# Shifting a 'complex' circle

1. Mar 10, 2013

### ericm1234

We typically have z=r*e^(i*theta). But let's say I want a circle centered at 2i.
Is it valid to write z=r*e^(i*theta)+2i ?

I ask this because I don't want to have abs(z-2i)=r; I want to solve for z.

2. Mar 10, 2013

### tiny-tim

hi ericm1234!

(try using the X2 button just above the Reply box )
yes, a circle centre a and radius b is z - a = be, for all values of θ

3. Mar 10, 2013

### ericm1234

Awesome.

4. Mar 12, 2013

### HallsofIvy

Staff Emeritus
Note that, taking z= x+ iy, $|z- 2i|= |x+ (y-2)i|= r$ is the same as $\sqrt{x^23+ (y- 2)^2}= r$ so that $x^2+ (y- 2)^2= r^2$, the circle of radius r with center at (0, 2) or 2i in the complex plane.