 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2010/analysis/5/c/solution

For

∣ z ∣ < 1

, differentiate the geometric series to obtain

(1 - t)^{- 2} = \sum_{k \geq 0} (k + 1) t^{k}

. Since

(z - i)^{2} = - (1 - z / i)^{2}

\frac{1}{z ( z - i ) ^{2}} = - k = 0 \sum \infty (k + 1) (- i)^{k} z^{k - 1} = - \frac{1}{z} + 2 i + 3 z - 4 i z^{2} - 5 z^{3} + \dots .

The convergence region is precisely

0 < ∣ z ∣ < 1

, the specified punctured disk; the other pole is on its boundary at

i

 New to topics? Read the docs here!