Laurent expansion for a complex function

[Dixanadu]

Homework Statement

Expand $f(z)=\frac{1}{z-4}$ in a laurent series valid for (a) $|z|<4$ and (b) $|z|>4$

Homework Equations

The formula for laurent expansion...
$\sum_{n=-∞}^{+∞}a_{n}(z-z_{0})^{n}$
where
$a_{n}=\frac{1}{2\pi i}\oint_c \frac{f(z)}{(z-z_{0})^{n+1}}dz$

The Attempt at a Solution

umm...I have no idea..help please T_T

 

[brmath]

f(z) = 1/(z-4) = (1/z)$\frac{1}{1 - 4/z}$. The second factor of that product is the sum of a geometric series. So expand that factor and multiply by 1/z. That gives you a Laurent expansion -- it is convergent in a circle out to the first pole which is where?

The |z| > 4 part is easier because the function is analytic for those values. How does one expand an analytic function?

It is always possible to get the values of the Laurent coefficients by evaluating the appropriate integral, which I have found mostly to be a pain. But often there is an easier way (as per above) .

 

[Dixanadu]

I'm sorry i really don't get what you did with the $(1/z)\frac{1}{1-4/z}$ thing...but as far as the situation is for |z|>4, is it just a taylor expansion around $z_{0}=4$ since it is analytic after the singularity at z=4?

 

[vela]

No, it's not a Taylor expansion about $z_0 = 4$. You're being asked to find the Laurent series about the point $z_0 = 0$. There's a singularity at $z=4$, which is a distance 4 from $z_0=0$, so the complex plane is divided into two regions: $\lvert z-z_0 \rvert < 4$ and $\lvert z-z_0 \rvert >4$. You'll get different series for the two regions.

This is a really basic problem. It's almost certainly covered in your textbook. If you truly have no idea how to do it, it wouldn't hurt for you to go back and reread your textbook and notes so you'll have some idea about how to tackle it.

 

[brmath]

I'm sorry i really don't get what you did with the $(1/z)\frac{1}{1-4/z}$ thing...but as far as the situation is for |z|>4, is it just a taylor expansion around $z_{0}=4$ since it is analytic after the singularity at z=4?

Do you know what a geometric series is and what its sum is? That is at the bottom of this approach.

Sorry I said some things wrong. The function is analytic everywhere for |z| < 4, and the geometric series should come out to the Taylor's series in that area. While the function is also analytic for |z|> 4 , if you expand it about z = 0 you will get a Laurent series with an exponent for the 1/z term, because of the singularity at z = 4. In both cases you use a geometric sum -- which one depends on whether |z| is > or < 4.

As Zela says, this is standard material, so do check your textbook or notes.

 

[Dixanadu]

I think i get it now...thanks a bunch guys! Yea its my first shot at this stuff...but i realize that you kinda have to convert them to binomial expansions like (1+z)^n where |z| is determined by the region you are looking at. Thanks again guys!