Laurent series (Complex)

**scorpion007** · May 14th 2010, 05:48 PM

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?

**Drexel28** · May 14th 2010, 05:58 PM

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 $f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2}$

**scorpion007** · May 14th 2010, 07:47 PM

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.

**scorpion007** · May 14th 2010, 08:36 PM

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

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

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

**Drexel28** · May 14th 2010, 08:38 PM

Originally Posted by scorpion007

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

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

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

Hint: $\frac{1}{(1+z)^2}=-\left(\frac{1}{1+z}\right)'$ ...

**scorpion007** · May 14th 2010, 08:45 PM

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...

**Drexel28** · May 14th 2010, 08:48 PM

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...

$\left(\frac{1}{1-x}\right)'=\left(\sum_{n\in\mathbb{N}}x^n\right)'= \sum_{n\in\mathbb{N}}nx^{n-1}$

Adapt that.

**scorpion007** · May 14th 2010, 09:10 PM

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...

**Opalg** · May 15th 2010, 12:23 AM

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 $(1+z)^{-2}$ .

**scorpion007** · May 15th 2010, 01:03 AM

Ah, I figured it out guys.

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

Thread: Laurent series (Complex)

LinkBack

Thread Tools

Display

Laurent series (Complex)

Similar Math Help Forum Discussions

Complex Analysis: Laurent series

Complex Analysis: Laurent Series

complex analysis: Laurent series

Laurent Series - Complex Numbers

Laurent Series-Complex

Search Tags