 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ib/paper-4/8b/b/solution

In the middle annulus, expand the pole at one in negative powers and the pole at two in positive powers:

- \frac{1}{z - 1} = - \sum_{n = 0}^{\infty} z^{- n - 1}, \frac{1}{z - 2} = - \sum_{n = 0}^{\infty} 2^{- n - 1} z^{n} .

Their geometric series convergence conditions are

∣ z ∣ > 1

and

∣ z ∣ < 2

, respectively. The required Laurent series is

f (z) = - n = 0 \sum \infty z^{- n - 1} - n = 0 \sum \infty 2^{- n - 1} z^{n}, 1 < ∣ z ∣ < 2.

 New to topics? Read the docs here!