JEE Corner — Complex Numbers and Quadratic Equations for JEE Main and Advanced

Why This Section Exists

The rationalised textbook stops at the Argand plane — but the JEE Main syllabus explicitly includes the argument (polar form), the triangle inequality, and quadratic equations in the real and complex number system with relations between roots and coefficients. This section restores the full JEE toolkit:

1. Argument and polar form — $z = r(\cos\theta + i\sin\theta)$;
2. Quadratic equations — complex roots, nature of roots, root-coefficient relations;
3. Square roots of a complex number — the three-equation method;
4. Cube roots of unity — $\omega$ and its identities;
5. The triangle inequality and modulus bounds.

The worked examples and practice questions here are modelled on the JEE Main and Advanced pattern and difficulty; they are practice questions in the exam style, not reproductions of specific past papers.

[Board Note] CBSE students: polar form and quadratic equations were removed from the rationalised Class 11 syllabus. Treat this section as JEE preparation — though the quadratic-roots material also silently returns in later chapters.

Argument and Polar Form

A non-zero z = x + iy sits at distance $r = |z| = \sqrt{x^2 + y^2}$ from the origin, in the direction making angle $\theta$ with the positive real axis. Then $x = r\cos\theta$, $y = r\sin\theta$, giving the polar form:

$$z = r(\cos\theta + i\sin\theta)$$

$\theta$ is called the argument (arg z); the value chosen in $(-\pi, \pi]$ is the principal argument.

Finding the argument safely: compute the acute reference angle $\alpha = \tan^{-1}\left|\frac{y}{x}\right|$, then place it by quadrant: Q I: $\theta = \alpha$; Q II: $\pi - \alpha$; Q III: $-(\pi - \alpha)$; Q IV: $-\alpha$ (principal values).

Standard conversions to know cold:

• $1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$
• $-1 + \sqrt{3}i = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$
• $-1 = \cos\pi + i\sin\pi$; $\quad -i$: r = 1, $\theta = -\frac{\pi}{2}$

Why polar form matters: moduli multiply and arguments ADD — $z_1z_2$ has modulus $r_1r_2$ and argument $\theta_1 + \theta_2$. Multiplication becomes rotation-plus-scaling, which is why $\times i$ rotates by 90°.

[JEE Tip] Never write arg z = $\tan^{-1}\frac{y}{x}$ blindly — that formula alone cannot tell Q I from Q III. Quadrant first, reference angle second. For $-1 - i$: reference $\frac{\pi}{4}$, Q III, principal argument $-\frac{3\pi}{4}$ (not $\frac{\pi}{4}$!).

Quadratic Equations Over the Complex Numbers

For $ax^2 + bx + c = 0$ with real coefficients and $a \neq 0$, the roots are

$$x = \frac{-b \pm \sqrt{D}}{2a}, \qquad D = b^2 - 4ac$$

• D > 0: two distinct real roots. D = 0: equal real roots $-\frac{b}{2a}$.
• D < 0: no real roots — but now $\sqrt{D} = i\sqrt{-D}$, giving the complex conjugate pair $x = \frac{-b \pm i\sqrt{-D}}{2a}$.

Model: $x^2 + x + 1 = 0$ has D = $-3$ and roots $\frac{-1 \pm \sqrt{3}i}{2}$.

Root-coefficient relations

If $\alpha, \beta$ are the roots: $$\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$$

and conversely the equation with roots $\alpha, \beta$ is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$. Power sums follow: $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$, $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$.

Key Point: With REAL coefficients, complex roots always arrive as conjugate pairs $p \pm iq$ — one root reveals the other for free. (With complex coefficients this fails: $x^2 - ix = 0$ has roots 0 and i.)

[JEE Tip] "Form the equation whose roots are…" = compute sum and product, plug into $x^2 - Sx + P = 0$. For conjugate pairs $p \pm iq$: S = 2p, P = $p^2 + q^2$ — both real, always.

Square Roots, Cube Roots of Unity, and the Triangle Inequality

Square roots of a complex number

To solve $(x + iy)^2 = a + ib$: match parts and add the modulus as a third equation:

$$x^2 - y^2 = a, \qquad 2xy = b, \qquad x^2 + y^2 = \sqrt{a^2 + b^2}$$

Solve the first and third for $x^2, y^2$; the SIGN of b fixes whether x and y share signs. Model: $\sqrt{-8 + 6i} = \pm(1 + 3i)$.

