Complex Numbers & Quadratic Equations
Master the algebra and geometry of complex numbers, De Moivre's theorem, roots of unity, and the complete theory of quadratic equations — a high-weightage JEE chapter.
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01
Complex Number Algebra
Algebraic operations, modulus, argument, conjugate, and polar form of complex numbers.
Fundamentals of Complex Numbers
Definition
A complex number $z = a + ib$ where $a, b \in \mathbb{R}$ and $i = \sqrt{-1}$. Here $a = \text{Re}(z)$ and $b = \text{Im}(z)$.
- Modulus: $|z| = \sqrt{a^2 + b^2}$
- Conjugate: $\bar{z} = a - ib$. Key: $z\bar{z} = |z|^2$
- Argument: $\arg(z) = \theta$ where $z = |z|(\cos\theta + i\sin\theta)$. Principal argument $\theta \in (-\pi, \pi]$.
- Polar form: $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$ (Euler's formula)
Essential Properties
$|z_1 z_2| = |z_1||z_2|$, $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$
$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$, $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$
Triangle Inequality: $|z_1 + z_2| \leq |z_1| + |z_2|$ and $|z_1 - z_2| \geq \big||z_1| - |z_2|\big|$
JEE Trick: $|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2)$ — parallelogram law.
$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$, $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$
Triangle Inequality: $|z_1 + z_2| \leq |z_1| + |z_2|$ and $|z_1 - z_2| \geq \big||z_1| - |z_2|\big|$
JEE Trick: $|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2)$ — parallelogram law.
📝 Example
If $|z - 3 + 2i| = 4$, identify the locus of $z$.
Let $z = x + iy$. Then $|z - (3 - 2i)| = 4$ represents a circle with centre $(3, -2)$ and radius $4$. In Cartesian form: $(x-3)^2 + (y+2)^2 = 16$.
Geometry in the Argand Plane
JEE Important Loci
$|z - z_1| = |z - z_2|$: perpendicular bisector of segment joining $z_1, z_2$.
$|z - z_1| + |z - z_2| = k$ (where $k > |z_1 - z_2|$): ellipse with foci $z_1, z_2$.
$\arg\left(\frac{z - z_1}{z - z_2}\right) = \alpha$: arc of a circle through $z_1, z_2$.
$|z - z_1| + |z - z_2| = k$ (where $k > |z_1 - z_2|$): ellipse with foci $z_1, z_2$.
$\arg\left(\frac{z - z_1}{z - z_2}\right) = \alpha$: arc of a circle through $z_1, z_2$.
Rotation Formula
If $z_2 - z_0 = (z_1 - z_0) \cdot e^{i\theta} \cdot \frac{|z_2 - z_0|}{|z_1 - z_0|}$, then $z_2$ is obtained by rotating $z_1$ about $z_0$ by angle $\theta$ and scaling.
Special case (same distance): $z_2 - z_0 = (z_1 - z_0)e^{i\theta}$.
Special case (same distance): $z_2 - z_0 = (z_1 - z_0)e^{i\theta}$.
📝 Example
If $z = x + iy$ and $\arg\left(\frac{z-1}{z+1}\right) = \frac{\pi}{4}$, find the locus of $z$.
Let $w = \frac{z-1}{z+1}$. The condition $\arg(w) = \pi/4$ means $w$ lies on the ray from origin at angle $\pi/4$. Writing $\frac{(x-1)+iy}{(x+1)+iy}$ and rationalising, we get $\arg = \pi/4$ implies the imaginary and real parts satisfy $\text{Im}/\text{Re} = 1$. After simplification: $x^2 + y^2 - 2y - 1 = 0$, i.e., $x^2 + (y-1)^2 = 2$. This is an arc of a circle with centre $(0,1)$ and radius $\sqrt{2}$ (the arc in the upper half-plane).
02
De Moivre's Theorem & Roots of Unity
Powers and roots of complex numbers, $n$-th roots of unity, and their elegant geometric properties.
De Moivre's Theorem
De Moivre's Theorem
For any integer $n$: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$.
Equivalently: $(e^{i\theta})^n = e^{in\theta}$.
Equivalently: $(e^{i\theta})^n = e^{in\theta}$.
$n$-th Roots of Unity
The solutions of $z^n = 1$ are $z_k = e^{2\pi i k/n} = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}$ for $k = 0, 1, \ldots, n-1$.
Let $\omega = e^{2\pi i/n}$. Then the roots are $1, \omega, \omega^2, \ldots, \omega^{n-1}$.
