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.
Best Answer
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 $z\bar{z}=1$.