Given that $z=\cos\theta+i\sin\theta$, show that $Re\left(\frac{z-1}{z+1}\right)=0, \quad z\ne-1$ 
Given that  $z=\cos\theta+i\sin\theta$, show that
  $Re\left(\frac{z-1}{z+1}\right)=0, \quad z\ne-1$

For this question I had to show that the real part of $\frac{z-1}{z+1}=0$ 
To find that I first substituted $z$ with $\cos\theta+i\sin\theta$ to get
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}$$
I then multiplied by the conjugate of the denominator 
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}\cdot\frac{\cos\theta-i\sin\theta+1}{\cos\theta-i\sin\theta+1}$$
Which, when expanded, gives me
$$\frac{\cos^2\theta+\sin^2\theta+2i\sin\theta-1}{\cos^2\theta+\sin^2\theta+2i\cos\theta+1}$$
After that, 
$$\frac{2i\sin\theta}{2\cos\theta+2}=\frac{i\sin\theta}{\cos\theta+1}$$
How do I proceed?
Edit: Just realized (after being told by a commenter) that $\frac{i\sin\theta}{\cos\theta+1}$ is imaginary. I won't delete the question. Ínstead I'll leave it here, as a testament to my stupidity, for everyone to see. May God have mercy on my soul during my exam.
 A: A complex number $w$ is purely imaginary (that is, its real part is $0$) if and only if $w+\bar{w}=0$.
You're given that $|z|=1$; then the number plus its conjugate is
$$
\frac{z-1}{z+1}+\frac{\bar{z}-1}{\bar{z}+1}=
\frac{z\bar{z}-\bar{z}+z-1+z\bar{z}+\bar{z}-z-1}{(z+1)(\bar{z}+1)}=0
$$
because $z\bar{z}=1$.
A: You are done indeed
$$\frac{i\sin\theta}{\cos\theta+1}=i\cdot \frac{\sin\theta}{\cos\theta+1}$$
is purely imaginary.
As an alternative since $z\bar z=|z|^2=\cos^2 \theta+\sin^2 \theta=1$ we have that
$$\frac{z-1}{z+1}=\frac{z-1}{z+1}\frac{\bar z+1}{\bar z+1}=\frac{z\bar z+z-\bar z-1}{z\bar z+z+\bar z+1}=i\cdot \frac{\,\Im(z)}{\Re(z)+1}$$
which is again purely imaginary.
A: Yet another way:   $|z| = 1 \iff z\bar z=1 \iff \bar z = \dfrac{1}{z}\,$, then using that $\,|\operatorname{Re}(z)| \le |z| \le 1\,$:
$$
\left(\frac{z-1}{z+1}\right)^2=\frac{z^2-2z+1}{z^2+2z+1}=\frac{z-2+\dfrac{1}{z}}{z+2+\dfrac{1}{z}} = \frac{z+\bar z - 2}{z+\bar z +2} = \frac{2\operatorname{Re}(z)-2}{2\operatorname{Re}(z)+2} = \frac{\operatorname{Re}(z)-1}{\operatorname{Re}(z)+1} \;\le\; 0
$$
Since the square of $\,\dfrac{z-1}{z+1}\,$ is a non-positive real, $\,\dfrac{z-1}{z+1}\,$ is either $\,0\,$, or purely imaginary.
A: Note that we can resort to trigonometric formulae.
$$\frac{z-1}{z+1}=\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}=\frac{1-2\sin^2\frac{\theta}{2}+2i\cos\frac{\theta}{2}\sin\frac{\theta}{2}-1}{2\cos^2\frac{\theta}{2}-1+2i\cos\frac{\theta}{2}\sin\frac{\theta}{2}+1}=\frac{2i\sin\frac{\theta}{2}\cdot e^{i\theta/2}}{2\cos\frac{\theta}{2}\cdot e^{i\theta/2}}=i\tan\frac{\theta}{2}\implies\Re\left(\frac{z-1}{z+1}\right)=0$$
