aligned} &=\tan ^{-1}\left[\frac{\frac{y_1}{x_1}-\frac{y_2}{x_2}}{1+\left(\frac{y_1}{x_1}\right)\left(\frac{y_2}{x_2}\right)}\right]=\tan ^{-1}\left(\frac{x_2 y_1-x_1 y_2}{x_1 x_2+y_1 y_2}\right) \\ &=\tan ^{-1}\left(\frac{x_2 y_1-x_1 y_2}{0}\right)=\tan ^{-1} \infty=\frac{\pi}{2} \qquad \textbf{Proved.} \end{aligned} $$arg$: (z_1)-arg : (z_2)=\frac{\pi}{3}$

Example 13. Find the complex number$z$if$\arg(z+1)=\frac{\pi}{6}$and$\arg(z-I)=\frac{2 \pi}{3}$. (M.U. 2009, 2000, 01,02, 03)

Solution. Let$\quad z=x+i y$, \par$\therefore \hspace{1in}\quad z+1=(x+1)+i y$We have that$$ \begin{aligned} \operatorname{Arg}(z+1) &=\tan ^{-1}\left(\frac{y}{x+1}\right)=\frac{\pi}{6} \\ \therefore \qquad\frac{y}{x+1}&=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\ \therefore \qquad\sqrt{3} y &=x+1 \\ \text { Now }\qquad z - 1 &= (x - 1) + iy \qquad\text { [From (1)] } \end{aligned} $$and$$ \begin{aligned} \tan^{-1}\left(\frac{y}{x-1}\right) &=\frac{2\pi}{3} \\ &\Rightarrow \qquad\frac{y}{x-1}=\tan 120^{\circ}\qquad\Rightarrow \quad \qquad\frac{y}{x-1}=-\cot 30^{\circ}=-\sqrt{3} \\ &\therefore \quad -y=\sqrt{3} x-\sqrt{3} \\ &\Rightarrow \quad -\sqrt{3} y=3 x-3 \\ & \end{aligned} $$Adding (2) and (3), we get$$ \begin{aligned} 0&=4 x-2 \Rightarrow 4 x=2 \Rightarrow x=\frac{1}{2} \end{aligned} $$Putting$x=\frac{1}{2}$in (2), we get$$ \begin{aligned} \sqrt{3} y&=\frac{1}{2}+1 \Rightarrow \sqrt{3} y=\frac{3}{2} \Rightarrow y=\frac{\sqrt{3}}{2} \end{aligned} $$Putting the values of$x$and$y$in (1), we get$$ \begin{aligned} z&=\frac{1}{2}+\frac{\sqrt{3}}{2} i \qquad \textbf{Ans.} \end{aligned} $$

Example 14. Prove that (i)$\left|z_1+z_2\right| \leq \left|z_1\right|+\left|z_2\right|$(ii)$\left|z_1-z_2\right| \geq \left|z_1\right|-\left|z_2\right|$

Solution. \hspace{.5in}(a) \hspace{.5in}(By Geometry) \hspace{
.5in}Let \hspace{1in}$z_1=x_1+iy_1$\hspace{.5in}and \hspace{5in}$z_2=x_2+iy_2$are the two complex numbers shown in the figure$\hspace{1.5in}\left|z_1\right|=O P, \quad \left|z_1\right|=O Q$

(i) Since any side is less than the
sum of the other two in a triangle.
In$triangle O P R, \quad O R (b) By Algebra.$\quad z_1+z_2=x_1+i y_1+x_2+i y_2=\left(x_1+x_2 \right)+i \left(y_1+y_2 \right)$Proved.

Example 15. Express in polar form :$1-\sqrt{2}+i$

Solution. Let$(1-\sqrt{2})+i=r(\cos \theta+i \sin \theta)$\hspace{1.5in}therefore \quad$$ \begin{aligned} 1-\sqrt{2} & =r \cos \theta \qquad \dots (1) \\ 1 & =r \sin \theta \qquad \dots (2) \end{aligned} $$Squaring and adding (1) and (2), we get$$ \begin{aligned} r^2\left(\cos ^2 \theta+\sin ^2 \theta \right)) & =(1-\sqrt{2})^2+1^2 \\ &\Rightarrow r^2 =1-2 \sqrt{2}+2+1 \qquad\Rightarrow r=\sqrt{4-2 \sqrt{2}} \end{aligned} $$Putting the value of$r$in (1) and (2), we get
\par$$ \hspace{1.9in}\begin{aligned} \cos \theta&=\frac{1-\sqrt{2}}{\sqrt{4-2 \sqrt{2}}} \quad \text { and } \quad \sin \theta=\frac{1}{\sqrt{4-2 \sqrt{2}}} \end{aligned} Hence, the polar form is $$$$ \begin{aligned} \sqrt{4-2 \sqrt{2}} \left{\frac{1-\sqrt{2}}{\sqrt{4-2 \sqrt{2}}}+i \frac{1}{\sqrt{4-2 \sqrt{2}}}\right} \end{aligned} $$

Example 16. Find the smallest positive integer n for which (Nagpur University, Winter 2004)$\begin{aligned} (\frac{1+i}{1-i})^n&=1. \end{aligned}$

Solution.$$ \begin{aligned} {\left[\frac{1+i}{1-i}\right]^n } & =1 \\ {\left[\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right]^n } & =1 \Rightarrow \left(\frac{1-1+2 i}{1+1}\right)^n=1 \\ (i)^n & =1=(i)^4 \Rightarrow n=4 \qquad \textbf{Ans.} \end{aligned} $$

Example 17.Find the square root of the complex number$5+12 i$

Solution. Let $\sqrt{5+12 i}=x+i y$\hspace{4.6in}\dots(1)

Squaring both sides of$(1)$, we get$5+12 i=(x+dot{y})^2=\left(x^2-y^2 \right)+i 2 x y$\hspace{2.2in}\dots(2)
Equating real and imaginary parts of (2), we get
and$$ \begin{aligned} x^2-y^2 & =5 \qquad\dots(3) \\ 2 x y & =12 \qquad\dots(4) \end{aligned} $$Now, \hspace{.3in}$$ \begin{aligned} x^2+y^2 & =\sqrt{\left(x^2-y^2 \right)^2+4 x^2 y^2}=\sqrt{(5)^2+(12)^2} \\ & =\sqrt{25+144}=\sqrt{169}=13 \end{aligned} $$\Rightarrow \hspace{.3in} x^2+y^2=13$\hspace{5.2in}\dots(5) \par Adding (3) and (5), we get$2 ^2=13-5=8 \Rightarrow x=\sqrt{\frac{18}{2}}=\sqrt{9}= \pm 3$\par Subtracting (3) from (5), we get$2 y^2=13-5=8 \Rightarrow y=\sqrt{\frac{8}{2}}=\sqrt{4}= \pm 2$\par. Since$x y$is positive,$x$and$y$are the same sign. Hence,$x= \pm 3, y= \pm 2$\par$therefore \hspace{1in} \sqrt{5+12 i}= \pm 3 \pm 2 i$i.e.$(3+2 i)$or$-(3+2 i) Ans.

Example 18. Prove that if the sum and product of two complex numbers are real then the two numbers must be either real or conjugate.$(M U .2008)$

Solution. Let$z_1$and$z_2$be the two complex numbers.

We are given that$z_1+z_2=a($real)

and \hspace{1in}$\quad z_1 \cdot z_2=b($real$)$

If the sum and product of the roots of a quadratic equation are given. Then the equation becomes
\hspace{1in}
\hspace{0in}$x^2$- (sum of the roots)$x+$product of the roots$=0$

\hspace{1.4in}$x^2-a x+b=0$

\hspace{1in}Root$=x=\frac{a \pm \sqrt{a^2-4 b}}{2}$

Case I. If$a^2>4 b \quad$Then both the roots are real

Case II. If$a^2<4 b$

Then$$ \begin{aligned} \text { one root } & =\frac{a}{2}+i \frac{\sqrt{4 b-a^2}}{2} \text { Second root } & =\frac{a}{2}-i \frac{\sqrt{4 b-a^2}}{2} \end{aligned} $$

These roots are conjugate to each other. Proved.

Example 19. Express$\begin{aligned} \frac{(\cos \theta+i \sin \theta)^8}{(\sin \theta+i \cos \theta)^4} \text{in the form} (x+i y). \end{aligned}$

Solution.$$ \begin{aligned} & \frac{(\cos \theta+i \sin \theta)^8}{(\sin \theta+i \cos \theta)^4}=\frac{(\cos \theta+i \sin \theta)^8}{(i)^4\left(\cos \theta+\frac{1}{i} \sin \theta \right))^4} \\ = & \frac{(\cos \theta+i \sin \theta)^8}{(\cos \theta-i \sin \theta)^4}=\frac{(\cos \theta+i \sin \theta)^8}{[\cos (-\theta)+i \sin (-\theta)]^4} \\ = &\frac{(\cos : \theta+i : \sin \theta)^8}{[(\cos : \theta+i: \sin : \\ \theta)^{-1}]^4} &= \frac{(\cos : \theta+i : \sin \theta)^8}{[(\cos : \theta+i: \sin : \theta)]^4} = (\cos : \theta+i: \sin : \theta)^{12} \\ = &\cos 12 : \theta+i \sin 12 : \theta \qquad \textbf{Ans}. \end{aligned} $$

Example 20. Prove that$(1+\cos \theta+i \sin \theta)^n+(1+\cos \theta-i \sin \theta)^n=2^{n+1} \cos ^n \frac{\theta}{2} \cos \frac{n \theta}{2}$

Solution.L.H.S.$=(1+\cos \theta+i \sin \theta)^n+(1+\cos \theta-i \sin \theta)^n$$$ \begin{aligned} &\qquad={\left[1+2 \cos ^2 \frac{\theta}{2}-1+2 i \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right]^n+\left[1+2 \cos ^2 \frac{\theta}{2}-1-i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right]^n } \\ &\qquad=\left[2 \cos ^2 \frac{\theta}{2}+2 i \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right]^n+\left[2 \cos ^2 \frac{\theta}{2}-2 i \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right]^n \\ &\qquad=\left(2 \cos \frac{\theta}{2}\right)^n \left[\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right]^n+\left(2 \cos \frac{\theta}{2}\right)^n \left[\cos \frac{\theta}{2}-i \sin \frac{\theta}{2}\right]^n \\ &\qquad=2^n \cos ^n \frac{\theta}{2}\left[\cos \frac{n \theta}{2}+i \sin \frac{n \theta}{2}\right]+2^n \cos ^n \frac{\theta}{2}\left[\cos \frac{n \theta}{2}-i \sin \frac{n \theta}{2}\right] \\ &\qquad= 2^n \cos ^n \frac{\theta}{2}\left[\cos \frac{n \theta}{2}+i \sin \frac{n \theta}{2}+\cos \frac{n \theta}{2}-i \sin \frac{n \theta}{2}\right] \\ &\qquad=2^n \cos ^n \frac{\theta}{2}\left(2 \cos \frac{n \theta}{2}\right)=2^{n+1} \cos ^n \frac{\theta}{2} \cos \frac{n \theta}{2}=\text { R.H.S. }\qquad \textbf{Proved.} \end{aligned} $$