Cube roots of unity

Solving $x^3 = 1$: $x^3 - 1 = (x - 1)(x^2 + x + 1) = 0$ gives 1 and the roots of $x^2 + x + 1 = 0$:

$$\omega = \frac{-1 + \sqrt{3}i}{2}, \qquad \omega^2 = \frac{-1 - \sqrt{3}i}{2}$$

Key Point (the $\omega$ identities): $\omega^3 = 1$, $\quad 1 + \omega + \omega^2 = 0$, $\quad \omega^2 = \bar{\omega} = \frac{1}{\omega}$. Powers cycle with period 3: $\omega^{3k} = 1$, $\omega^{3k+1} = \omega$, $\omega^{3k+2} = \omega^2$.

Every $\omega$ computation reduces by these: $\omega^{100} = \omega^{99}\omega = \omega$; $(1 + \omega)^3 = (-\omega^2)^3 = -\omega^6 = -1$.

The triangle inequality

$$|z_1 + z_2| \leq |z_1| + |z_2| \qquad \text{and} \qquad |z_1 + z_2| \geq \big||z_1| - |z_2|\big|$$

Equality on the left iff the numbers point the same way. Consequence bounds: for $|z| = r$, the value $|z + a|$ lies between $|r - |a||$ and $r + |a|$.

[JEE Tip] Factorisations via $\omega$: $a^2 + ab + b^2 = (a - b\omega)(a - b\omega^2)$ and $a^3 + b^3 = (a + b)(a + b\omega)(a + b\omega^2)$ — they turn intimidating JEE products into one-line evaluations.

Solved Examples

Example 1: Polar form conversions

Express in polar form: (i) $1 + i$ (ii) $-1 + \sqrt{3}i$ (iii) $-2 - 2i$.

Solution:

Step 1 — (i) Modulus, quadrant, reference angle. $r = \sqrt{1 + 1} = \sqrt{2}$; the point (1, 1) is in Q I with reference angle $\frac{\pi}{4}$, so $\theta = \frac{\pi}{4}$: $\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$.

Step 2 — (ii) Same ritual. $r = \sqrt{1 + 3} = 2$; Q II with reference $\frac{\pi}{3}$: $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$: $2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$.

Step 3 — (iii) Watch the principal placement. $r = \sqrt{4 + 4} = 2\sqrt{2}$; Q III with reference $\frac{\pi}{4}$: principal $\theta = -\left(\pi - \frac{\pi}{4}\right) = -\frac{3\pi}{4}$: $2\sqrt{2}\left[\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right]$.

Takeaway: Modulus, quadrant, reference angle, principal placement — the four-step ritual, every time.

Example 2: Argument arithmetic

Find the principal argument of (i) $z = i(1 + i)$ (ii) $z = \frac{1 + i}{1 - i}$, using argument addition.

Solution:

Step 1 — (i) Add the arguments. arg z = arg i + arg(1 + i) = $\frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.

Step 2 — Verify the quadrant. $i(1 + i) = -1 + i$, in Q II with reference $\frac{\pi}{4}$ — argument $\frac{3\pi}{4}$ ✓.

Step 3 — (ii) Subtract for a quotient. arg = arg(1 + i) − arg(1 − i) = $\frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}$.

Step 4 — Verify. The quotient is i, whose argument is indeed $\frac{\pi}{2}$ ✓.

Takeaway: Arguments add over products and subtract over quotients — then verify the quadrant.

Example 3: A quadratic with complex roots

Solve $x^2 + 3x + 9 = 0$ over the complex numbers.

Solution:

Step 1 — Compute the discriminant. $D = 3^2 - 4(1)(9) = 9 - 36 = -27$.

Step 2 — Write $\sqrt{D}$ in i-form. $\sqrt{-27} = i\sqrt{27} = 3\sqrt{3}i$.

Step 3 — Apply the formula. $x = \frac{-3 \pm 3\sqrt{3}i}{2}$ — a conjugate pair, as real coefficients demand.

Takeaway: $\sqrt{-27} = 3\sqrt{3}i$ — simplify the surd inside the i-form; the roots are conjugates.

Example 4: Nature of roots without solving

Without solving, classify the roots of: (i) $2x^2 - 7x + 3 = 0$ (ii) $x^2 - 4x + 4 = 0$ (iii) $x^2 + x + 2 = 0$.

Solution:

Step 1 — (i) Discriminant. $D = 49 - 24 = 25 > 0$: two distinct real roots — and since 25 is a perfect square, they are rational.

Step 2 — (ii) Discriminant. $D = 16 - 16 = 0$: equal real roots ($x = 2$ twice).

Step 3 — (iii) Discriminant. $D = 1 - 8 = -7 < 0$: a conjugate pair of non-real roots.

Takeaway: D answers the classification question in one subtraction — solving is never required for "nature of roots".

