====== Question 4, Exercise 1.4 ======
Solutions of Question 4 of Exercise 1.4 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 4=====
If $\dfrac{1+z}{1-z}=\cos 2 \theta+i \sin 2 \theta$, show that $z=i \tan \theta$

** Solution. **

Given 
\begin{align}&\dfrac{1+z}{1-z}=\cos 2 \theta+i \sin 2 \theta\\
\implies &\dfrac{1+z}{1-z}=e^{i2\theta}\\
\implies &(1+z)=(1-z)e^{i2\theta}\\
\implies &z+z e^{i2\theta}=e^{i2\theta}-1\
\end{align}
\begin{align}
z & =\dfrac{e^{i2\theta}-1}{e^{i2\theta}+1}\\
& =\dfrac{\cos 2\theta+i\sin 2\theta-1}{\cos 2\theta+i\sin 2\theta+1}\\
& =\dfrac{\cos^2\theta-\sin^2\theta +i2\sin\theta \cos\theta -1}{\cos^2\theta-\sin^2\theta +i2\sin\theta \cos\theta +1}\\
& =\dfrac{-(1-\cos^2\theta)-\sin^2\theta +i2\sin\theta \cos\theta}{\cos^2\theta+1-\sin^2\theta +i2\sin\theta \cos\theta}\\
& =\dfrac{-\sin^2\theta-\sin^2\theta +i2\sin\theta \cos\theta}{\cos^2\theta+\cos^2\theta +i2\sin\theta \cos\theta}\\
& =\dfrac{-2\sin^2\theta+i2\sin\theta \cos\theta}{2\cos^2\theta +i2\sin\theta \cos\theta)}\\
& =\dfrac{2\sin^\theta(-\sin\theta+i\cos\theta}{2\cos\theta(\cos\theta +i2\sin\theta)}\\
& =\tan\theta\left(\dfrac{i^2\sin\theta+i\cos\theta}{\cos\theta +i\sin\theta}\right)\\
& =i\tan\theta\left(\dfrac{\cos\theta+i\sin\theta}{\cos\theta +i\sin\theta}\right)\\
\end{align}
This implies
$$
z=i\tan \theta.
$$
====Go to ====
<text align="left"><btn type="primary">[[math-11-nbf:sol:unit01:ex1-4-p3|< Question 3]]</btn></text>
<text align="right"><btn type="success">[[math-11-nbf:sol:unit01:ex1-4-p5|Question 5 >]]</btn></text>