Example 21. Evaluate$\begin{aligned} \left(\frac{1+\sin \alpha+i \cos \alpha}{1+\sin \alpha-i \cos \alpha}\right)^n \quad \end{aligned}$(MU. 2001, 2004, 2005)

Solution. We know that,$$ \begin{aligned} & 1=\sin ^2 \alpha+\cos ^2 \alpha \\ \Rightarrow & 1=\sin ^2 \alpha-i^2 \cos ^2 \alpha \\ \Rightarrow &1=(\sin \alpha+i \cos \alpha)(\sin \alpha-i \cos \alpha) \\ & \text { Adding } \sin \alpha+i \cos \alpha \text { both sides of }(1), \text { we get } \\ & \qquad1+\sin \alpha+i \cos \alpha=(\sin \alpha+i \cos \alpha)(\sin \alpha-i \cos \alpha)+(\sin \alpha+i \cos \alpha) \\ &\qquad=(\sin \alpha+i \cos \alpha)(\sin \alpha-i \cos \alpha+1) \\ & therefore \qquad\quad \frac{1+\sin \alpha+i \cos \alpha}{1+\sin \alpha-i \cos \alpha}=\sin \alpha+i \cos \alpha \\ &\qquad=\cos \left(\frac{\pi}{2}-\alpha \right))+i \sin \left(\frac{\pi}{2}-\alpha \right)) \\ & \Rightarrow \quad \left|(\frac{1+\sin \alpha+i \cos \alpha}{1+\sin \alpha-i \cos \alpha}\right)^n=\left{\cos \left(\frac{\pi}{2}-\alpha \right))+i \sin \left(\frac{\pi}{2}-\alpha \right))\right}^n \\ & \qquad \quad=\cos n \left(\frac{\pi}{2}-\alpha \right))+i \sin n \left(\frac{\pi}{2}-\alpha \right)) \qquad \textbf{Ans.} \end{aligned} $$

Example 22. Prove that the general value of$\theta$which satisfies the equation$\begin{aligned} (\cos: \theta)+i: \sin : \theta).(\cos : 2 \theta)+i: \sin 2: \theta)...(\cos : n \theta)+(i : \sin : n : \theta)&=\text{1 is} : \frac{4 : m : \pi}{n(n+1)},: \text{where m is any integer} \end{aligned}$

Solution.$\quad(\cos \theta+i \sin \theta)(\cos 2 \theta+i \sin 2 \theta) \ldots(\cos n \theta+i \sin n \theta)=1$\par$$ \begin{aligned} (\cos \theta+i \sin \theta)(\cos \theta+i \sin \theta)^2 \ldots(\cos \theta+i \sin \theta)^n=1 & \\ (\cos \theta+i \sin \theta)^{1+2 \ldots+n}=1 & \\ \cos \theta+i \sin \theta)^{\frac{s(n+1)}{2}}=(\cos 2 m \pi+i \sin 2 m \pi) & \\ \frac{n(n+1)}{2} \theta+i \sin \frac{n(n+1)}{2} \theta=\cos 2 m \pi+i \sin 2 m \pi & \qquad\text { \textbf{Proved}. } \end{aligned} $$

Example 23. If$\left(a_1+i b_1 \right) \cdot \left((a_2+i b_2 \right) \ldots \left(a_n+i b_n \right)=A+i B$, then prove that (i)$\tan ^{-1} \frac{b_1}{a_1}+\tan ^{-1} \frac{b_2}{a_2}+\ldots+\tan ^{-1} \frac{b_{mathrm{s}}}{a_{mathrm{n}}}=\tan ^{-1} \frac{B}{A}$(ii)$\begin{aligned} \left(a_2^2+b_1^2 \right)\left(a_2^1+b_2^1 \right) \quad \ldots \quad \left|(a_n^2+b_n^2 \right)&=A^2+B^2 \end{aligned}$

Solution.

Let
[$$ \begin{aligned} a_1 &= r_1cosalpha_1, &\qquad b_1 &= r_1sinalpha_1, \\ a_2 &= r_2cosalpha_2, &\qquad b_2 &= r_2sinalpha_2, \\ &vdotswithin{=} & &vdotswithin{=} \\ a_n &= r_ncosalpha_n, &\qquad b_n &= r_nsinalpha_n, \\ A &= R\cos \theta, &\qquad B &= R\sin \theta. \end{aligned} $$]

Given,
[
(a_1+ib_1)(a_2+ib_2)\cdots(a_n+ib_n)=A+iB.
]

Therefore,
[$$ \begin{aligned} &r_1(cosalpha_1+isinalpha_1) \\ r_2(cosalpha_2+isinalpha_2) \\ \cdots \\ r_n(cosalpha_n+isinalpha_n) \\ &\qquad=R(\cos \theta+i\sin \theta). \end{aligned} $$]

Using De Moivre's theorem,
[$$ \begin{aligned} &r_1r_2cdots r_n \\ \left[ \\ \cos(alpha_1+alpha_2+\cdots+alpha_n) \\ +isin(alpha_1+alpha_2+\cdots+alpha_n) \\ \right] \\ &\qquad=R(\cos \theta+i\sin \theta). \end{aligned} $$]

Comparing the moduli of both sides,
[
r_1r_2cdots r_n=R.
]

Hence,
[$$ \begin{aligned} R^2 \\ &=r_1^2r_2^2cdots r_n^2 \\ &=(a_1^2+b_1^2)(a_2^2+b_2^2)\cdots(a_n^2+b_n^2). \end{aligned} $$]

Since
[
R^2=A^2+B^2,
]
we get
[
\boxed{
(a_1^2+b_1^2)(a_2^2+b_2^2)\cdots(a_n^2+b_n^2)
=A^2+B^2
}.
]

Also, comparing the arguments,
[
alpha_1+alpha_2+\cdots+alpha_n=\theta.
]

Therefore,
[
\boxed{
\tan^{-1}\frac{b_1}{a_1}
+\tan^{-1}\frac{b_2}{a_2}
+\cdots+
\tan^{-1}\frac{b_n}{a_n}
=
\tan^{-1}\frac{B}{A}
}
]

[
\boxed{Proved.}
]

Example 24. If$\cos \alpha+\cos \beta+\cos \gamma=0=\sin \alpha+\sin \beta+\sin \gamma$, then prove that (i) $\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=\frac{3}{2}$(ii)$\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=0$(iii)$\cos (\alpha+\beta)+\cos (\beta+\gamma)+\cos (\gamma+\alpha)=0$(iv)$\sin (\alpha+\beta)+\sin (\beta+\gamma)+\sin (\gamma+\alpha)=0$(M.U. 2009)

Solution. Here, we have \par$\hspace{1in}(\cos \alpha+\cos \beta+\cos \gamma)+i(\sin \alpha+\sin \beta+\sin \gamma)=0$\par$\hspace{.7in}(\cos \alpha+i \sin \alpha)+(\cos \beta+i \sin \beta)+(\cos \gamma+i \sin \gamma)=0$therefore \hspace{1in} a+b+c=0$say \par where,$a=\cos \alpha+i \sin \alpha, b=\cos \beta+i \sin \beta$and$c=\cos \gamma+i \sin \gamma$Also, we can write$\hspace{.3in}(\cos \alpha-i \sin \alpha)+(\cos \beta-i \sin \beta)+(\cos \gamma-i \sin \gamma)=0$$ \begin{aligned} & \Rightarrow \quad(\cos \alpha+i \sin \alpha)^{-1}+(\cos \beta+i \sin \beta)^{-1}+(\cos \gamma+i \sin \gamma)^{-1}=0 \\ & \Rightarrow \qquad\quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0 \\ & \Rightarrow \qquad\quad \frac{b c+c a+a b}{a b c}=0 \Rightarrow a b+b c+c a=0 \\ & \text { But } \qquad\quad(a+b+c)^2=\left(a^2+b^2+c^2 \right)+2(a b+b c+c a) \qquad \dots(2) \\ & \qquad0=\left(a^2+b^2+c^2 \right)+0 \qquad \text{[From (1) and (2)]} \\ & \Rightarrow \qquad \quad a^2+b^2+c^2=0 \\ & \Rightarrow \qquad\quad(\cos \alpha+i \sin \alpha)^2+(\cos \beta+i \sin \beta)^2+(\cos \gamma+i \sin \gamma)^2=0 \\ & \Rightarrow \quad(\cos 2 \alpha+i \sin 2 \alpha)+(\cos 2 \beta+i \sin 2 \beta)+(\cos 2 \gamma+i \sin 2 \gamma)=0 \\ & \Rightarrow \qquad\quad(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma)+i(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma)=0 \\ & \Rightarrow \qquad\quad \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=0 \text {. } \qquad \dots(3) \\ & \Rightarrow \qquad\quad 2 \cos ^2 \alpha-1+2 \cos ^2 \beta-1+2 \cos ^2 \gamma-1=0 \\ & \Rightarrow \qquad\quad \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=\frac{3}{2} \qquad \dots(4) \\ \text{Further} \qquad 1-\sin ^2 \alpha+1-\sin ^2 \beta+1-\sin ^2 \gamma&=\frac{3}{2} \\ \Rightarrow \qquad\quad \sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma&=\frac{3}{2} \qquad \dots(5) \end{aligned} $$\hspace{.1in} Again consider \hspace{1.9in} a b+b c+c a=0 \hspace{1.8in} [From (2)]$$ \begin{aligned} \Rightarrow(\cos \alpha+i \sin \alpha)(\cos \beta+i \sin \beta)+(\cos \beta+i \sin \beta)(\cos \gamma+i \sin \gamma) \end{aligned} $$\hspace{4in}+(\cos \gamma+i \sin \gamma)(\cos \alpha+i \sin \alpha)=0$\par$\hspace{.1in} \Rightarrow[\cos (\alpha+\beta)+i \sin (\alpha+\beta)]+[\cos (\beta+\gamma)+i \sin (\beta+\gamma)]$\par$\hspace{4.3in} +[\cos (\gamma+\alpha)+i \sin (\gamma+\alpha)]=0

Equating real and imaginary parts, we get$$ \begin{aligned} & \qquad\cos (\alpha+\beta)+\cos (\beta+\gamma)+\cos (\gamma+\alpha)=0 \\ & \qquad\sin (\alpha+\beta)+\sin (\beta+\gamma)+\sin (\gamma+\alpha)=0 \end{aligned} $$ \hspace{1.8in}\text { Proved. }

Example 25. Solve$x^4+i=0$(M U .2008)

Solution. Here, we have.$$ \begin{aligned} x^4 & =-i=\cos \frac{\pi}{2}-i \sin \frac{\pi}{2} \\ x^4 & =\cos \left(2 n \pi+\frac{\pi}{2}\right)-i \sin \left(2 n \pi+\frac{\pi}{2}\right) \\ \Rightarrow \quad x & =\left[\cos \left(2 n \pi+\frac{\pi}{2}\right)-i \sin \left(2 n \pi+\frac{\pi}{2}\right)\right]^{\frac{1}{4}} \\ & =\cos (4 n+1) \frac{\pi}{8}-i \sin (4 n+1) \frac{\pi}{8} \end{aligned} $$Putting$n=0,1,2,3$we get the roots as$$ \begin{aligned} & x_1=\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}, \qquad\quad x_2=\cos \frac{5 \pi}{8}-i \sin \frac{5 \pi}{8} \\ & x_3=\cos \frac{9 \pi}{8}-i \sin \frac{9 \pi}{8}, \quad x_4=\cos \frac{13 \pi}{8}-i \sin \frac{13 \pi}{8} \end{aligned} $$ Ans.

