# Complex Analysis - Finding the equation of a circle

1. Mar 18, 2012

### NewtonianAlch

1. The problem statement, all variables and given/known data
If $\frac{z}{z + 3}$ is purely imaginary, show that z lies on a certain circle and find the equation of that circle.

3. The attempt at a solution

So,

$\frac{z}{z + 3}$ = $\frac{x + iy}{x + iy + 3}$

Multiplying by the complex conjugate (and simplifying), we get,

$\frac{x^{2} + y^{2} + 3x + 3iy}{x^{2} + y^{2} + 6x + 9}$

Since we're only interested in the imaginary part here, I take,

$\frac{3iy}{x^{2} + 6x + 9 + y^{2}}$

I am not too sure what to do from here, also...does "z lies on a certain circle" mean on the boundary line or anywhere in that enclosed zone including the boundary?

2. Mar 18, 2012

### awkward

You're not "only interested in the imaginary part". You are told z/(z+3) is pure imaginary; this means the real part is zero. That's probably where you want to start.

A circle is the set of points equidistant from its center; the circle does not include its interior. So "on a circle" means on the curvy line.

3. Mar 18, 2012

### NewtonianAlch

So if the real part is going to be zero, it means the numerator can only be zero. From that we can get an equation for a circle, which I believe is (x + $\frac{3}{2}$)$^{2}$ + $y^{2}$ = $\frac{9}{4}$ - but is this the circle I'm looking for?

If so, why am I getting an equation for a circle from the real part, when the expression in question is supposedly purely imaginary?

4. Mar 18, 2012

### awkward

It's _because_ the expression is pure imaginary that the real part is zero, which gives you the equation you are after.

5. Mar 19, 2012

### NewtonianAlch

Ah I see, that makes it more clear now.