 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2022/9/a/iii/solution

For

z \neq = - 1

\overline{z} + 1

Re \frac{z - 1}{z + 1} = \frac{∣ z ∣ ^{2} - 1}{∣ z + 1 ∣ ^{2}} .

Thus its zero locus is

z = e^{i θ}, θ \neq \equiv π (mod 2 π) .

It is the unit circle with the point

- 1

removed. That point cannot be restored by the zero numerator obtained after clearing a denominator: the original expression is undefined there.

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