Finding the Laurent Series of e^(1/(1-z)) for Residue Calculation

[Nikitin]

Homework Statement

Hi! I need to find the laurent series of ##e^{1/(1-z)}## to get the residue at ##z=1##. Can somebody help me?

The Attempt at a Solution

https://scontent-a-ams.xx.fbcdn.net/hphotos-frc3/q71/s720x720/1461607_10201796752217165_1002449331_n.jpg

I tried using the taylor series of e^x around x=1, but I seem to have failed.. what am I to do? I can't use the maclaurin series of e^x, right?

 

[pasmith]

Homework Statement

Hi! I need to find the laurent series of ##e^{1/(1-z)}## to get the residue at ##z=1##. Can somebody help me?

The Attempt at a Solution

https://scontent-a-ams.xx.fbcdn.net/hphotos-frc3/q71/s720x720/1461607_10201796752217165_1002449331_n.jpg

I tried using the taylor series of e^x around x=1, but I seem to have failed.. what am I to do? I can't use the maclaurin series of e^x, right?

You can and should. [itex]e^x = \displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}[/itex] is a definition of [itex]e^x[/itex]. Setting [itex]x = (1 - z)^{-1}[/itex] is clearly going to give you a sum of powers of [itex](z - 1)[/itex], which is after all what you want.

 

[Nikitin]

But why can you do that? Shouldn't the taylor series used to find the laurent series be expanded about the same value as the laurent series is being expanded about (in this case z=1)?

Could you remind me of the rules for stuff?

 

[pasmith]

But why can you do that? Shouldn't the taylor series used to find the laurent series be expanded about the same value as the laurent series is being expanded about (in this case z=1)?

Why should it?

If the power series [itex]g(z) = \sum a_n z^n[/itex] has radius of convergence [itex]R[/itex], then if [itex]|f(z)| < R[/itex] we have [itex]g(f(z)) = \sum a_n f(z)^n[/itex].

If [itex]f(z)[/itex] is not within the radius of convergence then one generally has to find a different power series for [itex]g[/itex]. But that would point to expanding [itex]g[/itex] in a Taylor series about [itex]f(z_0)[/itex], not [itex]z_0[/itex] itself.

Conveniently the power series for exp, which is in fact the definition of the exp function on the complex plane, has infinite radius of convergence, so
[tex]
\exp (f(z)) = \sum_{n= 0}^{\infty} \frac{f(z)^n}{n!}
[/tex]
for any [itex]f(z)[/itex].

 

[Nikitin]

Hmm, indeed. But still, what happens if f(z) has multiple poles? How would you find the residues of them?