# Thread: Circles in a complex plane

1. ## Circles in a complex plane

If z = x + iy show that the general equation for a circle in the complex plane is |z|^2 + 2Re(az) > - |a|^2

where a is an arbitrary complex number.

2. Well, first, that's not correct. Was that supposed to be "= ", not ">"? I'm going to assume "="

You know, I hope, that the equation of a circle with center (a,b) and radius r is $(x-a)^2+ (y-b)^2= r^2$.

Now look at what you are given: |z|^2+ 2Re(az)= -|a|^2. Write z= x+ yi and a= a+ bi (I'm using a to mean to different things but we aren't going to use them in the same equation).

|z|^2= (x+ yi)(x- yi)= x^2+ y^2

az= (a+ bi)(x+ yi)= ax- by+ i(bx+ ay) so 2Re(az)= ax- by.

Finally, |a|^2= (a+bi)(a-bi)= a^2+ b^2.

Sp |z|^2+ 2Re(az)= -|a|^2 becomes x^2+ y^2+ ax- by= -a^2- b^2 which is the same as x^2+ ax+ a^2+ y^2- by+ b^2= 0

Hmmm, (x+a)^2+ (y-b)^2= 0 is NOT the equation of a circle, it is a single point, (-a, b). And keeping the ">" also does not give a circle, it gives every point in the plane except (-a, b).

3. Indeed there's a mistake, and here is the way I would derive the correct equation:

The equation of the circle of center $a\in\mathbb{C}$ and radius $r$ is obviously $|z-a|=r$.

Squaring this equation gives $r^2=|z-a|^2=(z-a)(\bar{z}-\bar{a})=|z|^2-a\bar{z}-z\bar{a}+|a|^2=|z|^2-2\Re(\bar{a}z)+|a|^2$ (I used the formulas $|z|^2=z\bar{z}$ and $z+\bar{z}=2\Re(z)$).

Hence the equation is $|z|^2-2\Re(\bar{a}z)=r^2-|a|^2$.

More generally, we deduce that for any $a\in\mathbb{C}$ and $M\geq -|a|^2$,

$|z|^2+2\Re(az)=M$

is the equation of a circle (of center $-\bar{a}$ and radius $\sqrt{M+|a|^2}$), which is perhaps what you meant (but wrote unproperly).