Complex Numbers and Quadratic Equations

Complex numbers reward a change of viewpoint more than any other algebra topic. Treated as a + ib they are an algebra problem and the work is heavy; treated as points in a plane with a modulus and an argument they become geometry, and most JEE questions collapse into a line or two. |z − a| = r is a circle, |z − a| = |z − b| is a perpendicular bisector, and arg((z−a)/(z−b)) = θ is an arc. The quadratic half of the chapter is more familiar, but it hides one trap worth naming early: the relation between roots and coefficients holds only when the leading coefficient is genuinely non-zero, and questions are built on candidates forgetting to check that.

Modulus, Argument and Geometry

For z = x + iy the modulus is |z| = √(x² + y²), the distance from the origin, and the argument is the angle the vector makes with the positive real axis. In polar form z = r(cos θ + i sin θ) = re^iθ, and multiplication becomes multiply the moduli, add the arguments — which is the whole reason polar form saves time. Two identities follow: |z₁z₂| = |z₁||z₂| and |z^n| = |z|^n.

Reading a locus straight off the equation
Condition	Locus
\|z − a\| = r	Circle of radius r centred at a
\|z − a\| = \|z − b\|	Perpendicular bisector of the segment joining a and b
\|z − a\| + \|z − b\| = k	Ellipse with foci a and b, if k > \|a − b\|
\| \|z − a\| − \|z − b\| \| = k	Hyperbola with foci a and b
arg(z − a) = θ	A ray from a at angle θ
arg((z−a)/(z−b)) = π/2	Circle on ab as diameter, excluding a and b

🎯 Why the triangle inequality has a lower bound as well as an upper one

Since complex numbers add like vectors, |z₁ + z₂| ≤ |z₁| + |z₂|, with equality only when the two point the same way — that is, when their arguments are equal. Equally, |z₁ + z₂| ≥ | |z₁| − |z₂| |, with equality when they point in opposite directions. So a sum of two complex numbers of fixed moduli is bounded on both sides, and questions asking for the maximum or minimum of |z + 1/z| or of |z − 3| given |z| = 2 are asking you to align or oppose the vectors. Drawing two arrows and asking "same way or opposite" replaces a page of algebra with a sketch.

De Moivre's theorem states (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ, which gives the n-th roots of unity as e^2πik/n for k = 0 to n − 1. Those roots sit at the vertices of a regular n-gon on the unit circle, they sum to zero and their product is (−1)ⁿ⁺¹. For cube roots specifically, 1 + ω + ω² = 0 and ω³ = 1, and those two facts alone dispose of most questions involving ω — any power of ω reduces modulo 3, so ω¹⁰⁰ = ω.

Quadratic Equations — Roots and Coefficients

For ax² + bx + c = 0 with a ≠ 0, the sum of the roots is −b/a and the product is c/a. The discriminant D = b² − 4ac decides their nature: D > 0 gives two distinct real roots, D = 0 a repeated root, D < 0 a conjugate pair of complex roots. If a, b and c are rational and D is a perfect square the roots are rational; if not, they are irrational conjugates.

⚠️ Check that the leading coefficient is non-zero before using −b/a

A question that gives (k − 1)x² + kx + 1 = 0 and asks for the value of k making the roots equal is not only a discriminant problem. If k = 1 the equation degenerates into the linear equation x + 1 = 0, which has one root and no "equal roots" to speak of — so k = 1 must be excluded before D = 0 is even solved. The same care applies whenever a coefficient contains a parameter: state the condition a ≠ 0 first, solve second. Examiners include the degenerate value among the options precisely because it satisfies the algebra while violating the premise.

Conditions on the roots, translated into algebra
Requirement	Conditions to impose
Both roots positive	D ≥ 0, sum > 0 and product > 0
Both roots negative	D ≥ 0, sum < 0 and product > 0
Roots of opposite signs	Product < 0 alone — D > 0 follows automatically
Both roots greater than k	D ≥ 0, af(k) > 0 and −b/2a > k
k lies between the roots	af(k) < 0 — nothing else is needed

🧠 Use the sign of af(k) instead of solving for the roots

A parabola af(x) opens upwards, so it is negative only strictly between its two roots. That single observation means af(k) < 0 is exactly the condition that k lies between the roots — no need to find the roots, no need even to compute the discriminant, because a sign change forces two real roots to exist. Similarly af(k) > 0 with the vertex on the correct side puts both roots on one side of k. This turns a whole family of "find the range of the parameter" questions into evaluating one expression and reading its sign, which is far faster and far less error-prone than solving the quadratic in terms of a parameter.

Two structural facts complete the chapter. If a quadratic with real coefficients has one complex root, the conjugate is automatically the other — so a real quadratic can never have exactly one complex root, and a question offering that as a possibility is offering something impossible. And if two quadratics have both roots common their coefficients are proportional, a₁/a₂ = b₁/b₂ = c₁/c₂; if exactly one root is common, eliminating x² between them gives that root as (c₁a₂ − c₂a₁)/(a₁b₂ − a₂b₁).

