Question:
If |z + 2| = |z – 2|, then the locus of z is ____________.
Solution:
|z + 2| = |z – 2| for z = x + iy
i.e |x + iy + 2| = |x + iy – 2|
i.e |(x + 2) + iy| = |(x – 2) + iy|
Square both sides,
|(x + 2)| + iy|2 = |(x – 2)| + iy|2
i.e (x + 2)2 + y2 = (x – 2)2 + y2
i.e $x^{2}+4+4 x+y^{2}=x^{2}+4-4 x+y^{2}$
i.e $f x=0$
i.e $x=0$
Hence, locus is perpendicular bisector of the segment joining (–2, 0) and (2, 0).