Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 7 / e

Table of contents
- Solution e

Solution

Put

ζ = z - 3

and apply the geometric series:

\frac{1}{z - 2 i} = \frac{1}{3 - 2 i} \frac{1}{1 + ζ / ( 3 - 2 i )} = \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n}}{( 3 - 2 i ) ^{n + 1}} (z - 3)^{n} .

Thus

a_{n} = \frac{( - 1 ) ^{n}}{( 3 - 2 i ) ^{n + 1}}, ∣ z - 3∣ < 13, R = 13 .

The radius is the distance from the expansion centre

3

to the nearest singularity

2 i

, not the distance of that pole from the origin. On the boundary circle the series terms all have magnitude

1/ 13

, so they do not approach zero: this particular series diverges at every boundary point.

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