Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 12F (Complex Analysis OR Complex Methods)

Table of contents
- a 12F
  - Solution a
- b 12F
  - Solution b
- c 12F
  - i c
    - Solution i
  - ii c
    - Solution ii
  - iii c
    - Solution iii

a

Solution

The classification of isolated singularities can be stated through the Laurent expansion of

f

about

a

:

The singularity is removable when $f$ extends holomorphically to $a$ , equivalently when every negative-power Laurent coefficient is zero.
It is a pole of order $k$ when $(z - a)^{k} f (z)$ extends holomorphically to a function nonzero at $a$ . Equivalently, the Laurent series starts with a nonzero $(z - a)^{- k}$ term and has no more negative power.
It is essential when it is neither removable nor a pole, equivalently when infinitely many negative-power coefficients are nonzero.

b

Solution

Write the Laurent expansion on the punctured disc as

f (z) = \sum_{n = - \infty}^{\infty} a_{n} z^{n} .

For every integer

k \geq 1

, the Laurent coefficient formula and the Cauchy-Schwarz inequality give

∣ a_{- k} ∣ = \frac{r ^{k}}{2 π} \int_{0}^{2 π} f (r e^{i θ}) e^{ik θ} d θ \leq \frac{r ^{k}}{2 π} (\int_{0}^{2 π} ∣ f (r e^{i θ}) ∣^{2} d θ)^{1/2} (\int_{0}^{2 π} 1 d θ)^{1/2} \leq \frac{r ^{k}}{2 π} .

Letting

r ↓ 0

shows that

a_{- k} = 0

. Every coefficient in the principal part vanishes, so the uniform L2 circle bound for a removable singularity proves that

f has a removable singularity at 0 .

c

i

Solution

Since

tan z = sin z / cos z

, its singularities occur at the simple zeros of the denominator,

a_{k} = \frac{( 2 k + 1 ) π}{2}, k \in Z .

At each one,

sin a_{k} \neq = 0

and

(cos z)^{'} ∣_{a_{k}} = - sin a_{k} \neq = 0

, so all are simple poles. There are no other finite singularities:

tan z is meromorphic with simple poles at a_{k} .

ii

Solution

Put

z_{k} = \frac{1}{a _{k}} = \frac{2}{( 2 k + 1 ) π}, w = z - z_{k} .

Near

a_{k}

δ = 1/ z - a_{k}

gives

tan (a_{k} + δ) = - cot δ = - \frac{1}{δ} + \frac{δ}{3} + O (δ^{3}) .

But

δ = \frac{1}{z} - \frac{1}{z _{k}} = - \frac{w}{z _{k} ( z _{k} + w )},

and hence, exactly,

- \frac{1}{δ} = \frac{z _{k} ( z _{k} + w )}{w} = \frac{z _{k}^{2}}{w} + z_{k} .

Since

δ = O (w)

, the first two Laurent terms are

tan \frac{1}{z} = \frac{z _{k}^{2}}{z - z _{k}} + z_{k} + O (z - z_{k}) .

Equivalently, substitute

z_{k} = 2/ ((2 k + 1) π)

in this expression.

iii

Solution

At every

z_{k}

tan (1/ z)

has a simple pole. The exponential of a pole is an essential singularity, so

z_{k} = \frac{2}{( 2 k + 1 ) π} (k \in Z)

are essential singularities of

exp (tan (1/ z))

.

These points accumulate at

0

. Therefore no punctured neighbourhood of zero is a domain of holomorphy for the function: zero is a non-isolated singularity, specifically an accumulation point of essential singularities, rather than an isolated essential singularity. There are no other singularities in

C

.

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