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?
Best Answer
As $z\ne0,$ with $|z|=1$
$$\dfrac{z^2-1}{z^2+1}=\dfrac{z-\dfrac1z}{z+\dfrac1z}$$
Now $\dfrac1z=\dfrac1{\cos\theta+i\sin\theta}=\cos\theta-i\sin\theta$