 Solution

ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2014/3/c/ii/solution

h (z) = e^{π z /2}

is a conformal bijection from the width-two strip to the upper half-plane, and

h (i) = i

. Its inverse there is

h^{- 1} (w) = 2 L (w) / π

Write

e^{π β /2} = A + i B

, where

B > 0

. Conjugate the preceding affine half-plane map by

h

g_{β} (z) = \frac{2}{π} L (A + B e^{π z /2}) .

Every argument of

L

lies in the upper half-plane. The formula is therefore a conformal bijection of the strip onto itself, and

g_{β} (i) = \frac{2}{π} L (A + i B) = \frac{2}{π} L (e^{π β /2}) = β .

The final equality uses

0 < Im (π β /2) < π

, so no multiple of

2 πi

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