Example 26. Solve$x^5=1+i$and find the continued product of the roots. (M U .2005,2004)

Solution. \hspace{.3in}$x^5=1+i$[$$ \begin{aligned} x^5 \\ &=\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \\ \Rightarrow \quad \\ x \\ &=2^{\frac{1}{10}} \\ \left[ \\ cos \left(\frac{\pi}{4}+2kpi \right)\frac{1}{5} \\ +isin \left(\frac{\pi}{4}+2kpi \right)\frac{1}{5} \\ \right] \\ &=2^{\frac{1}{10}} \\ \left[ \\ \cos \frac{(8k+1)\pi}{20} \\ +i \sin \frac{(8k+1)\pi}{20} \\ \right]. \end{aligned} $$]

The roots are obtained by putting$k=0,1,2,3,4$.

[$$ \begin{aligned} x_1&=2^{\frac{1}{10}} \\ \left[\cos \frac{\pi}{20}+i \sin \frac{\pi}{20}\right], \\ & \\ x_2&=2^{\frac{1}{10}} \\ \left[\cos \frac{9pi}{20}+i \sin \frac{9pi}{20}\right], \\ [4pt] \\ x_3&=2^{\frac{1}{10}} \\ \left[\cos \frac{17pi}{20}+i \sin \frac{17pi}{20}\right], \\ & \\ x_4&=2^{\frac{1}{10}} \\ \left[\cos \frac{5pi}{4}+i \sin \frac{5pi}{4}\right], \\ [4pt] \\ x_5&=2^{\frac{1}{10}} \\ \left[\cos \frac{33pi}{20}+i \sin \frac{33pi}{20}\right]. \end{aligned} $$]

[$$ \begin{aligned} x_1x_2x_3x_4x_5 \\ &=\left(2^{\frac{1}{10}}\right)^5 \\ \left(\cos \frac{\pi}{20}+i \sin \frac{\pi}{20}\right) \\ \left(\cos \frac{9pi}{20}+i \sin \frac{9pi}{20}\right) \\ &quadtimes \\ \left(\cos \frac{17pi}{20}+i \sin \frac{17pi}{20}\right) \\ \left(\cos \frac{5pi}{4}+i \sin \frac{5pi}{4}\right) \\ &quadtimes \\ \left(\cos \frac{33pi}{20}+i \sin \frac{33pi}{20}\right) \\ &=2^{frac12} \\ \left[ \\ cos \left( \\ \frac{\pi}{20} \\ +\frac{9pi}{20} \\ +\frac{17pi}{20} \\ +\frac{5pi}{4} \\ +\frac{33pi}{20} \\ \right) \\ \right. \\ &qquadq\quad \left|. \\ +isin \left( \\ \frac{\pi}{20} \\ +\frac{9pi}{20} \\ +\frac{17pi}{20} \\ +\frac{5pi}{4} \\ +\frac{33pi}{20} \\ \right) \\ \right] \\ &=\sqrt{2} \\ \left[ \\ \cos \frac{17pi}{4} \\ +i \sin \frac{17pi}{4} \\ \right] \\ &=\sqrt{2} \\ \left[ \\ cos \left(4pi+\frac{\pi}{4}\right) \\ +isin \left(4pi+\frac{\pi}{4}\right) \\ \right] \\ &=\sqrt{2} \\ \left[ \\ \cos \frac{\pi}{4} \\ +i \sin \frac{\pi}{4} \\ \right] \\ &=\sqrt{2} \\ \left[ \\ \frac{1}{\sqrt{2}} \\ +ifrac{1}{\sqrt{2}} \\ \right] \\ &=1+i. \end{aligned} $$]

[
\boxed{therefore x_1x_2x_3x_4x_5=1+i}
qquadtextbf{Ans.}
]

Example 27. If$\alpha, \alpha^2, \alpha^3, \alpha^4$, are the roots of$x^5-1=0$find them and show that$(1-\alpha)\left(1-\alpha^2 \right)\left(1-\alpha^3 \right)\left(1-\alpha^4 \right)=5$(M U .2007)

$$ \begin{aligned} x^5-1&=(x-1)(x-\alpha)(x-\alpha^2)(x-\alpha^3)(x-\alpha^4). \end{aligned} $$Therefore,$$ \begin{aligned} \frac{x^5-1}{x-1} \\ &=(x-\alpha)(x-\alpha^2)(x-\alpha^3)(x-\alpha^4). \end{aligned}$But $$\frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1.$$Hence,$$(x-\alpha)(x-\alpha^2)(x-\alpha^3)(x-\alpha^4)
=x^4+x^3+x^2+x+1.$$Putting (x=1), we get$$$$ \begin{aligned} (1-\alpha)(1-\alpha^2)(1-\alpha^3)(1-\alpha^4) \\ &=1+1+1+1+1 \\ &=5. \end{aligned} $$$$\boxed{(1-\alpha)(1-\alpha^2)(1-\alpha^3)(1-\alpha^4)=5}$$$$thereforequad \boxed{\text{Proved.}}$$

Example 28. If$\omega$is a cube root of unity, prove that$\begin{aligned} (1-\omega)^6&=-27 \qquad(MU. 2003) \end{aligned}$

$$ \begin{aligned} &\Rightarrow \quad 1+\omega^2=-\omega. \end{aligned} $$Now,$$ \begin{aligned} (1-\omega)^6 \\ &=\left[(1-\omega)^2 \right]^3 \\ &=\left[1-2omega+\omega^2 \right]^3 \\ &=\left[1-2omega-\omega-1 \right]^3 \\ qquadtext{[Since }1+\omega^2&=-omegatext{]} \\ &=(-3omega)^3 \\ &=-27omega^3 \\ &=-27. \end{aligned}$Since (\omega^3=1), $$\boxed{(1-\omega)^6=-27}$$$$thereforequad \boxed{Proved.}

Example 29. Use De Movie's theorem to solve the equation$x^4-x^2+x^2-x+1=0$.

$$\textbf{Solution.}\qquad x^4-x^3+x^2-x+1=0$$Multiplying both sides by (x+1), we get$$ \begin{aligned} (x+1)(x^4-x^3+x^2-x+1)&=0 \\ \Rightarrow \quad x^5+1&=0 \\ \Rightarrow \quad x^5&=-1. \end{aligned}$Now, $$-1=cospi+isinpi
=\cos(2n+1)\pi+isin(2n+1)\pi.$$Therefore,$$$$ \begin{aligned} x \\ &=\left[\cos(2n+1)\pi+isin(2n+1)pi \right]^{1/5} \\ &=\cos \frac{(2n+1)\pi}{5} \\ +i \sin \frac{(2n+1)\pi}{5}. \end{aligned} $$Putting (n=0,1,2,3,4), we obtain the five roots of (x^5+1=0):$$ \begin{aligned} &\cos \frac{\pi}{5}+i \sin \frac{\pi}{5}, \\ &\cos \frac{3pi}{5}+i \sin \frac{3pi}{5}, \\ &cospi+isinpi, \\ &\cos \frac{7pi}{5}+i \sin \frac{7pi}{5}, \\ &\cos \frac{9pi}{5}+i \sin \frac{9pi}{5}. \end{aligned}$But $$cospi+isinpi=-1.$$Since (x=-1) is the root of (x+1=0), it is not a root of the given equation. Hence, it is rejected. Therefore, the required roots are$$\boxed{
\cos \frac{\pi}{5}+i \sin \frac{\pi}{5},
\cos \frac{3pi}{5}+i \sin \frac{3pi}{5},
\cos \frac{7pi}{5}+i \sin \frac{7pi}{5},
\cos \frac{9pi}{5}+i \sin \frac{9pi}{5}
}$$$$\boxed{Ans.}v

Example 30. Prove that$(\cosh x-\sinh x)^n=\cosh n x-\sinh n x$. (MU. 2001, 2002)

Solution. L.HS.$=(\cosh x-\sinh x)^n$\par$$ \begin{aligned} {1.2in} &=\left[\frac{e^x+e^{-x}}{2}-\frac{e^x-e^{-x}}{2}\right]^5=\left[\frac{2 e^{-x}}{2}\right]^n=\left(e^{-x}\right)^n=e^{-n x} {1.8in} \dots(1) \end{aligned} $$

{.7in}R.H.S.$=\cosh n x-\sinh n x$$$ \begin{aligned} {1.2in} &=\left(\frac{e^{n z}+e^{-x z}}{2}-\frac{e^{n z}+e^{-n z}}{2}\right) \quad=\frac{2 e^{-n z}}{2}=e^{-n z} {2.2in} \dots(2) \end{aligned} $$From (1) and (2), we have
{1in}L.H.S. = R.H.S. Proved.

