Complex Analysis: Locus Sketching

altcmdesc
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Homework Statement



Sketch the locus of |z-2i|=z+3 in C

2. The attempt at a solution

Let z=x+iy, then |z-i|=|x+iy-2i)|=|x+i(y-2)|=(x^2+(y-2)^2)^(1/2)=z+3

The problem is that I can't tell what this means geometrically. Is it a spiral?
 
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altcmdesc said:
Sketch the locus of |z-2i|=z+3 in C

Hi altcmdesc! :smile:

Hint: |z - 2i| is real, so z + 3 must be real, so … ? :wink:
 
So it must be the real line?
 
altcmdesc said:
So it must be the real line?

It must be part of the real line.
 
I'm thinking of this as the simpler problem z=|z-i| first, but I'm having a hard time believing this could be true for any z because of the triangle inequality.
 
altcmdesc said:
I'm thinking of this as the simpler problem z=|z-i| first, but I'm having a hard time believing this could be true for any z because of the triangle inequality.

That's right! :smile:

But how about z+1=|z-i| ? :wink:
 
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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