Question:
If |z + 2i| = |z – 2i|, then the locus of z is ____________.
Solution:
Given |z + 2i| = |z – 2i|
for z = |x + iy|
|x + iy + 2i| = |x + iy – 2i|
Squaring both sides, |x + i(y + 2)|2 = |x + i (y – 2)|2
i.e x2 + (y + 2)|2 = x2 + (y – 2)|2
i.e x2 + y2 + 4 + 4y = x2 + y2 + 4 – 4y
i.e 8y = 0
i.e y = 0
∴ locus is perpendicular bisector of the segment joining (0, –2) and (0, 2)