<s>[INST] 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.
 [/INST] Note that we can resort to trigonometric formulae.
$$\frac{z-1}{z+1}=\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}=\frac{1-2\sin^2\frac{\theta}{2}+2i\cos\frac{\theta}{2}\sin\frac{\theta}{2}-1}{2\cos^2\frac{\theta}{2}-1+2i\cos\frac{\theta}{2}\sin\frac{\theta}{2}+1}=\frac{2i\sin\frac{\theta}{2}\cdot e^{i\theta/2}}{2\cos\frac{\theta}{2}\cdot e^{i\theta/2}}=i\tan\frac{\theta}{2}\implies\Re\left(\frac{z-1}{z+1}\right)=0$$
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