Prove: if $|z|=1$ than $\frac{z+1}{z-1}$ is an imaginary number

Prove: if $|z|=1$ than $\frac{z+1}{z-1}$ is an imaginary number

I have tried to look at $$\frac{z+1}{z-1}=\frac{z+1}{z-1}\cdot \overline{\frac{z-1}{\overline{z-1}}}=\frac{z+1}{z-1}\cdot \frac{\overline{z}-1}{\overline{z}-1}$$

But did not get far

Solution 1:

Hint

$$\frac{z+1}{z-1}\cdot\frac{\overline{z}-1}{\overline{z}-1}=\frac{z\cdot\overline{z}+\overline{z}-z-1}{z\cdot\overline{z}-\overline{z}-z+1}$$

but $$z\cdot \overline{z}=|z|^2=1$$ $$z-\overline{z}=2i\text{ Im}(z)$$ $$z+\overline{z}=2\text{ Re}(z)$$

Can you finish?

Solution 2:

A number is imaginary if and only if it is the opposite of its conjugate. Now we have : $$\overline{\left(\frac{z+1}{z-1}\right)}=\frac{\overline{z+1}}{\overline{z-1}}=\frac{\overline{z}+1}{\overline{z}-1}=\frac{z\overline{z}+z}{z\overline{z}-z}=\frac{1+z}{1-z}=-\frac{z+1}{z-1}.$$