Example 31.If :$\tan (\theta+i \phi)=\cos \alpha+i \sin \alpha$prove that:$\theta=\frac{npi}{2}+\frac{\pi}{4}$and$\phi=\frac{1}{2} \log \tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right) \quad$(Nagpur University, Summer 2002, Writer 200I)

Solution. We have,$\tan (\theta+iphi)=\cos \alpha+i \sin \alpha$\par$therefore \hspace{1in}\tan (\theta-iphi)=\cos \alpha-i \sin \alpha$$$ \begin{aligned} & \text { But }\qquad \quad \tan 2 \theta=\tan [(\theta+i \phi)+(\theta-i \phi)] \\ & =\frac{\tan (\theta+i \phi)+\tan (\theta-i \phi)}{1-\tan (\theta+i \phi) \tan (\theta-i \phi)}=\frac{\cos \alpha+i \sin \alpha+\cos \alpha-i \sin \alpha}{1-(\cos \alpha+i \sin \alpha)(\cos \alpha-i \sin \alpha)} \\ & \qquad=\frac{2 \cos \alpha}{1-\left(\cos ^2 \alpha+\sin ^2 \alpha \right))}-\frac{2 \cos \alpha}{1-1}=\infty-\tan \frac{\pi}{2} \\ & \\ therefore \quad 2 \theta & =\frac{\pi}{2} \text { or for general values, } \\ 2theta & =n \pi+\frac{\pi}{2} \Rightarrow \theta=\frac{n \pi}{2}+\frac{\pi}{4} \\ & \text { Again, } \tan (2 i \phi)=\tan [(\theta+i \phi)-(\theta-i \phi)]=\frac{\tan (\theta+i \phi)-\tan (\theta-i \phi)}{1+\tan (\theta+i \phi) \tan (\theta-i \phi)} \\ & \qquad=\frac{\cos \alpha+i \sin \alpha-(\cos \alpha-i \sin \alpha)}{1+(\cos \alpha+i \sin \alpha)(\cos \alpha-i \sin \alpha)} \\ & \qquad=\frac{2 i \sin \alpha}{1+\cos ^2 \alpha+\sin ^2 \alpha}=\frac{2 i \sin \alpha}{1+1}=\frac{2 i \sin \alpha}{2}=i \sin \alpha \\ \Rightarrow & i \tanh 2 \phi=i \sin \alpha \\ (because \tan i x&=i \tanh x) \\ \Rightarrow & \tanh 2 \phi=\sin \alpha \\ \text { i.e., } \frac{e^{2 \phi}-e^{-2 \phi}}{e^{2 \phi}+e^{-2 \phi}}&=\frac{\sin \alpha}{1} \\ therefore \frac{e^{2 \phi}-e^{-2 \phi}+e^{2 \phi}+e^{-2 \phi}}{\left(e^{2 \phi}+e^{-2 \phi}\right)-\left(e^{2 \phi}-e^{-2 \phi}\right)}&=\frac{1+\sin \alpha}{1-\sin \alpha}\qquad\text{(Componendo and dividendo)} \\ i.e. \frac{2 e^{2 \phi}}{2 e^{-2 \phi}}\qquad&=\frac{1-\cos \left(\frac{\pi}{2}+\alpha \right))}{1+\cos \left(\frac{\pi}{2}+\alpha \right))} \qquadbecause \cos \left(\frac{\pi}{2}+\theta \right))=-\sin \theta \\ \Rightarrow & e^{4 \phi}=\frac{2 \sin ^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)} \\ \Rightarrow & e^{4 \phi}=\tan ^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right) \Rightarrow e^{2 \phi}=\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right) \end{aligned} $$Hence,$\quad 2 \phi=\log _e \tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right) \Rightarrow \phi=\frac{1}{2} \log _e \tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)$ Proved.

Example 32. If$\cosh x=\sec \theta$, prove that:$(i) \theta=\frac{\pi}{2}-2 \tan ^{-1}\left(e^{-x}\right)$(ii)$\tanh \frac{\pi}{2}=\tan \frac{\theta}{2}$(M.U. 2003, 2005)v

Solution. (i) Let$\tan ^{-1} e^{-x}=\alpha$\par$\Rightarrow \hspace{.3in}e^{-x}=\tan : \alpha : \text{and} : \alpha=\tan^{-1}(e^{-1})$\hspace{4in} \dots(4) \par$\Rightarrow \hspace{.3in} \quad e^x=\cot \alpha$\hspace{5.2in} \dots(5)
Now,\hspace{.3in}$\quad \sec \theta=\cosh x=\frac{e^x+e^{-x}}{2}$\hspace{1.8in}(3) (Given)
Putting the values of$e^{-x}$and$e^x$from (1) and (2) in (3), we get$$ \begin{aligned} &\qquad \sec \theta=\frac{\cot \alpha+\tan \alpha}{2} \\ & therefore \qquad\quad 2 \sec \theta=\cot \alpha+\tan \alpha=\frac{\cos \alpha}{\sin \alpha}+\frac{\sin \alpha}{\cos \alpha}=\frac{\cos ^2 \alpha+\sin ^2 \alpha}{\sin \alpha \cos \alpha} \\ &\qquad=\frac{2}{2 \sin \alpha \cos \alpha} \\ &\qquad=\frac{2}{\sin 2 \alpha}\qquad therefore \quad \left|[because \cos ^2 \alpha+\sin ^2 \alpha=1 \right] \\ & \Rightarrow \qquad\quad \cos \theta=\sin 2 \alpha \\ & therefore \qquad\quad \cos \theta=\cos \left(\frac{\pi}{2}-2 \alpha \right)) \\ & therefore \qquad\quad \theta=\frac{\pi}{2}-2 \alpha=\frac{\pi}{2}-2 \tan ^{-1}\left(e^{-x}\right) \quad \text { [From (1)] \qquad \textbf{Proved}. } \end{aligned} $$(ii) We have,$$ \begin{aligned} \cosh : x : &= \sec : \theta \qquad\text{(Given)} \\ & \Rightarrow & \frac{e^x+e^{-x}}{2} & =\sec \theta \qquad [therefore \cosh : x=\frac{e^x+e^{-x}}{2}] \\ & therefore & e^x-2 \sec \theta+e^{-x} & =0 \\ & therefore & \left(e^x \right)^2-2 e^x \sec \theta+1 & =0 \end{aligned} $$Solving the quadratic equation in$e^x$.$$ \begin{array}{rlrl} & \hspace{.3in} e^x =\frac{2 \sec \theta \pm \sqrt{4 \sec ^2 \theta-4}}{2} \Rightarrow & \hspace{.3in}e^x =\sec \theta \pm \sqrt{\sec ^2 \theta-1} \Rightarrow & \hspace{.3in}e^x =\sec \theta \pm \tan \theta \hspace{4.4in} \dots(4) \text { Now, } & \hspace{.3in}\tanh \frac{x}{2} =\frac{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}{e^{\frac{x}{2}}+e^{-\frac{x}{2}}}=\frac{e^x-1}{e^x+1} \hspace{3.9in} \dots(5) \par \end{array} $$Putting the value of$e^x$from (4) in (5), we get$$ \begin{aligned} \tanh \frac{x}{2} & =\frac{\sec \theta+\tan \theta-1}{\sec \theta+\tan \theta+1} \\ & =\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}=\frac{(1-\cos \theta)+\sin \theta}{(1+\cos \theta)+\sin \theta} \qquad[Using (1)] \\ & =\frac{2 \sin ^2 \frac{\theta}{2}+2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}+2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\tan \frac{\theta}{2} \quad \qquad\text { \textbf{Proved}. } \end{aligned} $$

Example 33. Separate the following into real and imaginary parts: (i)$\sin (x+i y)$(ii)$\cos (x+i y)$(iii)$\tan (x+i y)$

Solution. (i)$\sin (x+i y)=\sin x \cos i y+\cos x \sin (i y)=\sin x \cosh y+i \cos x \sinh y$.
(ii)$\cos (x+i y)=\cos x \cos (i y)-\sin x \sin (i y)=\cos x \cosh y-i \sin x \sinh y$.
(iii)$$ \begin{aligned} \tan (x+i y)&=\frac{\sin (x+i y)}{\cos (x+i y)}=\frac{2 \sin (x+i y) \cos (x-i y)}{2 \cos (x+i y) \cos (x-i y)} \\ &=\frac{\sin 2 x+\sin (2 i y)}{\cos 2 x+\cos 2 i y}=\frac{\sin 2 x+i \sinh 2 y}{\cos 2 x+\cosh 2 y} \end{aligned} $$$$ \left{\begin{array}{l} because \quad 2 \sin A \cdot \cos B=\sin (A+B)+\sin (A-B) \text { and } 2 \cos A \cdot \cos B=\cos (A+B)+\cos (A-B) \end{array}\right} $$

Example 34. If$\tan (A+i B)=x+i y$, prove that$\begin{aligned} \tan 2 A&=\frac{2 x}{1-x^2-y^2} \text{} 2 B=\frac{2 x}{1+x^2+y^2} \quad \end{aligned}$(Nagpur University, Summer 2000)

Solution.$\tan (A+i B)=x+i y ; \tan (A-i B)=x-i y$
and \tanh
\hspace{.3in}$\tan 2 A=\tan (A+i B+A-i B)$$$ \begin{aligned} & =\frac{\tan (A+i B)+\tan (A-i B)}{1-\tan (A+i B) \tan (A-i B)} \\ \tan 2 A & =\frac{(x+i y)+(x-i y)}{1-(x+i y)(x-i y)}=\frac{2 x}{1-\left(x^2+y^2 \right)}=\frac{2 x}{1-x^2-y^2} \end{aligned} $$Again \hspace{.4in}$$ \begin{aligned} \tan 2 i B&=\tan (A+i B-A+i B)=\frac{\tan (A+i B)-\tan (A-i B)}{1+\tan (A+i B) \tan (A-i B)} \\ & \tan 2 i B=\frac{(x+i y)-(x-i y)}{1+(x+i y)(x-i y)}=\frac{(2 y) i}{1+x^2+y^2} \\ & \tanh 2 B=\frac{2 y}{1+x^2+y^2} \quad \qquad\tan i x=i \tanh x \quad \qquad\text { \textbf{Proved}. } \end{aligned} $$

Example 35. If$\sin (\alpha+i \beta)=x+i y$, prove that (a)$\frac{x^2}{\cosh ^2 \beta}+\frac{y^2}{\sinh ^2 \beta}=1$(b)$\frac{x^2}{\sin ^2 \alpha}-\frac{y^2}{\cos ^2 \alpha}=1$

Solution. (a)$x+i y=\sin (\alpha+i \beta)=\sin \alpha \cosh \beta+i \cos \alpha \sinh \beta$
Equating real and imaginary parts, we get$$ \begin{aligned} x & =\sin \alpha \cosh \beta, y=\cos \alpha \sinh \beta \\ \sin \alpha & =\frac{x}{\cosh \beta} \text { and } \cos \alpha=\frac{y}{\sinh \beta} \end{aligned} $$Squaring and adding,$\sin ^2 \alpha+\cos ^2 \alpha=\frac{x^2}{\cosh ^2 \beta}+\frac{y^2}{\sinh ^2 \beta}$
\hspace{.1in}$\Rightarrow \hspace{.6in}\quad 1=\frac{x^2}{\cosh ^2 \beta}+\frac{y^2}{\sinh ^2 \beta}$ Proved.

