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

[kpizzano]

Homework Statement

Determine the coefficients c_n of the Laurent series expansion

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

that is valid for |z| > 1.

Homework Equations

none

The Attempt at a Solution

I found expansions valid for |z|>1 and |z|<1:

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

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

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 |z|>1, so I only need

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

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

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

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!