# Thread: find Laurent representation of f

1. ## find Laurent representation of f

I have to find Laurent representation of a function $f(x)= \frac{1}{(z-1)(z-2)}$ , $|z|<1$

So

$f(x)= \frac{1}{(z-1)(z-2)} = \frac{1}{z-2}- \frac{1}{z-1}$

$\frac{1}{z-2} = \frac{1}{z} \cdot \frac{1}{1- \frac{2}{z} } = \sum_{0}^{n} \frac{2^{n}}{z^{n+1}}$

$\frac{1}{z-1}= - \frac{1}{1-z} = \sum_{0}^{n} - z^{n}$

Finally

$\frac{1}{(z-1)(z-2)} = \sum_{0}^{n} \frac{2^{n}}{z^{n+1}} + \sum_{0}^{n} z^{n}$

Good?

2. ## Re: find Laurent representation of f

I don't think this is quite correct. Your second expansion looks good - the 1/(z-1).

But the second one should be done in a similar way as the first I think. The way you have done it, your series converges iff |2/z|<1 iff |z|>2. Which is not what you need, you need
|z|<1. Try taking out 1/2 in the 1/(z-2) expansion instead of 1/z. I get the following answer doing that:

1/(z-1)(z-2) = sum_n=0^inf [1-2^(-1-n)]z^n, |z|<1

Wolfram also gets that:

1&#47;&#40;&#40;z-1&#41;&#40;z-2&#41;&#41; - Wolfram|Alpha

Hope this helps.
PS what is the quick way of using latex on this forum? lol

3. ## Re: find Laurent representation of f

I don't think this is quite correct. Your second expansion looks good - the 1/(z-1).

But the second one should be done in a similar way as the first I think. The way you have done it, your series converges iff |2/z|<1 iff |z|>2. Which is not what you need, you need
|z|<1. Try taking out 1/2 in the 1/(z-2) expansion instead of 1/z. I get the following answer doing that:

1/(z-1)(z-2) = sum_n=0^inf [1-2^(-1-n)]z^n, |z|<1

Wolfram also gets that:

1/((z-1)(z-2)) - Wolfram|Alpha

Hope this helps.
PS what is the quick way of using latex on this forum? lol

4. ## Re: find Laurent representation of f

Thanks, obviously you're right

so there is no principal part in this series?

and if our contour is defined by $|z|<1$ integral equals to $0$ , because there are no poles.

5. ## Re: find Laurent representation of f

Yes, sounds right to me