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find the laurent series for 1/(z^2-1)^2 valid in 0 < |z-1| < 2 and |z + 1| > 2

we know that f(z) has poles of order 2 at 1 and -1...

In the first region, there are no poles (since z=-1 isn't a part of it). We can write the equation as a product of 1/(z-1)^2 and 1/(z+1)^2. The laurent series for the latter is simply itself, so is the whole equation's expansion simply itself as well? That doesn't seem right to me, because my reasoning for the second region is that we find the expansion for 0 < |z+1| < 1 and subtract it from the first. However, if it too is its own laurent series, then we will get 0.

Any help?