Key Properties:
$1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0$
$\omega^n = 1$
The roots form a regular $n$-gon inscribed in the unit circle.
Let $\omega = e^{2\pi i/n}$. Then the roots are $1, \omega, \omega^2, \ldots, \omega^{n-1}$.
Key Properties:
$1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0$
$\omega^n = 1$
The roots form a regular $n$-gon inscribed in the unit circle.
JEE Trick — Cube Roots of Unity
$\omega = \frac{-1 + i\sqrt{3}}{2}$, $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$, $\omega^3 = 1$, $1 + \omega + \omega^2 = 0$.
Shortcut: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega)$.
Shortcut: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega)$.
📝 Example
Find $(1 + i)^{20}$.
$1 + i = \sqrt{2}\,e^{i\pi/4}$. By De Moivre: $(1+i)^{20} = (\sqrt{2})^{20} \cdot e^{i \cdot 20\pi/4} = 2^{10} \cdot e^{i \cdot 5\pi} = 1024 \cdot e^{i\pi} = 1024(-1) = \boxed{-1024}$.
03
Quadratic Equations
Discriminant, nature of roots, Vieta's formulas, and solving quadratic equations — a JEE staple.
Discriminant & Nature of Roots
Quadratic Equation
$ax^2 + bx + c = 0$ ($a \neq 0$). Roots: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Discriminant $D = b^2 - 4ac$.
- $D > 0$: two distinct real roots. If $D$ is a perfect square and $a,b,c \in \mathbb{Q}$, roots are rational.
- $D = 0$: two equal real roots (repeated root $= -b/(2a)$).
- $D < 0$: two complex conjugate roots.
Vieta's Formulas
If $\alpha, \beta$ are roots of $ax^2 + bx + c = 0$:
$\alpha + \beta = -\frac{b}{a}$, $\quad \alpha\beta = \frac{c}{a}$.
JEE Trick: To form a quadratic with roots $\alpha, \beta$: $x^2 - (\alpha+\beta)x + \alpha\beta = 0$.
$\alpha + \beta = -\frac{b}{a}$, $\quad \alpha\beta = \frac{c}{a}$.
JEE Trick: To form a quadratic with roots $\alpha, \beta$: $x^2 - (\alpha+\beta)x + \alpha\beta = 0$.
JEE Shortcut — Symmetric Functions
$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$
$\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$
$|\alpha - \beta| = \frac{\sqrt{D}}{|a|}$
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = -\frac{b}{c}$
$\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$
$|\alpha - \beta| = \frac{\sqrt{D}}{|a|}$
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = -\frac{b}{c}$
📝 Example
If $\alpha, \beta$ are roots of $x^2 - 5x + 6 = 0$, find $\alpha^4 + \beta^4$.
$\alpha + \beta = 5$, $\alpha\beta = 6$.
$\alpha^2 + \beta^2 = 25 - 12 = 13$.
$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = 169 - 72 = \boxed{97}$.
$\alpha^2 + \beta^2 = 25 - 12 = 13$.
$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = 169 - 72 = \boxed{97}$.
04
Relation Between Roots & Coefficients
Higher-degree polynomials, common roots, and transformation of equations — JEE Advanced level.
Higher Degree & Common Roots
Cubic & Quartic — Vieta's Extensions
For $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$:
$\alpha + \beta + \gamma = -b/a$
$\alpha\beta + \beta\gamma + \gamma\alpha = c/a$
$\alpha\beta\gamma = -d/a$
$\alpha + \beta + \gamma = -b/a$
$\alpha\beta + \beta\gamma + \gamma\alpha = c/a$
$\alpha\beta\gamma = -d/a$
Common Root Condition
If $a_1x^2 + b_1x + c_1 = 0$ and $a_2x^2 + b_2x + c_2 = 0$ have a common root $\alpha$, then:
$(a_1c_2 - a_2c_1)^2 = (a_1b_2 - a_2b_1)(b_1c_2 - b_2c_1)$
JEE Method: Subtract the two equations to find the common root directly.
$(a_1c_2 - a_2c_1)^2 = (a_1b_2 - a_2b_1)(b_1c_2 - b_2c_1)$
JEE Method: Subtract the two equations to find the common root directly.
Transformation of Equations
To find the equation whose roots are $f(\alpha), f(\beta)$: replace $x$ by $f^{-1}(x)$ in the original equation.
Roots are $\alpha^2, \beta^2$: replace $x$ by $\sqrt{x}$ and square.
