Originally Posted by

**dougwm** For the question 2, about the Laurent expansions I have:

z/(z^2 - 1)

by partial fractions this becomes:

1/2(z - 1) + 1/2(z + 1)

so for:

0 < |z - 1| < 2

1/2(z + 1) = 1/2(2 + z - 1) = 1/4(1 + (z - 1)/2)

= 1/4 x (sum from n to infinity) (-1)^n . [|z - 1|/2]^n

this proves |z - 1| < 2

not too sure how to prove that |z - 1| > 0 though.

That's correct. You have found the series expansion of 1/(2(z+1)) (in powers of z-1) which converges when |z - 1| < 2. But don't forget to include the other partial fraction 1/(2(z-1)). This is a negative power of z-1, so it is only defined when z≠1, or in other words when |z - 1| > 0.

Originally Posted by

**dougwm** for:

|z + 1| > 2 i have:

1/2(z - 1) = 1/2(-2 + z + 1) = 1/2 . 1/(z + 1) . 1/(1 - 2/(z + 1))

= 1/2 . 1/(z + 1) . (sum from n to infinity) [ 2/|z + 1|]^n

which proves |z + 1| > 2

Again, the calculation is correct, but it's not appropriate to say that it "proves" |z + 1| > 2. What you should asy is that this series *converges* when |z + 1| > 2. (Here too, don't forget to include the other part of the partial fraction expression.)

For the third part of the problem, in the region |z| > 1, you need to get series expansions of both parts of the partial fraction expression:

$\displaystyle \frac1{2(z - 1)} + \frac1{2(z + 1)} = \frac1{2z}\left(\frac1{1-\frac1z} + \frac1{1+\frac1z}\right) = \ldots.$