Expanding f(z) in a Laurent Series

[Pouyan]

Homework Statement

Expand the function f(z)=1/z(z-2) in a Laurent series valid for the annual region 0<|z-3|<1

Homework Equations

I know 1/z(z+1) = 0.5(1/(z-2)) - 0.5(1/z)

Taylor for 0.5(1/(z-2)) is : ∑(((-1)^k/2) * (z-3)^k) (k is from 0 to ∞)For the second 0.5(1/z) the answer is a Taylor : ∑((1/6)*(-1/3)^k * (z-3)^k)

But why the answer for 0.5(1/z) is not 1/2(z-3) *∑((-1)^k*(3/(z-3))^k) (k is from 0 to ∞) ?

The correct answers for both are from Taylor but I thought for the second is a Laurent

 

[vela]

Did you consider the annular region you were given?

 

[Pouyan]

Did you consider the annular region you were given?

Yes but I can't draw it in this page !

 

[vela]

You don't need to draw it.

What's the significance of the annular region? Why did the problem bother giving it to you? If you understand that, it's the answer to your original question.