# Complex Numbers Involving a Circle

## Homework Statement

Show that the equation $$|z - z_0| = R$$ of a circle, centered at $$z_0$$ with radius R, can be written

$$|z|^2 - 2Re(z\bar{z_0}) + |z_0|^2 = R^2$$.

## The Attempt at a Solution

Honestly, I have no clue where to start with this problem. I know that I need to reduce the given equation to the basic equation of a circle but I do not know where to start.
I also know that the two equations are almost exact except for the $$- 2Re(z\bar{z_0})$$ which should reduce to zero somehow I just do not know where to start.

I know $$Re(z) = Re(\bar{z}) = Re\frac{(z + \bar{z})}{2} = x$$. Is this where I start?

rock.freak667
Homework Helper

Thank you for that tidbit it really helped me almost solve it. So here is what I got and I am stumped:

$$2Re(z\bar{z_0}) = 2\frac{z\bar{z_0} + \overline{z\bar{z_0}}}{2} = z\bar{z_0} + \bar{z}z_0$$

$$z\bar{z_0} = (x+iy)(x_0 - iy_0) = (xx_0 - yy_0 + iyx_0 - xiy_0)$$

$$\bar{z}z_0 = (x - iy)(x_0 + iy_0) = (xx_0 - yy_0 - iyx_0 + xiy_0)$$

So $$z\bar{z_0} + \bar{z}z_0 = 2xx_0 - 2yy_0$$

Is there a method where I can simplify this anymore? because I'm clueless.

HallsofIvy
Homework Helper
$|z- z_0|= \sqrt{(z- z_0)(\overline{z}- \overline{z_0})}$ so $z-z_0|= R$ is the same as $|z- z_0|= \sqrt{(z- z_0)(\overline{z}- \overline{z_0})}= R$. square both sides of that. I would NOT go to "x+ iy".

Sorry about the nessed up Latex!

Last edited by a moderator:
$$|z- z_0|= \sqrt{(z- z_0)(\overline{z}- \overline{z_0})}$$ so $$|z-z_0|= R$$ is the same as $$|z- z_0|= \sqrt{(z- z_0)(\overline{z}- \overline{z_0})}= R$$. square both sides of that. I would NOT go to "x+ iy".

Thank you Halls but that seriously confuses me. I do not understand how I should apply it to my problem. Anyone else got anything in mind?

Neither would I use the "x+iy" route.
Recall the identity $$z\overline{z} = |z|^{2}$$

Thank you so much Fightfish. It was so easy to solve once I used that identity. Goes to show I need to learn the properties better.