Example 5: Root-coefficient relations at work

If $\alpha, \beta$ are the roots of $x^2 - 5x + 3 = 0$, find (i) $\alpha^2 + \beta^2$ (ii) $\frac{1}{\alpha} + \frac{1}{\beta}$ (iii) the equation with roots $\alpha^2, \beta^2$.

Solution:

Step 1 — Extract S and P. $\alpha + \beta = 5$ and $\alpha\beta = 3$.

Step 2 — (i) Power sum. $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 25 - 6 = 19$.

Step 3 — (ii) Reciprocal sum. $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5}{3}$.

Step 4 — (iii) New sum and product. Sum of new roots $= \alpha^2 + \beta^2 = 19$; product $= (\alpha\beta)^2 = 9$: equation $x^2 - 19x + 9 = 0$.

Takeaway: Every symmetric function of the roots is reachable from S and P — the roots themselves are never needed.

Example 6: Forming an equation from complex roots

Form the quadratic equation with real coefficients having $2 + 3i$ as a root.

Solution:

Step 1 — Invoke the conjugate-pair rule. Real coefficients force the other root to be $2 - 3i$.

Step 2 — Compute S and P. $S = (2 + 3i) + (2 - 3i) = 4$; $P = (2 + 3i)(2 - 3i) = 4 + 9 = 13$.

Step 3 — Write the equation. $x^2 - Sx + P = 0$: $x^2 - 4x + 13 = 0$.

Takeaway: One complex root + real coefficients = the conjugate comes free; S and P are automatically real.

Example 7: Square root of a complex number

Find $\sqrt{3 - 4i}$.

Solution:

Step 1 — Set up the three equations. $(x + iy)^2 = 3 - 4i$ gives $x^2 - y^2 = 3$, $2xy = -4$, and the modulus equation $x^2 + y^2 = |3 - 4i| = 5$.

Step 2 — Solve for the squares. Adding and halving: $x^2 = 4$; subtracting: $y^2 = 1$.

Step 3 — Fix the signs. $xy = -2 < 0$: opposite signs, so the roots are $\pm(2 - i)$.

Step 4 — Check. $(2 - i)^2 = 4 - 4i + i^2 = 3 - 4i$ ✓.

Takeaway: The sign of the middle coefficient b decides same-signs (b > 0) or opposite-signs (b < 0) — the only place students slip.

Example 8: An ω evaluation

Evaluate $\omega^{100} + \omega^{200} + \omega^{300}$, where $\omega$ is a non-real cube root of unity.

Solution:

Step 1 — Reduce every exponent mod 3. $100 = 3(33) + 1$, $200 = 3(66) + 2$, $300 = 3(100) + 0$.

Step 2 — Substitute. $\omega^{100} + \omega^{200} + \omega^{300} = \omega + \omega^2 + 1$.

Step 3 — Invoke the master identity. $1 + \omega + \omega^2 = 0$.

Takeaway: Reduce every exponent mod 3, then invoke $1 + \omega + \omega^2 = 0$ — the two-step that solves 90% of $\omega$ problems.

Example 9: The (1 + ω)-type powers

Evaluate $(1 + \omega - \omega^2)^7$, using the $\omega$ identities.

Solution:

Step 1 — Convert $1 + \omega$. From the sum identity, $1 + \omega = -\omega^2$, so the base is $-\omega^2 - \omega^2 = -2\omega^2$.

Step 2 — Raise to the 7th power. $(-2\omega^2)^7 = (-1)^7 2^7 \omega^{14} = -128\,\omega^{14}$.

Step 3 — Reduce the exponent mod 3. $14 = 3(4) + 2$: $\omega^{14} = \omega^2$, so the answer is $-128\omega^2$.

Takeaway: Convert $1 + \omega$ or $1 + \omega^2$ via the sum identity FIRST; powers then reduce mod 3.

Example 10: An ω product

Evaluate $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$.

Solution:

Step 1 — Collapse the first factor. $1 + \omega^2 = -\omega$, so $1 - \omega + \omega^2 = -\omega - \omega = -2\omega$.

Step 2 — Collapse the second. $1 + \omega = -\omega^2$, so $1 + \omega - \omega^2 = -\omega^2 - \omega^2 = -2\omega^2$.

Step 3 — Multiply. $(-2\omega)(-2\omega^2) = 4\omega^3 = 4$.

Takeaway: Both factors collapse before any multiplication — the identities do all the work.

Example 11: Triangle inequality bounds

If $|z| = 3$, find the maximum and minimum values of $|z + 4 - 3i|$ — the distance from z to the point $-4 + 3i$.

