# Thread: Taylor or Laurent series

1. ## Taylor or Laurent series

Hey!

Got a bit stuck with the series expansion of the following:

I am given a function f(z) = 1/(2z^2+(6z-4i)z-12i)

How do I work out the series expansion of f in the region |z1|<|z|<|z2| ?

I am also unsure as to how you would tell whether it is a Taylor or Laurent series?

Your help would be much appreciated

Thank you

2. 1) Are you certain you have written it correctly? It seems there may be an extra 'z' in the middle, there.

2) Solve for singularities. With that extra 'z' gone, I get z = -3 and z = 2i. Is that what you get?

3) Having established singularities, you get to decide what region to work with. You have one selected for you. That is good. Using the central region, one of your singularities is not really a singularity. Well. it still ist, we just don't care for the moment.

4) The partial fraction decomposition will then help you on your way. I get the delightfully straightforward:

$\frac{1}{2}\cdot\frac{1}{3+2i}\cdot\left(\frac{1}{ z-2i}-\frac{1}{z+3}\right)$

Now what?

3. Sorry my mistake, it should read:

f(z) = 1/(2z^2+(6-4i)z-12i)

I don't understand how you put it into partial fractions, mines not working out!

$

\frac{1}{2}\cdot\frac{1}{3+2i}\cdot\left(\frac{1}{ z-2i}-\frac{1}{z+3}\right)
$

How do you know whether it fits the Taylor or Laurent series?

And then how do you put it into either?

Thank you!

4. You must factor the denominator and follow the tried-and-true methods for partial fractions. I also pulled out some constants, so it would look simpler. Just try it again and be more careful.

Your assignment is to find the series expansion in the region between the singularities. Thus, only the singularity with lesser magnitude will need Laurent.

You have $|z-2i| = \sqrt{5}$ and $|z+3| = \sqrt{10}$. There, you can now order the washer-shaped region and carry on.

5. thanks

6. Let's see what you managed.