Laurent Series Expansion coefficient for f(z) = 1/(z-1)^2

[kpizzano]

Homework Statement

Determine the coefficients [itex]c_n[/itex] of the Laurent series expansion

[itex]\frac{1}{(z-1)^2} = \sum_{n = -\infty}^{\infty} c_n z^n[/itex]

that is valid for [itex]|z| > 1[/itex].

Homework Equations

none

The Attempt at a Solution

I found expansions valid for [itex]|z|>1[/itex] and [itex]|z|<1[/itex]:

[itex]\sum_{n = 0}^{\infty} \left(n-1\right)z^n, |z|>1[/itex] and

[itex]\sum_{n = 2}^{\infty} \left(n-1\right)z^{-n}, |z|<1[/itex]

I know that if I negate the n's in the second equation and change the index of the sum to go from -∞ to -2 I can add them together to get the sum from -∞ to ∞, but I don't know what to do about the missing n=1 term. Any suggestions?

 

[vela]

There are two different Laurent series for the function, each one being valid in different portions of the complex plane. You can't add them together because, at a particular value of z, only one will converge.

You might want to show us your work on how you found the series because they're not correct.

 

[kpizzano]

I just noticed that I missed the part in the problem statement that says valid for [itex] |z|>1 [/itex], so I only need

[itex] \sum_{n=0}^{\infty}\left(n+1\right)z^{-\left(n+2\right)}[/itex].

I got that by noticing that [itex] \frac{1}{\left(z-1\right)^2} = \frac{1}{z^2\left(1-\frac{1}{z}\right)^2}[/itex]

Using the geometric series expansion, [itex]= \frac{1}{z^2}\left(1 + \frac{1}{z} + \frac{1}{z^2} + \frac{1}{z^3} + ...\right)^2
= \frac{1}{z^2}\left(1 + \frac{2}{z} + \frac{3}{z^2} + \frac{4}{z^3} + ...\right)
= \frac{1}{z^2} + \frac{2}{z^3} + \frac{3}{z^4} + \frac{4}{z^5} + ...
=\sum_{n=0}^{\infty}\left(n+1\right)z^{-\left(n+2\right)}, for |z|>1[/itex]

Now that I have that, I'm not sure how I can extend the indices of the summation so that they match what we were given in the problem statement.

 

[vela]

You got it. You don't need to extend the summation. The "missing" terms aren't there because c_n=0 for those powers of z.

 

[kpizzano]

Ok, thanks so much for commenting!