Branch Cut for (lnz)^2: Determining Branch Points and Range

CrimsonFlash
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Homework Statement


It is simply the same as the one for lnz i.e. does it go from 0 to ∞?
Also, is there any proper way to figure out branch points of a function?

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CrimsonFlash said:

Homework Statement


It is simply the same as the one for lnz i.e. does it go from 0 to ∞?
Also, is there any proper way to figure out branch points of a function?
Do you mean, "Does z go from 0 to ∞?" Your question is unclear, which is why no one will answer.

Generally, branch cuts can be made fairly arbitrarily as long as they make sense. Though, I personally prefer to use Riemann surfaces because they are more fun for me.
 
Certainly if you have a domain where ln(z) is defined holomorphically, then ln(z)2 is also holomorphic on the same domain
 
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Ok thanks.
 
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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