Complex Numbers Involving a Circle

[DEMJ]

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.

Homework Equations

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]

Start with z=x+iy and z₀=x₀+iy₀

 

[DEMJ]

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]

|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!

 

[DEMJ]

|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?

 

[Fightfish]

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

 

[DEMJ]

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.