Show that the equation $|z-z_0|=R$ of a circle centered at $z_0$ of radius $R$ can be written as $|z|^2-2\text{Re}(z\bar{z_0})+|z_0|^2=R^2$. Show that the equation $|z-z_0|=R$ of a circle centered at $z_0$ of radius $R$ can be written as $|z|^2-2\text{Re}(z\bar{z_0})+|z_0|^2=R^2$.
I tried squaring both sides, but then what?

We have that
$$R^2=|z-z_0|^2=(z-z_0)\cdot \overline{(z-z_0)}=(z-z_0)\cdot (\overline{z}-\overline{z_0})\\
=z\cdot \overline{z}-z\cdot \overline{z_0}
-\overline{z}\cdot z_0  +z_0\cdot\overline{z_0}\\
=|z|^2-\left(z\cdot \overline{z_0}+\overline{z\cdot \overline{z_0}}\right)+|z_0|^2\\
=|z|^2-2\mbox{Re}\left(z\cdot \overline{z_0}\right)+|z_0|^2.$$
