 Solution

ID: past-exam-of-the-algebra-and-analysis-i-for-mathematics-and-philosophy-course-of-the-university-of-oxford/2013/analysis/4/c/solution

Write

z = x + i y

. Both moduli are nonnegative, so squaring preserves their strict inequality:

∣ z ∣ < ∣ z - i ∣ ⟺ x^{2} + y^{2} < x^{2} + (y - 1)^{2} ⟺ 2 y < 1.

Thus

c = \frac{1}{2} .

The boundary is the perpendicular bisector of

0

and

i

0

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