Roots are $1/\alpha, 1/\beta$: replace $x$ by $1/x$ and multiply by $x^2$.
Roots are $\alpha^2, \beta^2$: replace $x$ by $\sqrt{x}$ and square.
Roots are $1/\alpha, 1/\beta$: replace $x$ by $1/x$ and multiply by $x^2$.
📝 Example
Find the equation whose roots are squares of the roots of $x^2 - 3x + 1 = 0$.
Replace $x$ by $\sqrt{x}$: $x - 3\sqrt{x} + 1 = 0 \Rightarrow 3\sqrt{x} = x + 1 \Rightarrow 9x = (x+1)^2 = x^2 + 2x + 1$.
Answer: $x^2 - 7x + 1 = 0$. (Alternatively: $\alpha^2 + \beta^2 = 9 - 2 = 7$, $\alpha^2\beta^2 = 1$.)
Answer: $x^2 - 7x + 1 = 0$. (Alternatively: $\alpha^2 + \beta^2 = 9 - 2 = 7$, $\alpha^2\beta^2 = 1$.)
05
Location of Roots
Conditions for roots to lie in specific intervals — a favourite in JEE Advanced.
Location of Roots of a Quadratic
Both Roots Greater Than $k$
For $f(x) = ax^2 + bx + c$ with $a > 0$: both roots $> k$ iff
(i) $D \geq 0$, (ii) $f(k) > 0$, (iii) $-b/(2a) > k$.
(i) $D \geq 0$, (ii) $f(k) > 0$, (iii) $-b/(2a) > k$.
Both Roots in Interval $(p, q)$
$a > 0$: (i) $D \geq 0$, (ii) $f(p) > 0$, (iii) $f(q) > 0$, (iv) $p < -b/(2a) < q$.
Exactly One Root in $(p, q)$
$f(p) \cdot f(q) < 0$ (the sign change guarantees a root between $p$ and $q$).
JEE Trick — Root Bracketing
For a continuous function, if $f(a)$ and $f(b)$ have opposite signs, at least one root lies in $(a,b)$. For quadratics specifically, if $f(p) \cdot f(q) < 0$, exactly one root is in $(p,q)$.
📝 Example
Find the values of $k$ for which both roots of $x^2 - 6x + k = 0$ are greater than 2.
$a = 1 > 0$. Conditions:
(i) $D \geq 0$: $36 - 4k \geq 0 \Rightarrow k \leq 9$.
(ii) $f(2) > 0$: $4 - 12 + k > 0 \Rightarrow k > 8$.
(iii) Vertex $> 2$: $-(-6)/(2) = 3 > 2$. Always satisfied.
Answer: $k \in (8, 9]$.
(i) $D \geq 0$: $36 - 4k \geq 0 \Rightarrow k \leq 9$.
(ii) $f(2) > 0$: $4 - 12 + k > 0 \Rightarrow k > 8$.
(iii) Vertex $> 2$: $-(-6)/(2) = 3 > 2$. Always satisfied.
Answer: $k \in (8, 9]$.
★ Key Takeaways
- $z\bar{z} = |z|^2$ is the single most useful identity — use it to simplify divisions and modulus calculations.
- For cube roots of unity: $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$ — memorise these and the factorisation of $a^3 + b^3 + c^3 - 3abc$.
- Rotation in the Argand plane uses $e^{i\theta}$ — this is heavily tested in JEE Advanced geometry problems.
- Vieta's formulas extend to any degree polynomial — never expand when you can use sum/product of roots.
- Location of roots problems require checking discriminant, function values at boundaries, and vertex position — all three conditions simultaneously.
✎ Practice Problems
Problem 1
If $z = \frac{1+i\sqrt{3}}{1-i\sqrt{3}}$, find $z^6$.
Show Solution ▼
Multiply numerator and denominator by conjugate: $z = \frac{(1+i\sqrt{3})^2}{(1)^2 + 3} = \frac{1 + 2i\sqrt{3} - 3}{4} = \frac{-2 + 2i\sqrt{3}}{4} = \frac{-1 + i\sqrt{3}}{2}$. This is $e^{i \cdot 2\pi/3} = \omega$ (a cube root of unity). So $z^6 = \omega^6 = (\omega^3)^2 = 1$. Answer: $\boxed{1}$.
Problem 2
Find the minimum value of $|z - 1| + |z + 1| + |z - i|$ for $z \in \mathbb{C}$.
