 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/7/d/i/solution

z_{0} \neq = 0

, factor out

z_{0}

\frac{1}{z _{0} + z} = \frac{1}{z _{0}} \frac{1}{1 + z / z _{0}} = \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n}}{z _{0}^{n + 1}} z^{n}, ∣ z ∣ < ∣ z_{0} ∣.

Thus

a_{n} = \frac{( - 1 ) ^{n}}{z _{0}^{n + 1}}, R = ∣ z_{0} ∣.

z = - z_{0}

. The case

z_{0} = 0

must be treated separately: the function is

1/ z

, has no Taylor series about zero and is already its own Laurent series on

0 < ∣ z ∣ < \infty

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