Complex Numbers: Exam-Pattern Drill (2)

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Tags Complex Numbers

Question

If zz satisfies ∣z−2+i∣=∣z+3−2i∣|z - 2 + i| = |z + 3 - 2i|, find the locus of zz in the complex plane.

Solution — Step by Step

∣z−z1∣|z - z_1| is the distance from point zz to point z1z_1 in the complex plane.

The condition ∣z−z1∣=∣z−z2∣|z - z_1| = |z - z_2| means zz is equidistant from z1z_1 and z2z_2 — which traces the perpendicular bisector of the segment z1z2z_1 z_2.

z1=2−iz_1 = 2 - i (since ∣z−(2−i)∣|z - (2 - i)|).

z2=−3+2iz_2 = -3 + 2i (since ∣z−(−3+2i)∣|z - (-3 + 2i)|).

Midpoint of z1z2z_1 z_2: ((2−3)/2,(−1+2)/2)=(−1/2,1/2)((2 - 3)/2, (-1 + 2)/2) = (-1/2, 1/2).

Slope of z1z2z_1 z_2: from (2,−1)(2, -1) to (−3,2)(-3, 2), slope =(2−(−1))/(−3−2)=3/(−5)=−3/5= (2 - (-1))/(-3 - 2) = 3/(-5) = -3/5.

Slope of perpendicular bisector: negative reciprocal =5/3= 5/3.

Equation: y−1/2=(5/3)(x+1/2)y - 1/2 = (5/3)(x + 1/2)

3(y−1/2)=5(x+1/2)3(y - 1/2) = 5(x + 1/2)

3y−3/2=5x+5/23y - 3/2 = 5x + 5/2

5x−3y+4=05x - 3y + 4 = 0

Locus is the straight line 5x−3y+4=05x - 3y + 4 = 0.

Why This Works

Treating ∣z−z1∣=∣z−z2∣|z - z_1| = |z - z_2| geometrically (rather than algebraically expanding) saves significant time. The condition immediately gives “perpendicular bisector” — we just need the midpoint and the slope of the original segment.

Geometric interpretations to memorise:

  • ∣z−z1∣=r|z - z_1| = r → circle of radius rr centred at z1z_1
  • ∣z−z1∣=∣z−z2∣|z - z_1| = |z - z_2| → perpendicular bisector of segment z1z2z_1 z_2
  • ∣z−z1∣+∣z−z2∣=k|z - z_1| + |z - z_2| = k (constant > ∣z1−z2∣|z_1 - z_2|) → ellipse
  • ∣∣z−z1∣−∣z−z2∣∣=k||z - z_1| - |z - z_2|| = k (constant < ∣z1−z2∣|z_1 - z_2|) → hyperbola

Alternative Method — Algebraic Substitution

Let z=x+iyz = x + iy. Then:

∣x+iy−2+i∣=∣x+iy+3−2i∣|x + iy - 2 + i| = |x + iy + 3 - 2i|

(x−2)2+(y+1)2=(x+3)2+(y−2)2\sqrt{(x - 2)^2 + (y + 1)^2} = \sqrt{(x + 3)^2 + (y - 2)^2}

Square: (x−2)2+(y+1)2=(x+3)2+(y−2)2(x - 2)^2 + (y + 1)^2 = (x + 3)^2 + (y - 2)^2

Expand: x2−4x+4+y2+2y+1=x2+6x+9+y2−4y+4x^2 - 4x + 4 + y^2 + 2y + 1 = x^2 + 6x + 9 + y^2 - 4y + 4

Cancel and simplify: −4x+2y+5=6x−4y+13-4x + 2y + 5 = 6x - 4y + 13

10x−6y+8=010x - 6y + 8 = 0, or 5x−3y+4=05x - 3y + 4 = 0. Same answer.

The geometric method is faster, but algebraic is more reliable for beginners.

Common Mistake

Students often try to expand both sides without recognising the geometric meaning, ending up with a long chain of algebra prone to sign errors.

Another classic: confusing ∣z−2+i∣|z - 2 + i| with ∣z−2∣+∣z+i∣|z - 2| + |z + i|. The minus sign is grouped with the entire complex number, not split.

JEE Main loves locus questions. The four geometric templates above cover ~95% of locus problems. Master them and answer in 30 seconds rather than 3 minutes. JEE Advanced 2022 had a tricky locus combining two such conditions.

Want to master this topic?

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