(b) Again$$ \begin{aligned} \cosh \beta & =\frac{x}{\sin \alpha} \text { and } \sinh \beta=\frac{y}{\cos \alpha} \\ \cosh ^2 \beta-\sinh ^2 \beta & =\frac{x^2}{\sin ^2 \alpha}-\frac{y^2}{\cos ^2 \alpha} \\ 1 & =\frac{x^2}{\sin ^2 \alpha}-\frac{y^2}{\cos ^2 \alpha}\qquad \textbf{Proved}. \end{aligned} $$

Example 36. Separate the following into real and imaginary parts of hyperbolic functions. (a)$\sinh (x+i y)$(b)$\cosh (x+i y)$(c)$\tanh (x+i y)$

Solution.$(a) \sinh (x+i y)=\sinh x \cosh (i y)+\cosh x \sinh (i y)$\par$\hspace{1.8in}=\sinh x \cos y+i \sin y \cosh x \hspace{3.2in} \text{\textbf{Ans.}}$(b)$\cosh (x+i y)=\cosh x \cosh (i y)-\sinh x \sinh i y=\cosh x \cos y-i \sinh x \sin y$.parhspace{6.8in} Ans.
(c)$$ \begin{aligned} \tanh (x+i y)&=\frac{\sinh (x+i y)}{\cosh (x+i y)}=\frac{-i \sin i(x+i y)}{\cos i(x+i y)} \\ &\qquad =\frac{-i \sin (i x-y)}{\cos (i x-y)}=\frac{-i 2 \sin (i x-y) \cos (i x+y)}{2 \cos (i x-y) \cos (i x+y)} \quad \text { (Note this step) } \\ &\qquad =-i \frac{\sin 2 i x-\sin 2 y}{\cos 2 i x+\cos 2 y}=-i \frac{i \sinh 2 x-\sin 2 y}{\cosh 2 x+\cos 2 y}=\frac{\sinh 2 x+i \sin 2 y}{\cosh 2 x+\cos 2 y} \\ &\qquad =\frac{\sinh 2 x}{\cosh 2 x+\cos 2 y}+i \frac{\sin 2 y}{\cosh 2 x+\cos 2 y} \quad \qquad\text { \textbf{Ans}. } \end{aligned} $$

Example 37. If$\tan (x+i y)=\sin (u+i v)$, prove that$\begin{aligned} \frac{\sin 2 x}{\sinh 2 y}&=\frac{\tan u}{\tanh v} \end{aligned}$

Solution. Now$\tan (x+i y)=\sin (u+i v)$separating the real and imaginary parts of both sides,

we have$$ \begin{aligned} \frac{\sin 2 x}{\cos 2 x+\cosh 2 y}+\frac{i \sinh 2 y}{\cos 2 x+\cosh 2 y}&=\sin u \cosh v+i \cos u \sinh v \end{aligned} $$Equating real and imaginary parts, we get$$ \begin{aligned} \frac{\sin 2 x}{\cos 2 x+\cosh 2 y}&=\sin u \cosh \end{aligned} $$and \quad$$ \begin{aligned} \frac{\sinh 2 y}{\cos 2 x+\cosh 2 y}&=\cos u \sinh v \end{aligned} $$Dividing (1) by (2), we obtain$$ \begin{aligned} \frac{\sin 2 x}{\sinh 2 y}&=\frac{\sin u \cosh v}{\cos u \sinh v} \\ \Rightarrow \qquad \frac{\sin 2 x}{\sinh 2 y}&=\frac{\tan u}{\tanh v} \qquad \text{\textbf{Proved.}} \end{aligned} $$

Example 38. If$\sin (\theta+i \phi)=\tan \alpha+i \sec \alpha$, show that$\cos 2 \theta \cosh 2 \phi=3$

Solution.$\quad \sin (\theta+mathrm{i} \phi)=\tan \alpha+i \sec \alpha$
\hspace{.1in}$\sin \theta \cosh \phi+i \cos \theta \sinh \phi=\tan \alpha+i \sec \alpha$
Equating real and imaginary parts, we get$$ \begin{aligned} & \sin \theta \cosh \phi=\tan \alpha \qquad \dots(1) \\ & \cos \theta \sinh \phi=\sec \alpha \qquad \dots(2) \end{aligned} $$We know that$$ \begin{aligned} \quad & \qquad \sec ^2 \alpha-\tan ^2 \alpha=1 \\ & \cos ^2 \theta \sinh ^2 \phi-\sin ^2 \theta \cosh ^2 \phi=1 \qquad \text{[From (1) and (2)]} \\ & \qquad\left(\frac{1+\cos 2 \theta}{2}\right)\left(\frac{\cosh 2 \phi-1}{2}\right)-\left(\frac{1-\cos 2 \theta}{2}\right)\left(\frac{\cosh 2 \phi+1}{2}\right)=1 \\ & {[-1+\cosh 2 \phi-\cos 2 \theta+\cos 2 \theta \cosh 2 \phi]-[\cosh 2 \phi+1-\cos 2 \theta \cosh 2 \phi-\cos 2 \theta]=4} \\ \Rightarrow & -2+2 \cos 2 \theta \cosh 2 \phi=4 \\ \Rightarrow & \qquad\quad 2 \cos 2 \theta \cosh 2 \phi=6 \Rightarrow \cos 2 \theta \cosh 2 \phi=3 \qquad\text{\textbf{Proved.}} \end{aligned} $$

Example 39. If$e^z=\sin (u+i v)$and$z=x+i y$, prove that$\begin{aligned} & \quad 2 e^{2 x} & =\cosh 2 v-\cos 2 u \end{aligned}$

Solution We have \hspace{.5in}$$ \begin{aligned} e^z & =\sin (u+i v) \\ &\Rightarrow \qquad e^{x+iy} =\sin (u+i v) \\ &\Rightarrow \qquad e^x \cdot e^{i y} =\sin u \cos i v+\cos u \sin i v \\ &\Rightarrow \qquad e^x(\cos y+i \sin y) =\sin u \cosh v+i \cos u \sinh v \end{aligned} $$Equating real and imaginary parts, we get$$ \begin{aligned} e^x \cos y & =\sin u \cosh v \\ \text{and} \qquade^x \sin y & =\cos u \sinh v \end{aligned} $$

By squaring and adding, we get$$ \begin{aligned} & \qquad e^{2 x}\left(\cos ^2 y+\sin ^2 y \right) =\sin ^2 u \cosh ^2 v+\cos ^2 u \sinh ^2 v \\ &\Rightarrow \qquad e^{2 x} =\left(1-\cos ^2 u \right) \cosh ^2 v+\cos ^2 u \left(\cosh ^2 v-1 \right) \\ &\Rightarrow \qquad e^{2 x} =\cosh ^2 v-\cos ^2 mathrm{u} \\ &\Rightarrow \qquade^{2 x}=\frac{1}{2}(1+\cosh 2 v) -\frac{1}{2}(1+cos2u) \\ &\Rightarrow e^{2 x}=\frac{1}{2}(\cosh 2 v-\cos 2 u) \\ &\Rightarrow 2 e^{2 x}=\cosh 2 v-\cos 2 u \qquad\qquad \textbf{\textbf{Proved.}} \end{aligned} $$

Example 40. If$\sin (\theta+i \phi)=\cos \alpha+i \sin \alpha$, prove that$\begin{aligned} \cos ^4 \theta&=\sin ^2 \alpha=\sinh ^4 \phi \qquad(M.U. 2003, 2004) \end{aligned}$

Solution. Here, we have$$ \begin{aligned} \sin (\theta+mathrm{i} \phi)&=\cos \alpha+i \sin \alpha \end{aligned} $$\Rightarrow \sin \theta \cosh \phi+i \cos \theta \sinh \phi=\cos \alpha+i \sin \alpha$\par Equating real and imaginary parts, we get$$$ \begin{aligned} \sin \theta \cosh \phi&=\cos \alpha \Rightarrow \cosh \phi=\frac{\cos \alpha}{\sin \theta} \qquad \dots(1) \end{aligned} $$and \hspace{.1in}$\quad \cos \theta \sinh \phi=\sin \alpha \Rightarrow \sinh \phi=\frac{\sin \alpha}{\cos \theta}$\hspace{3.9in} \dots(2)
But \hspace{1.2in}$\cosh ^2 \phi-\sinh ^2 \phi=1$$$ \begin{aligned} & \Rightarrow \qquad\quad \frac{\cos ^2 \alpha}{\sin ^2 \theta}-\frac{\sin ^2 \alpha}{\cos ^2 \theta}=1 \qquad\text{[Using (1) and (2)]} \\ & \Rightarrow \qquad\quad \cos ^2 \alpha \cdot \cos ^2 \theta-\sin ^2 \alpha \cdot \sin ^2 \theta=\sin ^2 \theta \cos ^2 \theta \\ & \Rightarrow \qquad\quad \left|(1-\sin ^2 \alpha \right)) \cos ^2 \theta-\sin ^2 \alpha \cdot \sin ^2 \theta=\left(1-\cos ^2 \theta \right)) \cos ^2 \theta \\ & \Rightarrow \qquad\quad \cos ^2 \theta-\sin ^2 alpha \left(\cos ^2 \theta+\sin ^2 \theta \right))=\cos ^2 \theta-\cos ^4 thetaRightarrow \\ & \Rightarrow \qquad\quad \sin ^2 \alpha=\cos ^4 \theta \end{aligned} $$Again \hspace{1in}$$ \begin{aligned} \sin ^2 \theta+\cos ^2 \theta&=1 \qquad\text{[Using (1) and (2)]} \\ therefore \qquad \frac{\cos ^2 \alpha}{\cosh ^2 \phi}+\frac{\sin ^2 \alpha}{\sinh ^2 \phi}&=1 \\ & \Rightarrow \qquad \quad \qquad\cos ^2 \alpha \cdot \sinh ^2 \phi+\sin ^2 \alpha \cosh ^2 phihspace{.3in}=\sinh ^2 \phi \cosh ^2 \phi \\ & \Rightarrow \qquad \left(1-\sin ^2 \alpha \right)) \sinh ^2 \phi+\sin ^2 alpha \left(1+\sinh ^2 \phi \right))=\sinh ^2 phi \left(1+\sinh ^2 \phi \right)) \\ & \Rightarrow \qquad \quad \sinh ^2 \phi-\sin ^2 \alpha \sinh ^2 \phi+\sin ^2 \alpha+\sin ^2 \alpha \sinh ^2 \phi=\sinh ^2 \phi+\sinh ^4 \phi \\ & \Rightarrow \quad \sin ^2 \alpha=\sinh ^4 \phi \text {. } \end{aligned} $$From (3) and (4), we have \hspace{.4in}$\cos ^4 \theta=\sin ^2 \alpha=\sinh ^4 \phi$ Proved.

