JEE Maths›Complex Numbers & Quadratic Equations›Visual Solution
Visual SolutionPYQ 2025 · Jan 22 Shift 1Tricky
Question
Let z1,z2z_1, z_2 and z3z_3 be three complex numbers on the circle ∣z∣=1|z| = 1 with arg⁡(z1)=−π4\arg(z_1) = -\frac{\pi}{4}, arg⁡(z2)=0\arg(z_2) = 0 and arg⁡(z3)=π4\arg(z_3) = \frac{\pi}{4}. If ∣z1zˉ2+z2zˉ3+z3zˉ1∣2=α+β2|z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1|^2 = \alpha + \beta\sqrt{2}, α,β∈Z\alpha, \beta \in \mathbb{Z}, then the value of α2+β2\alpha^2 + \beta^2 is:
(A)2424
(B)2929
(C)4141
(D)3131
Solution Path
Unit circle → conjugate property → Euler products → ∣S∣2=5−22|S|^2 = 5 - 2\sqrt{2} → α2+β2=29\alpha^2 + \beta^2 = 29
01Read the Problem
1/6
Three complex numbers on the unit circle with arguments −π/4-\pi/4, 00, π/4\pi/4. Find α2+β2\alpha^2 + \beta^2 where ∣z1zˉ2+z2zˉ3+z3zˉ1∣2=α+β2|z_1\bar{z}_2 + z_2\bar{z}_3 + z_3\bar{z}_1|^2 = \alpha + \beta\sqrt{2}.
∣z∣=1|z| = 1, three points on unit circle
02Plot on the Argand Plane
2/6
All three points lie on the unit circle ∣z∣=1|z| = 1. In Euler form: z1=e−iπ/4z_1 = e^{-i\pi/4}, z2=1z_2 = 1, z3=eiπ/4z_3 = e^{i\pi/4}.
z1z_1 and z3z_3 are conjugates, symmetric about the real axis
03Conjugate Property on Unit CircleKEY INSIGHT
3/6
On the unit circle, the conjugate is just the inverse: zˉ=e−iθ\bar{z} = e^{-i\theta} for z=eiθz = e^{i\theta}.
∣z∣=1  ⟹  zˉ=1/z=e−iθ|z| = 1 \implies \bar{z} = 1/z = e^{-i\theta}
04Compute Each Product
4/6
Multiply using Euler form - just add exponents. The first two products turn out identical: both equal e−iπ/4e^{-i\pi/4}.
z1zˉ2=z2zˉ3=e−iπ/4z_1\bar{z}_2 = z_2\bar{z}_3 = e^{-i\pi/4},   z3zˉ1=eiπ/2\;z_3\bar{z}_1 = e^{i\pi/2}
05Sum & Modulus Squared
5/6
Sum S=2e−iπ/4+eiπ/2S = 2e^{-i\pi/4} + e^{i\pi/2}. Expand to Cartesian form and compute ∣S∣2|S|^2.
∣S∣2=(2)2+(1−2)2=5−22|S|^2 = (\sqrt{2})^2 + (1-\sqrt{2})^2 = 5 - 2\sqrt{2}
06Final Answer
6/6
Comparing with α+β2\alpha + \beta\sqrt{2}: α=5\alpha = 5, β=−2\beta = -2.
α2+β2=25+4=29\alpha^2 + \beta^2 = 25 + 4 = \boxed{29}
Concepts from this question4 concepts unlocked
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Unit Circle Conjugate Property

STANDARD

When |z| = 1, the conjugate equals the reciprocal: z̅ = 1/z

∣z∣=1  ⟹  zˉ=1z=e−iθ|z| = 1 \implies \bar{z} = \frac{1}{z} = e^{-i\theta}

Converts division/conjugate operations into simple exponent changes on the unit circle

Locus problemsModulus equationsProduct simplification
Practice (6 Qs) →

Euler Form Multiplication

EASY

Multiplying in Euler form = adding exponents

eiα⋅eiβ=ei(α+β)e^{i\alpha} \cdot e^{i\beta} = e^{i(\alpha + \beta)}

Turns complex multiplication into simple addition - saves 2-3 minutes per question

Roots of unityDe Moivre's applicationsProduct/sum problems
Practice (10 Qs) →

Conjugate Symmetry on Argand Plane

STANDARD

Conjugate pairs are reflections across the real axis

z=x+iy  ⟹  zˉ=x−iyz = x + iy \implies \bar{z} = x - iy

Recognizing conjugate pairs halves the computation - products simplify dramatically

Locus problemsGeometric interpretationSymmetric root problems
Practice (11 Qs) →

Modulus Squared via Components

EASY

|a + bi|^2 = a^2 + b^2 - no square root needed

∣z∣2=z⋅zˉ=x2+y2|z|^2 = z \cdot \bar{z} = x^2 + y^2

Final step in many JEE problems. Students waste time square-rooting when they only need |z|^2

Distance problemsSum/product modulus\alpha^2 + \beta^2 type problems
Practice (11 Qs) →
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