Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 4 / 8B (Complex Methods)

Table of contents
- a 8B
  - Solution a
- b 8B
  - Solution b
- c 8B
  - Solution c

a

Solution

Partial fractions give

f (z) = - 1/ (z - 1) + 1/ (z - 2)

. In the interior disk, expand both terms as geometric series:

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

Thus the Laurent series is a Taylor series in this region:

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

The nearer pole at

z = 1

determines the radius of convergence.

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.

c

Solution

Outside both poles, the two geometric series use negative powers:

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

Hence

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

The coefficient of

z^{- 1}

vanishes, as it must since the rational function starts with

z^{- 2}

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