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.

On the unit circle $z=1/\overline z$ where the overline in complex conjugation.  So
$\dfrac{z-1}{z+1}=\dfrac{1/\overline z-1}{1/\overline z+1}=\dfrac{1-\overline z}{1+\overline z}=-\dfrac{\overline z-1}{\overline z+1}$
Thus $\dfrac{z-1}{z+1}$ is the negative of its own complex conjugate, therefore pure imaginary.  The real part is zero.