Solution:

Step 1 — Read the geometry. z runs over the circle of radius 3 centred at O; $|z + 4 - 3i| = |z - (-4 + 3i)|$ is the distance from z to the fixed point $w = -4 + 3i$.

Step 2 — Locate the fixed point. $|w| = \sqrt{16 + 9} = 5$ — it lies at distance 5 from the centre.

Step 3 — Apply the along-the-line bounds. Maximum $= |w| + r = 5 + 3 = 8$; minimum $= |w| - r = 5 - 3 = 2$.

Takeaway: Distances from a circle to a point are read off the line through the centre: $\big||w| - r\big| \leq |z - w| \leq |w| + r$.

Example 12: A root-relation with a twist

If one root of $x^2 + px + q = 0$ (p, q real) is $1 - i$, find p and q.

Solution:

Step 1 — Conjugate partner. Real coefficients force the other root to be $1 + i$.

Step 2 — Sum gives p. $S = 2 = -p$: $p = -2$.

Step 3 — Product gives q. $P = (1 - i)(1 + i) = 1 + 1 = 2 = q$.

Step 4 — Check. $x^2 - 2x + 2 = 0$ has $D = 4 - 8 = -4$ and roots $\frac{2 \pm 2i}{2} = 1 \pm i$ ✓.

Takeaway: Read p and q straight from S and P — no substitution of the complex root needed.

Example 13: Polar multiplication in action

Using polar forms, compute the product $(1 + i)(\sqrt{3} + i)$, and verify by expansion.

Solution:

Step 1 — Convert each factor. $1 + i$: $r = \sqrt{2}$, $\theta = \frac{\pi}{4}$. $\sqrt{3} + i$: $r = 2$, $\theta = \frac{\pi}{6}$.

Step 2 — Multiply moduli, add arguments. Product modulus $= 2\sqrt{2}$; argument $= \frac{\pi}{4} + \frac{\pi}{6} = \frac{5\pi}{12}$ (75°).

Step 3 — Verify by expansion. $(1 + i)(\sqrt{3} + i) = \sqrt{3} + i + \sqrt{3}i - 1 = (\sqrt{3} - 1) + (\sqrt{3} + 1)i$.

Step 4 — Compare moduli. $\sqrt{(\sqrt{3} - 1)^2 + (\sqrt{3} + 1)^2} = \sqrt{4 - 2\sqrt{3} + 4 + 2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$ ✓.

Takeaway: "Moduli multiply, arguments add" — and the expansion cross-check even yields the exact values of cos 75° and sin 75° for free.

Example 14: A quadratic in disguise

Solve $x^2 - (3 - 2i)x + (5 - 5i) = 0$, and note that the roots are NOT conjugates of each other.

Solution:

Step 1 — Compute the discriminant (complex coefficients allowed). $D = (3 - 2i)^2 - 4(5 - 5i) = 9 - 12i - 4 - 20 + 20i = -15 + 8i$.

Step 2 — Find $\sqrt{D}$ by the three-equation method. $x^2 - y^2 = -15$, $2xy = 8$, $x^2 + y^2 = |{-15} + 8i| = 17$: $x^2 = 1$, $y^2 = 16$, same signs ($xy = 4 > 0$): $\sqrt{D} = \pm(1 + 4i)$.

Step 3 — Apply the formula. $x = \frac{(3 - 2i) \pm (1 + 4i)}{2}$: plus gives $\frac{4 + 2i}{2} = 2 + i$; minus gives $\frac{2 - 6i}{2} = 1 - 3i$.

Step 4 — Observe. The roots are NOT conjugates — the conjugate-pair guarantee needs REAL coefficients.

Takeaway: With complex coefficients, use the formula plus the square-root method; expect unpaired roots.

Example 15: ω factorisation payoff

Evaluate $(3 + 5\omega + 3\omega^2)^2 + (3 + 3\omega + 5\omega^2)^2$, for a non-real cube root of unity $\omega$.

Solution:

Step 1 — Peel multiples of the zero-sum. $3 + 5\omega + 3\omega^2 = 3(1 + \omega + \omega^2) + 2\omega = 0 + 2\omega = 2\omega$.

Step 2 — Same for the second base. $3 + 3\omega + 5\omega^2 = 3(1 + \omega + \omega^2) + 2\omega^2 = 2\omega^2$.

Step 3 — Square and add. $(2\omega)^2 + (2\omega^2)^2 = 4\omega^2 + 4\omega^4 = 4\omega^2 + 4\omega$ (since $\omega^4 = \omega$).

Step 4 — Finish with the sum identity. $4(\omega + \omega^2) = 4(-1) = -4$.

Takeaway: Peel off multiples of $(1 + \omega + \omega^2)$ from every bracket — what remains is tiny.