If $z=\cos\theta + i\sin \theta$ prove $\frac{z^2-1}{z^2+1}=i\tan\theta$

If $$z=\cos\theta + i\sin \theta$$ prove $$\frac{z^2-1}{z^2+1}=i\tan\theta$$

Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.

$$\frac{(\cos\theta + i\sin \theta)^2-1}{(\cos\theta + i\sin \theta)^2+1}$$ $$\frac{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)-1}{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)+1}$$ $$\frac{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta) -1}{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta)+1}$$ $$\frac{\cos2\theta + i\sin 2\theta -1}{\cos2\theta + i\sin 2\theta +1}$$

I understand how I can do it with using $$z=e^{i \theta}$$, however I want to solve it using double angle identities.

• Just using trig identities multiply top and bottom by $(\cos \theta -i\sin\theta).$ You could also use $(\cos 2\theta -i\sin2 \theta + 1)$ – Doug M Jan 14 at 21:58

Your approach will work with double angle formulae, but this is quicker: since $$z=\exp i\theta$$, $$\frac{z-1/z}{z+1/z}=\frac{2i\sin\theta}{2\cos\theta}$$.

$$z=e^{i\theta}$$, $$\frac{z^2-1}{z^2+1}=\frac{e^{2 i\theta}-1}{e^{2i\theta}+1}=\frac{e^{ i\theta}-e^{ -i\theta}}{e^{i\theta}+e^{ i\theta}}=\frac{2 i\, sin(\theta)}{2\,cos(\theta)}=itan(\theta)$$

$$\cos \theta +i\sin \theta =e ^{i \theta}$$. Then

\begin{aligned} \frac{z^2-1}{z^2+1}&=\frac{z-1/z}{z+1/z}\\ &=\frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}\\ &=\frac{2 i \sin \theta}{2 \cos \theta}\\ &=i\tan\theta \end{aligned}

If you divide the numerator and denominator of

$$\frac{(\cos\theta + i\sin \theta)^2-1}{(\cos\theta + i\sin \theta)^2+1}$$

by $$\cos^2\theta$$ and then use the identity $$\sec^2\theta=1+\tan^2$$ on the result you obtain

$$\frac{(1+i\,\tan\theta)^2-1-\tan^2\theta}{(1+i\,\tan\theta)^2+1+\tan^2\theta}$$

which simplifies to

$$\frac{i\,\tan\theta-\tan^2\theta}{1+i\,\tan\theta}$$

Then factor the numerator to get

$$\frac{i\,\tan\theta(1+i\,\tan\theta)}{1+i\,\tan\theta}=i\,\tan\theta$$

Note: This is an approach which can be taken if the student does not yet know that $$e^{i\,\theta}=\cos\theta+i\,\sin\theta$$

Now where you've left off,

use $$\cos2y=2\cos^2y-1=1-2\sin^2y\ \ \ \ (1)$$

to find $$\frac{\cos2\theta + i\sin 2\theta -1}{\cos2\theta + i\sin 2\theta +1}=\dfrac{-2\sin^2\theta+ i\sin 2\theta}{2\cos^2\theta+ i\sin 2\theta}=\dfrac{2i\sin\theta(\cos\theta+i\sin\theta)}{2\cos\theta(\cos\theta+i\sin\theta)}=?$$

Similarly using Using De Moivre's theorem . $$(1+i)^{100}$$, $$z^n=(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$$ use $$(1)$$ to establish $$\dfrac{z^n-1}{z^n+1}=2i\tan\dfrac{n\theta}2$$