Example 41. If$\operatorname{cosec}\left(\frac{\pi}{4}+i x \right)=u+i v$, prove that$\left(u^2+v^2 \right)^2=2\left(u^2-v^2 \right) (MU. 2009)$v

Solution. Here, we have$$ \begin{aligned} u+i v & =\operatorname{cosec}\left(\frac{\pi}{4}+i x \right) \\ & =\frac{1}{\sin \left(\frac{\pi}{4}+i x \right)} \Rightarrow :=\frac{1}{\sin \frac{\pi}{4} \cos i x+\cos \frac{\pi}{4} \sin i x} \\ & =\frac{1}{\frac{1}{\sqrt{2}} \cosh x+\frac{1}{\sqrt{2}} i \sinh x}=\frac{\sqrt{2}}{\cosh x+i \sinh x} \\ &=\frac{\sqrt{2}(\cosh x-i \sinh x)}{\cosh ^2 x+\sinh ^2 x}=\frac{\sqrt{2}(\cosh x-i \sinh x)}{\cosh 2 x} \end{aligned} $$Equating real and imaginary parts, we get$$ \begin{aligned} u&=\frac{\sqrt{2} \cosh x}{\cosh 2 x}, v=-\frac{\sqrt{2} \sinh x}{\cosh 2 x} \end{aligned} $$Squaring and adding, we get$$ \begin{aligned} &\qquadu^2+v^2 =\frac{2\left(\cosh ^2 x+\sinh ^2 x \right)}{\cosh ^2 2 x}=\frac{2 \cosh 2 x}{\cosh ^2 2 x} \\ &\Rightarrow \qquad\left(u^2+v^2 \right)^2 =\left(\frac{2}{\cosh 2 x}\right)^2=\frac{4}{\cosh ^2 2 x} \qquad \dots(1) \\ &\text { Also, } \quad \qquadu^2-v^2 =\frac{2}{\cosh ^2 2 x}\left(\cosh ^2 x-\sinh ^2 x \right)=\frac{2}{\cosh ^2 2 x} \hspace{2,8in} \dots(2) \end{aligned} $$From (1) and (2), we have$$ \begin{aligned} u^2-v^2&=2(u^2-v^2) \\ \text{\textbf{Proved.}} \end{aligned} $$

Example 42. Separate into real and imaginary parts: [ \left(\sqrt{i}\right)^{\sqrt{i}} ] (M.U. 2008)

Solution. We have,$$ \begin{aligned} \sqrt{i}&=i^{\frac{1}{2}}=\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)^{\frac{1}{2}}=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}} \end{aligned} $$Also, \hspace{.4in}$$ \begin{aligned} \sqrt{i}&=\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)^{\frac{1}{2}}=\left(e^{i \frac{\pi}{2}}\right)^{\frac{1}{2}}=e^{i \frac{\pi}{4}} \\ therefore \quad(\sqrt{i})^{\sqrt{i}} & =\left(e^{\left.i \frac{\pi}{4}\right)\left(\frac{1}{\sqrt{2}}+i \cdot \frac{1}{\sqrt{2}}\right)}=e^{i \frac{\pi}{4 \sqrt{2}}-\frac{\pi}{4 \sqrt{2}}}\right. \\ & =e^{-\frac{\pi}{4 \sqrt{2}}} \cdot e^{i \frac{\pi}{4 \sqrt{2}}}=e^{-\frac{\pi}{4 \sqrt{2}}}\left(\cos \frac{\pi}{4 \sqrt{2}}+i \sin \frac{\pi}{4 \sqrt{2}}\right) \\ therefore \quad \text { Real part } & =e^{-\frac{\pi}{4 \sqrt{2}}} \cos \left(\frac{\pi}{4 \sqrt{2}}\right) \\ \text { Imaginary part } & =e^{-\frac{\pi}{4 \sqrt{2}}} \sin \left(\frac{\pi}{4 \sqrt{2}}\right) \qquad \text{\textbf{Ans}.} \end{aligned} $$v

Example 43. Define the logarithm of a complex number.

Solution. If$z$and$w$are two complex numbers and$z=e^w$then$w=\log z$, and

\hspace{1in}if$w=\log z ;$then$z=e^w$
Here$\log z$is a many-valued function. The general value of$\log z$is defined by Log$z$, where$\log z Log z+2 n \pi i$.

Example 44. Separate \log$(x+i v)$into its real and imaginary parts

Solution. Let \hspace{.6in}$x=r \cos \theta$\hspace{4in} \dots(1)

and \hspace{1.4in}$y=r \sin \theta$\hspace{4in} \dots(2)
Squaring and adding (1) and (2) we have$x^2+y^2=r^2$$$ \begin{aligned} & therefore \quad \qquadr=\sqrt{x^2+y^2} \text {, } \\ & \text { We have, } \quad \qquad\tan \theta=\frac{y}{x} \Rightarrow \theta=\tan ^{-1}\left(\frac{y}{x}\right) \qquad\text{[Dividing (2) by (1)]} \\ & therefore \quad \qquad\log (x+i y)=\log [r(\cos \theta+i \sin \theta)] \\ & \qquad=[\log r+\log (\cos \theta+i \sin \theta)] \\ & \qquad\log (x+dot{y})=\log r+\log [\cos (2 n \pi+\theta)+i \sin (2 n \pi+\theta)] \\ & \\ & =\log r+\log e^{i(2 n \pi+\theta)}=\log r+i(2 n \pi+\theta) \\ \quad \log (x+i y) & =\log \sqrt{x^2+y^2}+i \left(2 n \pi+\tan ^{-1} \frac{y}{x}\right) \\ \text{and}\qquad\log (x+i y) & =\log \sqrt{x^2+y^2}+i \tan ^{-1} \frac{y}{x}\qquad \textbf{Ans.} \end{aligned} $$

Example 45. Show that$\begin{aligned} \log \frac{x+i y}{x-i y}&=2 i \tan ^{-1} \frac{y}{x} \end{aligned}$. (Nagpur University, Winter 2003)

Solution. Let$\log (x+i y)=\log (r \cos \theta+i r \sin \theta)=\log r e^{i \theta}$$$ \begin{aligned} &=\log r+i \theta \\ \left[\begin{array}{l} \\ x&=r \cos \theta \\ y&=r \sin \theta \\ \end{array}\right] \end{aligned} $$Similarly,$$ \begin{aligned} \log (x-i y) & =\log r-i \theta \\ \log \frac{x+i y}{x-i y} & =\log (x+i y)-\log (x-i y)=(\log r+i \theta)-(\log r-i \theta)=2 i \theta \\ & =2 i \tan ^{-1} \frac{y}{x} .\qquad \textbf{Proved.} \end{aligned} $$

Example 46. Show that for real values of$a$and$b$$ e^{2 a i \cot ^{-1} b}\left[\frac{b i-1}{b i+1}\right]^{-a}=1 (M.U. 2008) $

Solution. Consider$$ \begin{aligned} \frac{b i-1}{b i+1}&=\frac{b i+i^2}{b i-i^2}=\frac{b+i}{b-i} \\ & \Rightarrow \left(\frac{b i-1}{b i+1}\right)^{-a}\qquad=\left(\frac{b+i}{b-i}\right)^{-a} \\ & \log \left[\frac{b i-1}{b i+1}\right]^{-a}\qquad=\log \left(\frac{b+i}{b-i}\right)^{-a}=-a[\log (b+i)-\log (b-i)] \\ &\qquad =-a\left[\log \sqrt{b^2+1}+i \tan ^{-1} \frac{1}{b}-\log \sqrt{b^2+1}+i \tan ^{-1} \frac{1}{b}\right] \\ &\qquad =-2 a i \tan ^{-1} \frac{1}{b} \\ &\qquad \left(\frac{b i-1}{b i+1}\right)^{-a}=e^{-2 a i \tan \left(\frac{1}{b}\right)} \qquad \\ \left[\begin{array}{l} \\ \text{If }\cot \theta&=b, \tan \theta=\frac{1}{b} \\ \text{Since}\cot ^{-1} b&=\tan ^{-1}\left(\frac{1}{b}\right) \\ \end{array}\right] \\ & e^{2 a i \cot ^{-1} b}\left(\frac{b i-1}{b i+1}\right)^{-a}=\left[e^{2 a i \tan ^{-1}\left(\frac{1}{b}\right)}\right] \cdot \left([e^{-2 a i+ta n^{-1}\left(\frac{1}{b}\right)}\right]=1 \qquad \textbf{Proved.} \end{aligned} $$

Example 47. Prove that$\sinh ^{-1} x=\log \left(x+\sqrt{x^2+1}\right)$(M+2009)

Solution. Let$\sinh ^{-1} x=y \Rightarrow x=\sinh y$$$ \begin{aligned} x &=\frac{e^y-e^{-y}}{2} \\ \Rightarrow \qquad \quad e^y-e^{-y} &=2 x \\ \Rightarrow \qquad \quad e^{2 y}-2 x e^y-1 &=0 \end{aligned} $$This is quadratic in$e^y$$$ \begin{aligned} e^y & =\frac{2 x \pm \sqrt{4 x^2+4}}{2}=x \pm \sqrt{x^2+1} \\ y & =\log \left(x+\sqrt{x^2+1}\right) \qquad \text{(Taking positive sign only)} \\ \sinh ^{-1} x & =\log \left(x+\sqrt{x^2+1}\right) \qquad\text{\textbf{Proved.}} \end{aligned} $$

Example 48. Prove that$\cosh^{-1} x=\log \left(x+\sqrt{x^2-1}\right)$(ML. 2009)

Solution. Let$y=\cosh ^{-1} x \Rightarrow x=\cosh y$\par$$ \begin{aligned} & \qquadx=\frac{e^y+e^{-y}}{2} \Rightarrow 2 x=e^y+e^{-y} \qquad \text{(This is quadratic in$e^y$)} \\ \Rightarrow \qquad& \qquad e^{2 y}-2 x e^y+1 =0 \\ \Rightarrow \qquad& \hspace{,9in}e^y =\frac{2 x \pm \sqrt{4 x^2-4}}{2}=x \pm \sqrt{x^2-1} \\ \Rightarrow \qquad & \qquad y =\log \left(x+\sqrt{\left.x^2-1 \right)}\right. \qquad \text{(Taking positive sign only )} \\ \Rightarrow \qquad & \qquad \cosh ^{-1} x =\log \left(x+\sqrt{\left.x^2-1 \right)}\right. \qquad \textbf{Proved.} \end{aligned} $$

Example 49. Prove that$\operatorname{\operatorname{sech}}^{-1} x=\log \frac{1+\sqrt{1-x^2}}{x}$

Solution.Let \hspace{1in}$y=\operatorname{sech}^{-1} x \Rightarrow x=\operatorname{sech} y$$$ \begin{aligned} &x =\frac{2}{e^y+e^{-y}} \Rightarrow x=\frac{2 e^y}{e^{2 y}+1} \\ \Rightarrow \qquad\quad x e^{2 y}-2 e^y+x & =0 \quad \Rightarrow \quad e^y=\frac{2 \pm \sqrt{4-4 x^2}}{2 x}=\frac{1 \pm \sqrt{1-x^2}}{x} \end{aligned} $$We take only positive sign$$ \begin{aligned} e^y & =\frac{1+\sqrt{1-x^2}}{x} \Rightarrow y=\log \frac{1+\sqrt{1-x^2}}{x} \\ \operatorname{sech}^{-1} x & =\log \frac{1+\sqrt{1-x^2}}{x} \end{aligned} $$Similarly, \hspace{.3in}$\quad \operatorname{cosech}^{-1} x=\log \frac{1+\sqrt{1+x^2}}{x}$ Proved.

