 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2013/analysis/4/d/solution

F (z) = \frac{1}{z} .

U_{1}

by part (b). For

z \neq = 0

∣ F (z) + i ∣ = \frac{∣1 + i z ∣}{∣ z ∣} = \frac{∣ z - i ∣}{∣ z ∣} .

Part (c) now gives

z \in U_{1}

if and only if

∣ F (z) + i ∣ > 1

, so

F (U_{1}) \subseteq U_{2}

. Conversely,

0 \in / U_{2}

because

∣0 + i ∣ = 1

. For any

w \in U_{2}

, put

z = 1/ w

; the same identity gives

∣ z - i ∣ > ∣ z ∣

, so

z \in U_{1}

and

F (z) = w

. Finally, reciprocation is injective on nonzero complex numbers. This proves bijectivity, with inverse

F^{- 1} (w) = \frac{1}{w} .

The inverse is also holomorphic, so this is a biholomorphism.

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