Complex Analysis - Finding the equation of a circle

NewtonianAlch
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Homework Statement


If \frac{z}{z + 3} is purely imaginary, show that z lies on a certain circle and find the equation of that circle.

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?
 
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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.
 
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?
 
It's _because_ the expression is pure imaginary that the real part is zero, which gives you the equation you are after.
 
Ah I see, that makes it more clear now.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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