Example 50. If$x+i y=\cos (\alpha+i \beta)$or if$\cos ^{-1}(x+i y)=\alpha+i \beta$express$x$and$y$in terms of$\alpha$and$\beta$. Hence show that$\cos ^2 \alpha$and$\cosh ^2 \beta$are the roots of the equation$\lambda^2-\left(x^2+y^2+1 \right) \lambda+x^2=0$. (M.U. 2002, 2004)

Solution. Here, we have$$ \hspace{1in}\begin{gathered} \cos (\alpha+i \beta)=x+i y \Rightarrow \cos \alpha \cos i \beta-\sin \alpha \sin i \beta=x+i y \Rightarrow \cos \alpha \cosh \beta-i \sin \alpha \sinh \beta=x+i y \end{gathered} $$\hspace{1in}Equating real and imaginary parts, we get$\hspace{1.8in}\cos \alpha \cosh \beta=x \text { and } \sin \alpha \sinh \beta=-y$We want to find the equation whose roots are$\cos ^2 \alpha$and$\cosh ^2 \beta$.
Now,$$ \begin{aligned} x^2+y^2+1 & =\cos ^2 \alpha \cos h^2 \beta+\sin ^2 \alpha \sinh ^2 \beta+1 \\ & =\cos ^2 \alpha \cosh ^2 \beta+\left(1-\cos ^2 \alpha \right))\left(\cosh ^2 \beta-1 \right)+1 \\ & =\cos ^2 \alpha \cosh ^2 \beta+\cosh ^2 \beta-1-\cos ^2 \alpha \cosh ^2 \beta+\cos ^2 \alpha+1 \\ & =\cos ^2 \alpha+\cosh ^2 \beta \\ \text { Sum of the roots } & =\cos ^2 \alpha+\cosh ^2 \beta \\ & =x^2+y^2+1 \end{aligned} $$And product of the roots$\hspace{.1in}=\cos ^2 \alpha \cosh ^2 \beta$\hspace{1.8in}=x^2$Hence, the equation whose roots are$\cos ^2 \alpha, \cosh ^2 \beta$is$$$ \begin{aligned} \lambda^2-\left(x^2+y^2+1 \right) \lambda+x^2&=0 \qquad \textbf{Proved.} \end{aligned} $$

Example 51. Separate into real and imaginary part$\cos^{-1}\left(\frac{3 i}{4}\right)$(M.L. 2003$)$

Solution. Let$\cos ^{-1}\left(\frac{3 i}{4}\right)\hspace{.1in}=x+iy$$$ \begin{aligned} \Rightarrow \frac{3 i}{4}&=\cos (x+i y) \quad \Rightarrow \frac{3 i}{4}=\cos x \cosh y-i \sin x \sinh y \end{aligned} $$Equating real and imaginary parts, we get$therefore \hspace{.3in}\quad \cos x \cosh y=0 \Rightarrow \cos x=0 \Rightarrow x=\frac{\pi}{2}$\par$$ \begin{aligned} & \qquad\text { and } \quad-\sin : x : \sinh : y=\frac{3}{4} \\ & \qquad-1 \sinh y=\frac{3}{4} \qquad\quad \sin x=\sin \left(\frac{\pi}{2}\right)=1 \\ & therefore \qquad\quad \sinh y=-\frac{3}{4} \\ & \Rightarrow y=\log \left(\frac{-3}{4}+\sqrt{1+\frac{9}{16}}\right) \Rightarrow y=\log \left(\frac{-3}{4}+\frac{5}{4}\right)=-\log 2=\log \left(\frac{1}{2}\right) \\ & therefore \text { Real part }=\frac{\pi}{2} \text { and imaginary Part }=-\log 2 \qquad \textbf{Proved.} \end{aligned} $$

Example 52. Separate \tan${ }^{-1}$$(\cos \theta+isin \theta)$into real and imaginary parts.(M.C. 2009)

Solution. Let$\tan ^{-1}(\cos \theta+i \sin \theta)=x+i y$$$ \begin{aligned} & \Rightarrow \qquad\quad \cos \theta+i \sin \theta=\tan (x+dot{mathrm{iy}}) \\ & \text { Similarly, } \quad \cos \theta-i \sin \theta=\tan (x-i y) \\ & \qquad\tan 2 x=\tan [(x+i y)+(x-i y)]=\frac{\tan (x+i y)+\tan (x-i y)}{1-\tan (x+iy) \tan (x-i y)} \\ & =\frac{(\cos \theta+i \sin \theta)+(\cos \theta-i \sin \theta)}{1-(\cos \theta+i \sin \theta)(\cos \theta-i \sin \theta)}=\frac{2 \cos \theta}{1-\left(\cos ^2 \theta+\sin ^2 \theta \right))} \\ & \qquad=\frac{2 \cos \theta}{1-1}=\frac{2 \cos \theta}{0}=\infty=\tan \frac{\pi}{2} \\ & \qquad\tan 2 x=\tan \left(n \pi+\frac{\pi}{2}\right) \Rightarrow 2 x-n \pi+\frac{\pi}{2} \Rightarrow x=\frac{n \pi}{2}+\frac{\pi}{4} \\ & \end{aligned} $$Now, \tan$$ \begin{aligned} 2 mathrm{iy}&=\tan [(x+i y)-(x-dot{y})]=\frac{\tan (x+i y)-\tan (x-i y)}{1+\tan (x+i y) \tan (x-i y)} \end{aligned} $$=\frac{(\cos \theta+i \sin \theta)-(\cos \theta-i \sin \theta)}{1+(\cos \theta+i \sin \theta)(\cos \theta-i \sin \theta)}=\frac{2 i \sin \theta}{1+\left(\cos ^2 \theta+\sin ^2 \theta \right))}=\frac{2 i \sin \theta}{1+1}=i \sin \theta$$ \begin{aligned} i \tanh 2 y&=i \sin \theta \Rightarrow \frac{e^{2 y}-e^{-2 y}}{e^{2 y}+e^{-2 y}}=\frac{\sin \theta}{1} \end{aligned} $$By componendo and dividendo, we have$$ \begin{aligned} \frac{2 e^{2y}}{2 e^{-2 y}}=\frac{1+\sin \theta}{1-\sin \theta} \Rightarrow e^4y & =\frac{1+\cos \left(\frac{\pi}{2}-\theta \right))}{1-\cos \left(\frac{\pi}{2}-\theta \right))}=\frac{1+2 \cos ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)-1}{1-\left[1-2 \sin ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right]} \\ & =\frac{\cos ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{\sin ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\cot ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right) \Rightarrow e^{2y}=\cot \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \\ \Rightarrow \qquad\quad 2 y & =\log \cot \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \Rightarrow y=\frac{1}{2} \log \cot \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \end{aligned} $$%%%%%%%%%%%%%% page nnumber 38 %%%%%%%%%%$$ \begin{aligned} & \qquad\text { Imaginary part }=\frac{1}{2} \log \cot \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \\ & \qquad \text { Real part }=\frac{n \pi}{2}+\frac{\pi}{4} \\ & \tan ^{-1}(\cos \theta+i \sin \theta)=\frac{npi}{2}+\frac{\pi}{4}+\frac{i}{2} \log \cot \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \qquad \textbf{Ans}. \end{aligned} $$

Example 53. Separate$\tan ^{-1}(a+i b)$into real and imaginary parts. (Nagpur University, Summer 2008, 2004)

Solution. Lettan${ }^{-1}(a+i b)=x+i y$\hspace{4in} \dots(1)$therefore \hspace{.8in}\quad \tan (x+i y)=a+i b$On both sides for$i$write$-i$we get$\hspace{1in}therefore \tan (x-i y)=a-i d$Now, \hspace{.3in}$\quad \tan 2 x=\tan [(x+i y)+(x-i y)]$\begin{gathered}
\hspace{0in}=\frac{\tan (x+i y)+\tan (x-i y)}{1-\tan (x+i 4) \tan (x-i y)}=\frac{a+i b+a-i b}{1-(a+i b)(a-i b)}=\frac{2 a}{1-a^2-b^2}
\hspace{1.1in}2 x=\tan ^{-1}\left[\frac{2 a}{1-a^2-b^2}\right] \Rightarrow x=\frac{1}{2} \tan ^{-1}\left[\frac{2 a}{1-a^2-b^2}\right]\hspace{1.6in} \dots(2)
\end{gathered}$and$$$ \begin{aligned} \tan (2: y: i) & =\tan [(x+i y)-(x-i y)] \\ & =\frac{\tan (x+i y)-\tan (x-i y)}{1+\tan (x+i y) \tan (x-i y)}=\frac{a+b i-a+b i}{1+(a+b i)(a-b i)} \\ i \tanh 2 y & =\frac{2 b i}{1+a^2+b^2} \operatorname{so,} \tanh 2 y=\frac{2 b}{1+a^2+b^2} \\ 2 y & =\tanh ^{-1}\left[\frac{2 b}{1+a^2+b^2}\right] \\ \operatorname{so:y} & =\frac{1}{2} \tanh ^{-1}\left[\frac{2 b}{1+a^2+b^2}\right] \qquad \dots(3) \end{aligned} $$From (1), (2), and (3), we here$\tan ^{-1}(a+i dot{b})=\frac{1}{2} \tan ^{-1}\left[\frac{2 a}{1-a^2-b^2}\right]+\frac{i}{2} \tanh ^{-1}\left[\frac{2 b}{1+a^2+b^2}\right] \qquad \textbf{Ans.}$

Example 54. Show that$\tan ^{-1} i \left(\frac{x-a}{x+a}\right)=\frac{i}{2} \log \left(\frac{x}{a}\right).$(M.U. 2005, 3002)

Solution.: Let \hspace{.5in}$\tan ^{-1} i \left(\frac{x-a}{x+a}\right)=u+i v$$$ \begin{aligned} &\Rightarrow \tan (u+i v)=i \left(\frac{x-a}{x+a}\right) \text { and } \tan (u-i v)=-i \left(\frac{x-a}{x+a}\right) \\ &\qquad\tan 2 u=\tan [(u+iv)+(u-i v)]=\frac{\tan (u+i v)+\tan (u-i v)}{1-\tan (u+i v) \tan (u-i v)}=\frac{i x-i a-i x+i a}{x+a}=0 \end{aligned} $$\par$therefore \hspace{.3in}\quad \tan 2 u=0 \Rightarrow 2 u=0 \Rightarrow u=0$Putting the value of uin (1), we get$$ \begin{aligned} & therefore \quad \tan ^{-i} i \left(\frac{x-a}{x+a}\right)-i v \qquad therefore i( \frac{x-a}{x+a}) = \text{\tan} : i : v : =i: \text{\tanh}: v \\ & therefore \quad \frac{x-a}{x+a}=\tanh v=\frac{e^v-e^{-v}}{e^v+e^{-v}} \end{aligned} $$By Component and dividondo, we get$$ \begin{gathered} \frac{2 x}{2 a}=\frac{2 e^v}{2 e^{-v}} \Rightarrow \frac{x}{a}=e^{2 v} \Rightarrow v=\frac{1}{2} \log \left(\frac{x}{a}\right) therefore \tan ^{-1} i \left(\frac{x-a}{x+a}\right)=u+iv=0+\frac{i}{2} \log \frac{x}{a}=\frac{{i}}{2} \log \left(\frac{u}{iv}\right)\qquad \textbf{Proved.} \end{gathered} $$

Example 55. Prove that (i)$\cosh ^{-1} \sqrt{1+x^2}=\sinh ^{-1} x$(M.U 2007) (ii)$\cosh ^{-1} \sqrt{1+x^2}=\tanh ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$(M.U). 2002)

Solution (i) Let$\cosh ^{-1} \sqrt{1+x^2}=y$\hspace{4.2in}\dots (1)$\Rightarrow \hspace{0in}\quad \sqrt{1+x^2}=\operatorname{\cosh} y \hspace{5.2in}\dots (2)$On squaring both sides, we get$$ \begin{aligned} & \qquad1+x^2=\cosh ^2 y \\ & therefore \qquad\quad x^2=\cosh ^2 y-1 \quad \Rightarrow x^2=\sinh ^2 y \\ & \Rightarrow \qquad\quad x=\sinh : y \qquad\dots (3) \\ & \Rightarrow \qquad\quad y=\sinh ^{-1} x \\ & \Rightarrow \qquad\quad \cosh ^{-1} \sqrt{1+x^2}=\sinh ^{-1} x \text{[Using (1)] \qquad \textbf{Proved}.} \\ & \end{aligned} $$(ii) Dividing (3) by (2), we get$$ \begin{aligned} \frac{\sinh y}{\cosh y}&=\frac{x}{\sqrt{1+x^2}} \\ & \Rightarrow \qquad\quad \tanh y=\frac{x}{\sqrt{1+x^2}} \Rightarrow y=\tanh ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) \\ & \Rightarrow \qquad \quad \cosh ^{-1} \sqrt{1+x^2}=\tanh ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)[\text { Using (1)] } \qquad \textbf{Proved.} \\ & \end{aligned} $$

Chapter 1

Example 1. Simplify the following: (a)\(i^{49}\), (b)\(i^{103}\).

