Shorter proof for expression of a circle in the complex plane

Question

In Complex Variables and Applications, by Brown and Chyrchill (McGraw-Hill), there is the following exercise: $$\textit{Show that $|z-z_0|=R$ can be written as $|z|^2-2Re(z\bar{z_0})+|z_0|^2=R^2$}$$

I solved it this way:

1. Square both sides: $|z-z_0|^2 = R^2$
2. Define $w:=z-z_0$, then we have $w\bar{w}=|w|^2=R^2$
3. Compute $(z-z_0)(\bar{z}-\bar{z_0}) = z\bar{z}-z\bar{z_0}-z_0\bar{z}+z_0\bar{z_0}=|z|^2-z\bar{z_0}-z_0\bar{z}+|z_0|^2 = R^2$
4. Compute $-z_0\bar{z} = -x_0x+x_0yi-xy_0i-yy_0$ and $-z\bar{z_0}=-xx_0+xy_0i-x_0yi-yy_0$
5. Sum them $-z_0\bar{z}-z\bar{z_0}=-2(x_0x_0+yy_0)=-2Re(z\bar{z_0})$
6. Hence $|z|^2-2Re(z\bar{z_0})+|z_0|^2 = R^2$

But it looks quite tedious and long. I am sure there must be a shorter way to do this, can anyone think of some other way? I tried $|z-z_0|^2 = |(z-z_0)^2|=|z^2-2zz_0+z_0^2|=R^2$, but this leads me nowhere.

Thank you

Answer 1

What you did is good. It is quite tedious because you detailed the steps.

You should remember the formula: $$\vert a - b \vert^2 = \vert a \vert^2 - 2Re(a\bar{b}) + \vert b \vert^2$$ as it is constantly used. Then the equation of a circle can be find in a much straightforward manner.