How can a -1 exponent be manipulated in the Sinh series?

saybrook1
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Homework Statement


Hello, I'm not trying to solve this exact problem although mine is similar and I am confused on how they were able to get a -1 in the exponent from one step to another.

Homework Equations


I have attached a picture indicating the step that I am confused about. How are they able to manipulate the series and pull out that -1 into the exponent thereby finding the residue?

The Attempt at a Solution


Some sort of series manipulation that I can't figure out; any help is greatly appreciated, thank you guys!
 

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It comes from the reciprocal rule for exponents.

In general, ## \frac{1}{x^n} = x^{-n} ##. So as an example:

## \frac{1}{x^4-x^6} = \frac{1}{x^4(1-x^2)} = x^{-4}(1-x^2)^{-1}##
 
saybrook1 said:

Homework Statement


Hello, I'm not trying to solve this exact problem although mine is similar and I am confused on how they were able to get a -1 in the exponent from one step to another.

Homework Equations


I have attached a picture indicating the step that I am confused about. How are they able to manipulate the series and pull out that -1 into the exponent thereby finding the residue?

The Attempt at a Solution


Some sort of series manipulation that I can't figure out; any help is greatly appreciated, thank you guys!
The common factor -(z - πi)3 can be factored out of the series expression of sinh3 z before it is inverted. The series involves only alternating odd powers of the common factor. After that, for the inversion, one can apply mfiq's hint about using the law of exponents.

There's no magic here - just straightforward algebra.
 
Awesome, makes perfect sense. Thank you both.
 
If either of you can still see this, would you possibly be able to tell me how they then lose that -1 power on the next line down allowing them to find the residue? Thanks again.
 
saybrook1 said:
If either of you can still see this, would you possibly be able to tell me how they then lose that -1 power on the next line down allowing them to find the residue? Thanks again.
Nevermind, think I've got it 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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