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JEE Mains Maths · Complex Numbers

Quadratic Equations with Complex Roots

Quadratics whose roots, or whose coefficients, are complex: solving them, using the sum and product of the roots, and finding high powers of the roots.

All Complex Numbers notesJEE Mains Maths strategy

Why this matters

Seventeen PYQs, eleven of them multiple choice, and four from 2026. Ten solve a quadratic or use the sum and product of its roots; seven ask for a high power of the roots, found by a recurrence or by polar form. Two ideas cover the page.

Concept 1 of 2: Roots, sum and product

The quadratic formula works with complex coefficients too; the only new step is a square root of a complex number, found by solving (p+iq)2=a+ib(p+iq)^2=a+ib. The sum and product of the roots, −ba-\frac ba and ca\frac ca, hold whatever the coefficients. When the coefficients are real, non-real roots come in conjugate pairs; with complex coefficients they need not.

Definition

  • α+β=−ba\alpha+\beta=-\frac ba, αβ=ca\alpha\beta=\frac ca.
  • a+ib\sqrt{a+ib}: solve p2−q2=ap^2-q^2=a, 2pq=b2pq=b.
  • Real coefficients: if p+iqp+iq is a root, so is p−iqp-iq.
  • α2+β2=(α+β)2−2αβ\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta; α3+β3=(α+β)3−3αβ(α+β)\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta).
  • For monic PP with roots xix_i: ∏(c−xi)=P(c)\prod(c-x_i)=P(c).

Sum and product of the roots

α+β=−ba,αβ=ca\alpha+\beta=-\frac ba,\qquad\alpha\beta=\frac ca

Worked example

Solve z2−(3+i)z+(2+2i)=0z^2-(3+i)z+(2+2i)=0.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q51Moderate

Example 1 · Complex Numbers · Quadratic Equations with Complex Roots

Let a,b∈Ca,b \in C. Let α,β\alpha,\beta be the roots of the equation x2+ax+b=0x^{2}+ ax + b = 0. If β−α=11\beta-\alpha=\sqrt{11} and β2−α2=3i11\beta^{2}-\alpha^{2}= 3i\sqrt{11}, then (β3−α3)2\left( \beta^{3}-\alpha^{3} \right)^{2} is equal to:

Conjugate roots need real coefficients

For z2−(3+i)z+(2+2i)=0z^2-(3+i)z+(2+2i)=0 the roots are 22 and 1+i1+i, not a conjugate pair. Use the conjugate root only when every coefficient is real.
Drill 9 more on roots, sum and product

Concept 2 of 2: Powers of the roots

For αn+βn\alpha^n+\beta^n there are two routes. If the roots have equal modulus, as they do for a real quadratic with negative discriminant, write them as ρe±iθ\rho e^{\pm i\theta} and use De Moivre. Otherwise use the equation itself: α2=pα+q\alpha^2=p\alpha+q gives Sn+2=pSn+1+qSnS_{n+2}=pS_{n+1}+qS_n for Sn=αn+βnS_n=\alpha^n+\beta^n (and the same for αn−βn\alpha^n-\beta^n), and it also turns any power of α\alpha into a linear expression Aα+BA\alpha+B.

Definition

  • α2=pα+q\alpha^2=p\alpha+q gives Sn+2=pSn+1+qSnS_{n+2}=pS_{n+1}+qS_n.
  • Real coefficients with D<0D<0: α,β=ρe±iθ\alpha,\beta=\rho e^{\pm i\theta}, ρ=c/a\rho=\sqrt{c/a}, and αn+βn=2ρncos⁡nθ\alpha^n+\beta^n=2\rho^n\cos n\theta.
  • Any αk\alpha^k reduces to Aα+BA\alpha+B using the equation.

Recurrence for power sums

Sn+2=p Sn+1+q Sn(α2=pα+q)S_{n+2}=p\,S_{n+1}+q\,S_n\qquad(\alpha^2=p\alpha+q)

Worked example

α,β\alpha,\beta are the roots of x2−2x+2=0x^2-2x+2=0. Find α8+β8\alpha^8+\beta^8.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q154Moderate

Example 2 · Complex Numbers · Quadratic Equations with Complex Roots

Let α,β\alpha,\beta be the roots of the equation x2−2x+2=0x^{2}-\sqrt{2}x + 2 = 0. Then α14+β14\alpha^{14}+\beta^{14} is equal to

Which root is alpha

When a question fixes Im(α)>Im(β)\mathrm{Im}(\alpha)>\mathrm{Im}(\beta), expressions like αn−βn\alpha^n-\beta^n or αβ\frac\alpha\beta depend on the choice. Name the roots before computing.
Drill 6 more on powers of the roots

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Roots, sum and product

    Sum and product of the roots

    α+β=−ba,αβ=ca\alpha+\beta=-\frac ba,\qquad\alpha\beta=\frac ca
  • Powers of the roots

    Recurrence for power sums

    Sn+2=p Sn+1+q Sn(α2=pα+q)S_{n+2}=p\,S_{n+1}+q\,S_n\qquad(\alpha^2=p\alpha+q)

Watch out for (2)

Test yourself on Complex Numbers

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Complex Numbers chapter testDrill all 17 questions