Complex Analysis: Solving for P(z) When Z=a+bi

Daniel Monroy
Messages
5
Reaction score
0
also P(z)=0, if it is, how is it related to Z=a+bi??
 
Physics news on Phys.org
It certainly depends on what P you have.
 
Daniel Monroy said:
also P(z)=0, if it is, how is it related to Z=a+bi??
Well, if P(z)= 0 and z= a+bi then P(a+bi)= 0. Other than that, I have no idea what you are talking about! What is "P"?
 
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...
Back
Top