1. ## Laurent series (Complex)

Using term-by-term differentiation, expand $f(z)=\frac{1}{(z+3)^2}$ in a Laurent series valid for $|z| > 3$.

Ok, so, suppose I get a few derivatives:

$f'(z)=-2(z+3)^{-3}$

$f''(z)=6(z+3)^{-4}$

$f'''(z)=-24(z+3)^{-5}$
...

There's infinitely many. What next?

2. Originally Posted by scorpion007
Using term-by-term differentiation, expand $f(z)=\frac{1}{(z+3)^2}$ in a Laurent series valid for $|z| > 3$.

Ok, so, suppose I get a few derivatives:

$f'(z)=-2(z+3)^{-3}$

$f''(z)=6(z+3)^{-4}$

$f'''(z)=-24(z+3)^{-5}$
...

There's infinitely many. What next?
Pattern.

Do you have to do it this way? Note that $f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2}$

3. I guess I have to, since that's what the question asks.

I know the pattern is:

$f^{(n)}(z)=(n+1)!(-1)^n(z+3)^{-(n+2)}$

but I don't know what to do with it.

4. Doing it the other way, I suppose what happens is:

$
f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2} = \frac{1}{z^2}\left(\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\right)^2
$
for $|z|>3$.

But the whole series is squared. How am I suppose to write out the first few terms?

5. Originally Posted by scorpion007
Doing it the other way, I suppose what happens is:

$
f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2} = \frac{1}{z^2}\left(\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\right)^2
$
for $|z|>3$.

But the whole series is squared. How am I suppose to write out the first few terms?
Hint: $\frac{1}{(1+z)^2}=-\left(\frac{1}{1+z}\right)'$...

6. Ok, so do I subs that in one of the factors and write it in terms of d/dz ?

I'm not sure what do with that...

7. Originally Posted by scorpion007
Ok, so do I subs that in one of the factors and write it in terms of d/dz ?

I'm not sure what do with that...
$\left(\frac{1}{1-x}\right)'=\left(\sum_{n\in\mathbb{N}}x^n\right)'= \sum_{n\in\mathbb{N}}nx^{n-1}$

8. I guess you want me to take the deriv of both sides?

$f'(z) =\frac{2}{z^2}\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\cdot \sum_{n=0}^\infty n \left(-\frac{3}{z}\right)^{n-1}\cdot \frac{3}{z^2}$

is that right? I could simplify this and all, but I don't quite see where this is going...

9. Originally Posted by Drexel28
Hint: $\frac{1}{(1+z)^2}=-\left(\frac{1}{1+z}\right)'$...
Maybe it's easier to use Newton's binomial theorem to expand $(1+z)^{-2}$.

10. Ah, I figured it out guys.

The example I needed was on this page: Pauls Online Notes : Calculus II - Power Series and Functions