Sinh series manipulation

[saybrook1]

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!

 

[mfig]

It comes from the reciprocal rule for exponents.

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

$\frac{1}{x^4-x^6} = \frac{1}{x^4(1-x^2)} = x^{-4}(1-x^2)^{-1}$

 

[SteamKing]

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.

 

[saybrook1]

Awesome, makes perfect sense. Thank you both.

 

[saybrook1]

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]

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.