Example 2. Multiply 3 + 4i by 7 − 3i

Example 3. Find out the conjugate of a complex number 7 + 6i.

Example 4. Divide 1 + i by 3 + 4i.

Example:5 If\(\begin{aligned} a&=\cos \theta+i \sin \theta, \text{Prove that}: 1+a+a^2=(l+2 \cos \theta)(\cos \theta+i \sin \theta) \end{aligned}\).

Example 6. If\(a^2+b^2+c^2=1\)and\(b+ic=(1+a)z\), prove that\(\frac{a+ib}{1+c}=\frac{1+iz}{1-iz}\).

Example 7. If\(z=\cos \theta+i \sin \theta\), prove that\(\frac{2}{1+z}=1-i \tan \frac{\theta}{2}\).

Example 8. If\(x=\cos \theta+i \sin \theta, y=\cos \phi+i \sin \phi\), prove that\(\begin{aligned} \frac{x-y}{x+y}&=i \tan \left(\frac{\theta-\phi}{2}\right)\qquad(M.U. 2008) \end{aligned}\)

Example 9. Find the modulus and principal argument of the complex number

Example 10. Find the modulus and principal argument of the complex number : \( 1+\cos \alpha+i \sin \alpha \left(0<\alpha<\frac{\pi}{2}\right) \)

Example 11. If\(z_1\)and\(z_2\)are any two complex numbers, prove that\(\left|z_1+z_2\right|^2+\left|z_1-z_2\right|^2=2\left[\left|z_1\right|^2+\left|z_2\right|^2\right]\)

Example 12. If\(z_1\)and\(z_2\)are two complex numbers such that\(\left|z_1+z_2\right|=\left|z_1-z_2\right| \text {, prove that }\)\(arg \cdot z_1-arg \cdot z_2=\frac{\pi}{2}\)(M. 2002, 2007)

Example 13. Find the complex number\(z\)if\(\arg(z+1)=\frac{\pi}{6}\)and\(\arg(z-I)=\frac{2 \pi}{3}\). (M.U. 2009, 2000, 01,02, 03)

Example 14. Prove that (i)\(\left|z_1+z_2\right| \leq \left|z_1\right|+\left|z_2\right|\)(ii)\(\left|z_1-z_2\right| \geq \left|z_1\right|-\left|z_2\right|\)

Example 15. Express in polar form :\(1-\sqrt{2}+i\)

Example 16. Find the smallest positive integer n for which (Nagpur University, Winter 2004)\(\begin{aligned} (\frac{1+i}{1-i})^n&=1. \end{aligned}\)

Example 17.Find the square root of the complex number\(5+12 i\)

Example 18. Prove that if the sum and product of two complex numbers are real then the two numbers must be either real or conjugate.\((M U .2008)\)

Example 19. Express\(\begin{aligned} \frac{(\cos \theta+i \sin \theta)^8}{(\sin \theta+i \cos \theta)^4} \text{in the form} (x+i y). \end{aligned}\)

Example 20. Prove that\((1+\cos \theta+i \sin \theta)^n+(1+\cos \theta-i \sin \theta)^n=2^{n+1} \cos ^n \frac{\theta}{2} \cos \frac{n \theta}{2}\)

Example 21. Evaluate\(\begin{aligned} \left(\frac{1+\sin \alpha+i \cos \alpha}{1+\sin \alpha-i \cos \alpha}\right)^n \quad \end{aligned}\)(MU. 2001, 2004, 2005)

Example 25. Solve\(x^4+i=0\)(M U .2008)

Example 26. Solve\(x^5=1+i\)and find the continued product of the roots. (M U .2005,2004)

Example 27. If\(\alpha, \alpha^2, \alpha^3, \alpha^4\), are the roots of\(x^5-1=0\)find them and show that\((1-\alpha)\left(1-\alpha^2 \right)\left(1-\alpha^3 \right)\left(1-\alpha^4 \right)=5\)(M U .2007)

Example 28. If\(\omega\)is a cube root of unity, prove that\(\begin{aligned} (1-\omega)^6&=-27 \qquad(MU. 2003) \end{aligned}\)

Example 29. Use De Movie's theorem to solve the equation\(x^4-x^2+x^2-x+1=0\).

Example 30. Prove that\((\cosh x-\sinh x)^n=\cosh n x-\sinh n x\). (MU. 2001, 2002)

Example 31.If :\(\tan (\theta+i \phi)=\cos \alpha+i \sin \alpha\)prove that:\(\theta=\frac{npi}{2}+\frac{\pi}{4}\)and\(\phi=\frac{1}{2} \log \tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right) \quad\)(Nagpur University, Summer 2002, Writer 200I)

Example 32. If\(\cosh x=\sec \theta\), prove that:\((i) \theta=\frac{\pi}{2}-2 \tan ^{-1}\left(e^{-x}\right)\)(ii)\(\tanh \frac{\pi}{2}=\tan \frac{\theta}{2}\)(M.U. 2003, 2005)v

Example 33. Separate the following into real and imaginary parts: (i)\(\sin (x+i y)\)(ii)\(\cos (x+i y)\)(iii)\(\tan (x+i y)\)

Example 34. If\(\tan (A+i B)=x+i y\), prove that\(\begin{aligned} \tan 2 A&=\frac{2 x}{1-x^2-y^2} \text{} 2 B=\frac{2 x}{1+x^2+y^2} \quad \end{aligned}\)(Nagpur University, Summer 2000)

Example 35. If\(\sin (\alpha+i \beta)=x+i y\), prove that (a)\(\frac{x^2}{\cosh ^2 \beta}+\frac{y^2}{\sinh ^2 \beta}=1\)(b)\(\frac{x^2}{\sin ^2 \alpha}-\frac{y^2}{\cos ^2 \alpha}=1\)

Example 36. Separate the following into real and imaginary parts of hyperbolic functions. (a)\(\sinh (x+i y)\)(b)\(\cosh (x+i y)\)(c)\(\tanh (x+i y)\)

Example 37. If\(\tan (x+i y)=\sin (u+i v)\), prove that\(\begin{aligned} \frac{\sin 2 x}{\sinh 2 y}&=\frac{\tan u}{\tanh v} \end{aligned}\)

Example 38. If\(\sin (\theta+i \phi)=\tan \alpha+i \sec \alpha\), show that\(\cos 2 \theta \cosh 2 \phi=3\)

Example 39. If\(e^z=\sin (u+i v)\)and\(z=x+i y\), prove that\(\begin{aligned} & \quad 2 e^{2 x} & =\cosh 2 v-\cos 2 u \end{aligned}\)

Example 40. If\(\sin (\theta+i \phi)=\cos \alpha+i \sin \alpha\), prove that\(\begin{aligned} \cos ^4 \theta&=\sin ^2 \alpha=\sinh ^4 \phi \qquad(M.U. 2003, 2004) \end{aligned}\)

Example 41. If\(\operatorname{cosec}\left(\frac{\pi}{4}+i x \right)=u+i v\), prove that\(\left(u^2+v^2 \right)^2=2\left(u^2-v^2 \right) (MU. 2009)\)v

Example 42. Separate into real and imaginary parts: [ \left(\sqrt{i}\right)^{\sqrt{i}} ] (M.U. 2008)

Example 43. Define the logarithm of a complex number.

Example 44. Separate \log\((x+i v)\)into its real and imaginary parts

Example 45. Show that\(\begin{aligned} \log \frac{x+i y}{x-i y}&=2 i \tan ^{-1} \frac{y}{x} \end{aligned}\). (Nagpur University, Winter 2003)

Example 46. Show that for real values of\(a\)and\(b\)\( e^{2 a i \cot ^{-1} b}\left[\frac{b i-1}{b i+1}\right]^{-a}=1 (M.U. 2008) \)

Example 47. Prove that\(\sinh ^{-1} x=\log \left(x+\sqrt{x^2+1}\right)\)(M+2009)

Example 48. Prove that\(\cosh^{-1} x=\log \left(x+\sqrt{x^2-1}\right)\)(ML. 2009)

Example 49. Prove that\(\operatorname{\operatorname{sech}}^{-1} x=\log \frac{1+\sqrt{1-x^2}}{x}\)

Example 50. If\(x+i y=\cos (\alpha+i \beta)\)or if\(\cos ^{-1}(x+i y)=\alpha+i \beta\)express\(x\)and\(y\)in terms of\(\alpha\)and\(\beta\). Hence show that\(\cos ^2 \alpha\)and\(\cosh ^2 \beta\)are the roots of the equation\(\lambda^2-\left(x^2+y^2+1 \right) \lambda+x^2=0\). (M.U. 2002, 2004)

Example 51. Separate into real and imaginary part\(\cos^{-1}\left(\frac{3 i}{4}\right)\)(M.L. 2003\()\)

Example 52. Separate \tan\({ }^{-1}\)\((\cos \theta+isin \theta)\)into real and imaginary parts.(M.C. 2009)

Example 53. Separate\(\tan ^{-1}(a+i b)\)into real and imaginary parts. (Nagpur University, Summer 2008, 2004)

Example 54. Show that\(\tan ^{-1} i \left(\frac{x-a}{x+a}\right)=\frac{i}{2} \log \left(\frac{x}{a}\right).\)(M.U. 2005, 3002)

Example 55. Prove that (i)\(\cosh ^{-1} \sqrt{1+x^2}=\sinh ^{-1} x\)(M.U 2007) (ii)\(\cosh ^{-1} \sqrt{1+x^2}=\tanh ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\)(M.U). 2002)

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