Determine the real part of $\left(\cfrac{z-1}{z+1}\right)$ Determine the $Re\left(\cfrac{z-1}{z+1}\right)$ if $z = cos\theta + i \, sin\theta$.
I'm not quite sure whether the right approach would be to stick with the polar form and substitute it into $\left(\cfrac{z-1}{z+1}\right)$ and perhaps use $ 1 = sin^2(\theta) + cos^2(\theta) $. Or maybe I should use the Euler's form.

Note that $z=\cos\theta + i\sin\theta$ implies $|z|=1$. Now,
$$\frac{z-1}{z+1} = \frac{z\bar z-\bar z}{z\bar z + \bar z} = \frac{1-\bar z}{1+\bar z} = - \overline{ \left(\frac{z-1}{z+1}\right)},$$
so, $\frac{z-1}{z+1}$ is purely imaginary, i.e, the real part is $0$.
