 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2011/5/a/solution

The possible singular points are

i, - i, 1

. Near

i

, put

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

; then

f (z) = \frac{h ( z ) - 1/4}{z - i}, h (i) = \frac{1}{4} .

The numerator vanishes at

i

, so the apparent pole cancels and $i$ is removable. Its extension value is

h^{'} (i) = 1/4 + 3 i /8

, obtained from

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

- i

, the first term has a simple pole with nonzero residue

lim_{z \to - i} (z + i) f (z) = \frac{1}{( - 2 i ) ( - i - 1 ) ^{2}} = \frac{1}{4} .

The subtracted term is holomorphic there, so $- i$ is a simple pole. At

1

lim_{z \to 1} (z - 1)^{2} f (z) = \frac{1}{1 ^{2} + 1} = \frac{1}{2} \neq = 0,

so $1$ is a double pole. There are no other finite singularities.

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