Key takeaways

In polar form multiplication multiplies moduli and adds arguments — that is why it saves time.
|z − a| = r is a circle and |z − a| = |z − b| is a perpendicular bisector — read the locus off the form.
The triangle inequality is two-sided: | |z₁|−|z₂| | ≤ |z₁+z₂| ≤ |z₁|+|z₂|, with equality on alignment.
The n-th roots of unity sum to zero; for cube roots, 1 + ω + ω² = 0 and ω³ = 1.
Sum of roots is −b/a and product c/a — but only after confirming a ≠ 0.
Roots of opposite signs needs only product < 0; D > 0 then follows automatically.
af(k) < 0 is exactly the condition that k lies between the roots — no need to find them.
A real quadratic with one complex root always has its conjugate as the other root.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The locus of z satisfying |z − 1| = |z + 1| is:
1. A circle of radius 1
2. The imaginary axis
3. The real axis
4. A single point
Show answer
Answer: B — The imaginary axis
The condition says z is equidistant from 1 and −1, so it lies on the perpendicular bisector of the segment joining them — the line x = 0, which is the imaginary axis. Recognising |z − a| = |z − b| as a perpendicular bisector answers the question at sight; expanding into x and y gives the same result but takes several lines.
If ω is a non-real cube root of unity, then ω¹⁰⁰ equals:
1. 1
2. ω
3. ω²
4. 0
Show answer
Answer: B — ω
Since ω³ = 1, powers of ω repeat with period 3, so only the remainder of the exponent on division by 3 matters. As 100 = 3 × 33 + 1, we get ω¹⁰⁰ = ω¹ = ω. Reducing the exponent modulo 3 is the whole technique for any question of this kind, and it works equally for ω raised to enormous powers.
The roots of x² + px + q = 0 (p, q real) are of opposite signs if:
1. p > 0 only
2. q < 0
3. p² − 4q = 0
4. q > 0 and p < 0
Show answer
Answer: B — q < 0
The product of the roots is q, and a product is negative exactly when the two factors have opposite signs, so q < 0 is both necessary and sufficient. Note that no separate discriminant condition is needed: a negative product forces the roots to be real, since a conjugate pair of complex roots has product |z|² which is positive. Recognising that saves imposing a redundant condition.
For (k − 1)x² + kx + 1 = 0 to have equal roots, the admissible value of k is:
1. k = 1 or k = 2
2. k = 2 only
3. k = 1 only
4. No value of k works
Show answer
Answer: B — k = 2 only
Setting the discriminant k² − 4(k − 1) = 0 gives k² − 4k + 4 = 0, so k = 2 twice. The algebra also seems to permit k = 1, but at k = 1 the x² term vanishes and the equation becomes the linear x + 1 = 0, which has a single root rather than equal roots — so k = 1 must be excluded on the prior condition that the leading coefficient be non-zero. Checking a ≠ 0 before solving is what separates the two answers.
If |z| = 2, the maximum value of |z − 3| is:
1. 1
2. 2
3. 5
4. 6
Show answer
Answer: C — 5
The point z lies on a circle of radius 2 about the origin, and |z − 3| is its distance from the point 3 on the real axis; that distance is largest when z is diametrically opposite, at z = −2, giving 5, and smallest at z = 2, giving 1. Equivalently the triangle inequality bounds |z − 3| between |3| − |z| = 1 and |3| + |z| = 5, which is the same answer read algebraically.
The sum of all the n-th roots of unity, for n ≥ 2, is:
1. 1
2. 0
3. n
4. −1
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Answer: B — 0
The n-th roots of unity are the roots of xⁿ − 1 = 0, whose xⁿ⁻¹ coefficient is zero, so by the sum-of-roots relation their total is zero. Geometrically they are the vertices of a regular n-gon centred at the origin, and the vectors to a symmetric set of vertices cancel exactly — either argument gives the answer without listing the roots.
If 3 lies between the roots of x² − px + 2 = 0, then:
1. f(3) > 0
2. f(3) < 0, that is p > 11/3
3. p = 11/3
4. No such p exists
Show answer
Answer: B — f(3) < 0, that is p > 11/3
Since the leading coefficient is positive, the parabola is negative only strictly between its roots, so 3 lying between them means f(3) < 0. That is 9 − 3p + 2 < 0, giving p > 11/3. No separate discriminant condition is needed, because a value at which the parabola dips below the axis guarantees two real roots — which is exactly what makes the af(k) test so economical.
A quadratic with real coefficients has 2 + 3i as one root. Its other root is:
1. 2 + 3i again
2. 2 − 3i
3. −2 + 3i
4. It cannot be determined
Show answer
Answer: B — 2 − 3i
For a polynomial with real coefficients, complex roots always occur in conjugate pairs, so the other root must be 2 − 3i. This is why a real quadratic can never have exactly one complex root — the sum and product of the roots would then fail to be real, contradicting the coefficients being real. It also lets you reconstruct the equation immediately as x² − 4x + 13 = 0.