Discussion
Let \(z = x + iy\), where \(x, y \in \mathbb{R}\) and \(i^{2} = -1\). If \(|z - i| = |z - 1|\), then \(y =\)
Answer: E
The condition gives \(y = x\).
\(|z - i| = |z - 1|\) says z is equidistant from the points i = (0, 1) and 1 = (1, 0), so it lies on the perpendicular bisector of the segment joining them. Algebraically:
\[ \begin{aligned} x^{2} + (y - 1)^{2} &= (x - 1)^{2} + y^{2} \\ -2y + 1 &= -2x + 1 \Rightarrow y = x \end{aligned} \]
\(y = -x\) is the line through the two points themselves turned the wrong way, and the shifted lines do not pass through the midpoint \(\left(\frac{1}{2}, \frac{1}{2}\right)\).
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