I have to find Laurent representation of a function ,
So
Finally
Good?
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
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