Complex number problem $i \tan (\theta)$ proof.

Question

Given $z = \cos (\theta) + i \sin (\theta)$, prove $\dfrac{z^{2}-1}{z^{2}+1} = i \tan(\theta)$

I know $|z|=1$ so its locus is a circle of radius $1$; and so $z^{2}$ is also on the circle but with argument $2\theta$; and $z^{2}+1$ has argument $\theta$ (isosceles triangle) so it lies on a line through the origin and $z$.

$z^{2}-1$, $z^{2}$, and $z^{2}-1$ all lie on a horizontal line $i \sin (\theta)$.

On the Argand diagram I can show $z^{2}+1$ and $z^{2}-1$ are perpendicular so the result follows.

Can anyone give an algebraic proof?

Answer 1

As $z\ne0,$ with $|z|=1$

$$\dfrac{z^2-1}{z^2+1}=\dfrac{z-\dfrac1z}{z+\dfrac1z}$$

Now $\dfrac1z=\dfrac1{\cos\theta+i\sin\theta}=\cos\theta-i\sin\theta$

Answer 2

For $z = \cos \theta + i \sin \theta$ you have $$ \frac{z^2-1}{z^2+1} = \frac{ \cos^2 \theta - \sin^2 \theta + 2i \cos\theta \sin \theta -1}{\cos^2 \theta - \sin^2 \theta + 2i \cos\theta \sin \theta +1 } \, . $$ Now substitute $1 = \cos^2 \theta + \sin^2 \theta $ in both numerator and denominator, and collect terms: $$ \frac{-2 \sin^2 \theta + 2 i \cos\theta \sin\theta}{2 \cos^2\theta + 2 i \cos\theta \sin\theta} = \frac{- \sin\theta + i \cos \theta}{\cos \theta + i \sin\theta} \cdot \frac{\sin\theta }{\cos\theta} \, . $$ Finally convince yourself that the first factor is equal to $i$, and you are done.

Answer 3

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

Answer 4

$\dfrac{z^2-1}{z^2+1}$ reminds me of Componendo and Dividendo?

$\dfrac{z^2}1=(\cos\theta+i\sin\theta)^2=\dfrac{\cos\theta+i\sin\theta}{\cos\theta-i\sin\theta}$ as $\dfrac1{\cos\theta-i\sin\theta}=\cos\theta+i\sin\theta$

Now apply Componendo and Dividendo