A locus problem where squaring hides a branch
A student is asked to describe the locus of points z in the complex plane satisfying |z - 2| = |z + 2i|. They square both sides, write (z - 2)(z̄ - 2) = (z + 2i)(z̄ - 2i), expand, cancel z z̄, and obtain -2z - 2z̄ + 4 = 2i z̄ - 2i z + 4. Rearranging, they get z(2i - 2) + z̄(-2 - 2i) = 0, then divide by (1 - i) to conclude z = z̄, i.e. the locus is the real axis. A classmate checks z = 3: |3 - 2| = 1 but |3 + 2i| = √13, so z = 3 is not on the locus.
1. Find the error in the student's algebra and determine the correct locus.
Read the worked answer
The expansion of the right side is wrong. Correctly:
(z + 2i)(z̄ - 2i) = z z̄ - 2i z + 2i z̄ + 4.
The student wrote the cross terms with the wrong signs. Redoing the comparison:
z z̄ - 2z - 2z̄ + 4 = z z̄ - 2i z + 2i z̄ + 4.
Cancel z z̄ and 4: -2z - 2z̄ = -2i z + 2i z̄, i.e. z(-2 + 2i) + z̄(-2 - 2i) = 0, or (1 - i)z + (1 + i)z̄ = 0.
Write z = x + iy. Then (1-i)(x+iy) + (1+i)(x-iy) = 2x + 2y = 0, so y = -x: the locus is the line through the origin with slope -1.
Check: z = 1 - i gives |1 - i - 2| = |-1 - i| = √2 and |1 - i + 2i| = |1 + i| = √2. Correct. And z = 3 fails, as the classmate found.
2. Explain geometrically why the locus must be a straight line, and verify the answer without algebra.
Read the worked answer
|z - 2| is the distance from z to the point A = 2, and |z + 2i| = |z - (-2i)| is the distance from z to B = -2i. The locus is the set of points equidistant from A and B, which is the perpendicular bisector of segment AB.
Midpoint of AB: (2 + (-2i))/2 = 1 - i. Direction of AB: -2i - 2 = -2 - 2i, slope (−2)/(−2) = 1. The perpendicular bisector has slope -1 and passes through 1 - i:
y - (-1) = -1(x - 1) ⟹ y = -x.
This matches the corrected algebra. The geometric view also explains why the answer had to be a line: equal-distance loci between two distinct points are always perpendicular bisectors.
3. The student wants a general rule to avoid this kind of error. State the correct general form of |z - a|^2 and explain how squaring can introduce or lose solutions here.
Read the worked answer
For a complex constant a, |z - a|^2 = (z - a)(z̄ - ā), not (z - a)(z̄ - a). The conjugate must be applied to the constant as well, because |w|^2 = w w̄ and w̄ = z̄ - ā.
In this problem a = -2i, so ā = 2i, giving (z + 2i)(z̄ - 2i) - the student used the wrong conjugate sign on the constant, which is exactly the sign error in the cross terms.
Regarding squaring: both sides are non-negative reals, so squaring is reversible here and introduces no extraneous solutions. Errors of this kind come from conjugate slips, not from squaring. A safe habit is to substitute one test point back into the original modulus equation before committing to a locus.