Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 6 / b / i

The first region is the lens around zero bounded by the two circular arcs through

- 1

and

1

z = x + i y

, its inequalities combine to

x^{2} + y^{2} + 2∣ y ∣ < 1

. The second region is the open first quadrant, and the third is the open horizontal strip between imaginary parts zero and

π /2

t = (1 + z) / (1 - z) = u + i v

. The inverse is

z = (t - 1) / (t + 1)

, and

1 - ∣ z ∣^{2} = \frac{4 u}{∣ t + 1 ∣ ^{2}}, Im z = \frac{2 v}{∣ t + 1 ∣ ^{2}} .

Hence the lens condition is exactly

u > ∣ v ∣

. Multiplying by

1 + i

u - v

u + v

, both positive exactly under that condition. Therefore

M (z) = (1 + i) \frac{1 + z}{1 - z}

is a Möbius transformation mapping the lens bijectively to the first quadrant. Its pole at one lies on the boundary, and its inverse is

z = (w - (1 + i)) / (w + (1 + i))

, well-defined throughout the quadrant. The derivative

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

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