What is the modulus of $|\frac{1}{z}+\frac{z}{r^2}|$ on the circle of radius r?

Question

I know that $|\frac{1}{z}+\frac{z}{r^2}|=\frac{2Re(z)}{r^2}$ but I can't seem to show it.

Possible Proof: $|\frac{1}{z}+\frac{z}{r^2}|=|\frac{z^2+r^2}{r^2z}|=|r^2\frac{1+e^{i\theta}}{r^3}|$. Here is where im stuck any ideas?

Answer 1

Hint: $|a+b|^{2}=|a|^{2}+|b|^{2}+2\Re {a \overline {b}}$; also when $|z|=r$ we have $z \overline {z}=r^{2}$ so $\overline {z}=\frac {r^{2}} z$.