 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2010/analysis/4/iv/solution

E (z) = e^{z}

maps the strip bijectively and holomorphically to the upper half-plane: its modulus is

e^{ℜ z} > 0

and its argument is

ℑ z \in (0, π)

. Its inverse is the holomorphic logarithm

L (w) = lo g ∣ w ∣ + i Arg w

with

Arg w \in (0, π)

Part (iii) applies to both maps. The required inverse correspondences are

h \in H (Ω_{2}) ⟼ h \circ E \in H (Ω_{1}), k ⟼ k \circ L .

They compose to the identity because

L \circ E

and

E \circ L

do. This proves injectivity and surjectivity, not merely a map between the two sets.

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