Question: Given that $z= \cos \theta + i\sin \theta$, prove that $\Re\left\{\dfrac{z-1}{z+1}\right\}=0$

How would I do this?

Answer: 
$$\frac{z-1}{z+1}=\frac{z-1}{z+1}\frac{\overline z+1}{\overline z+1}=\frac{|z|^2-1+2i\,\text{Im}\,z}{|z+1|^2}\implies\text{Re}\,\left(\frac{z-1}{z+1}\right)=\frac{|z|^2-1}{|z+1|^2}$$

But we know that $\;z=\cos\theta+i\sin\theta\implies |z|=1\;$ , so...
If $w=a+ib,\bar w=a-ib\implies w+\bar w=2a$

So, $w$ will be purely imaginary iff $ w+\bar w=0$

Here $$\frac{z-1}{z+1}+\overline{\left(\frac{z-1}{z+1}\right)}$$
$$=\frac{z-1}{z+1}+\frac{\bar z-1}{\bar z+1}=\frac{z\bar z-\bar z+z-1+(z\bar z+\bar z-z-1)}{(z+1)(\bar z+1)}$$

$$=\frac{2(z\bar z-1)}{z\bar z+(z+\bar z)+1}$$

Now, $z\bar z=|z|^2=|\cos\theta+i\sin\theta|^2=1$