 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/11/a/solution

The original PDF places a complex modulus around the quotient; those bars are essential. Its denominator excludes

z = - i

. Using

∣ z^{*} - i ∣ = ∣ z + i ∣

, the case

λ = 1

becomes

∣ z - i ∣ = ∣ z + i ∣

. In Cartesian coordinates,

x^{2} + (y - 1)^{2} = x^{2} + (y + 1)^{2} ⟹ y = 0.

i

and

- i

in the complex plane. The excluded denominator point does not lie on it.

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