Show Solution ▼
$|z-1| + |z+1| \geq |1 - (-1)| = 2$ by triangle inequality. The minimum of $|z-1|+|z+1|$ is $2$, achieved when $z$ lies on $[-1, 1]$. For such $z = x$ ($-1 \leq x \leq 1$), $|z-i| = \sqrt{x^2+1}$, minimised at $x = 0$ giving $1$. Total minimum $= 2 + 1 = \boxed{3}$. Verify: Fermat point analysis confirms $z = 0$ is not exactly optimal, but $|z-1|+|z+1|=2$ forces $z \in [-1,1]$, and $\sqrt{x^2+1}$ is minimised at $x=0$. Minimum $= 3$.
Problem 3
If $\alpha, \beta$ are roots of $x^2 + px + q = 0$ and $\alpha^4, \beta^4$ are roots of $x^2 - rx + s = 0$, express $r$ in terms of $p$ and $q$.
Show Solution ▼
$\alpha + \beta = -p$, $\alpha\beta = q$.
$\alpha^2 + \beta^2 = p^2 - 2q$.
$\alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2 = (p^2-2q)^2 - 2q^2 = p^4 - 4p^2q + 4q^2 - 2q^2 = p^4 - 4p^2q + 2q^2$.
Since $\alpha^4 + \beta^4 = r$: $\boxed{r = p^4 - 4p^2q + 2q^2}$.
$\alpha^2 + \beta^2 = p^2 - 2q$.
$\alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2 = (p^2-2q)^2 - 2q^2 = p^4 - 4p^2q + 4q^2 - 2q^2 = p^4 - 4p^2q + 2q^2$.
Since $\alpha^4 + \beta^4 = r$: $\boxed{r = p^4 - 4p^2q + 2q^2}$.
Problem 4
Find all values of $k$ for which the equation $x^2 - 2(k+1)x + k^2 = 0$ has both roots positive.
Show Solution ▼
$D \geq 0$: $4(k+1)^2 - 4k^2 \geq 0 \Rightarrow 4(2k+1) \geq 0 \Rightarrow k \geq -1/2$.
Sum of roots $> 0$: $2(k+1) > 0 \Rightarrow k > -1$.
Product of roots $> 0$: $k^2 > 0 \Rightarrow k \neq 0$.
Combining: $k \in [-1/2, 0) \cup (0, \infty)$.
Sum of roots $> 0$: $2(k+1) > 0 \Rightarrow k > -1$.
Product of roots $> 0$: $k^2 > 0 \Rightarrow k \neq 0$.
Combining: $k \in [-1/2, 0) \cup (0, \infty)$.
Problem 5
If $\omega$ is a non-real cube root of unity, evaluate $(1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5$.
Show Solution ▼
Since $1 + \omega + \omega^2 = 0$: $1 - \omega + \omega^2 = -2\omega$ (using $1 + \omega^2 = -\omega$). Similarly $1 + \omega - \omega^2 = -2\omega^2$ (using $1 + \omega = -\omega^2$).
$(-2\omega)^5 + (-2\omega^2)^5 = -32(\omega^5 + \omega^{10}) = -32(\omega^2 + \omega) = -32(-1) = \boxed{32}$.
$(-2\omega)^5 + (-2\omega^2)^5 = -32(\omega^5 + \omega^{10}) = -32(\omega^2 + \omega) = -32(-1) = \boxed{32}$.
🎯 Interactive MCQs
1. If $|z_1| = |z_2| = 1$ and $z_1 z_2 \neq -1$, then $\frac{z_1 + z_2}{1 + z_1 z_2}$ is:
A A real number
B A purely imaginary number
C Always equal to $1$
D A complex number with $|w| = 1$
2. The number of complex numbers $z$ satisfying $|z| = 1$ and $\left|\frac{z}{\bar{z}} + \frac{\bar{z}}{z}\right| = 1$ is:
A $4$
B $8$
C Infinite
D $6$
3. If $\alpha$ and $\beta$ are roots of $x^2 - x + 1 = 0$, what is $\alpha^{12} + \beta^{12}$?
A $0$
B $1$
C $2$
D $-2$
4. If both roots of $ax^2 + bx + c = 0$ lie in $(0, 1)$, which of the following must be true?
A $a + b + c = 0$
B $a$, $c$, and $a+b+c$ all have the same sign
C $b > 0$ always
D $c > a$ always
5. If $z$ is a complex number with $|z| = 1$ and $z \neq 1$, then $\text{Re}\left(\frac{1}{1-z}\right)$ equals:
A $\frac{1}{2}$
B $0$
C $1$
D Depends on $\arg(z)$