Calculations with complex numbers So I encountered this one question today in my math book, and I don't know how to get the right answer even though it seems really easy, I just wanna know how to do it so i can get some sleep.
This is the question:

When the complex number $z=r(\cos\theta + i \sin\theta)$ show that $z-\frac{1}{z}=i(2r\sin\theta)$

I haven't done much for this question other than do this using the actual complex number:
$$\frac{z^2-1}{z}$$
Can anyone show me how to do prove this question, any help is appreciated.

The identity is not true, indeed we have that
$$z-\frac1z=z-\frac{\bar z}{|z|^2}=r(\cos\theta + i \sin\theta)-\frac1r(\cos\theta - i \sin\theta)=\left(r-\frac1r\right)\cos \theta+\left(r+\frac1r\right)i\sin \theta$$
the identity is true only for $\left(r-\frac1r\right)=0$ and $\left(r+\frac1r\right)=2r$ that is $r=1$ or $z=e^{i\theta}$.
