# 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.

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 $$z=i\tan \theta.$$