Overview
This chapter introduces complex numbers and their role in solving quadratic equations that have no real roots. Starting from the imaginary unit i (i^2 = -1), students learn the algebraic form a + bi, equality, and arithmetic operations (addition, subtraction, multiplication, division). The chapter covers geometric representation on the Argand plane, modulus and conjugate, polar representation (modulus and argument) and conversion between rectangular and polar forms. Key applications include finding square roots of complex numbers, using conjugates to simplify expressions, and understanding how complex roots arise in quadratic equations with real coefficients (including the conjugate-root property). Important ideas like the discriminant, nature of roots, relations between roots and coefficients, and geometric interpretations (distance, locus) are developed. This foundation is essential for higher algebra, trigonometry, calculus and physics (wave theory, electrical circuits), and it trains students to handle equations and transformations beyond the real number system.
Learning Objectives
- Define complex numbers, the imaginary unit i, real and imaginary parts and represent numbers in the form a + ib
- Explain the geometric representation of complex numbers on the Argand plane and interpret modulus and argument
- Compute modulus and argument of a complex number and use them to find distances and angles between points
- Perform addition, subtraction, multiplication and division of complex numbers and simplify the results
- Apply complex conjugation to rationalize denominators, compute reciprocals and prove modulus properties
- Solve quadratic equations with real coefficients, including cases with complex roots, using factorization and the quadratic formula
- Use the discriminant to determine the nature of roots (real distinct, real equal, complex) and justify conclusions
- Derive and use the relations between roots and coefficients (α + β = −b/a, αβ = c/a) to find unknown coefficients
Topics in this chapter
13 topics · tap a topic title to jump straight to it.
Introduction to Complex Numbers
Introduction to Complex Numbers
Key Point: i^2 = -1
What is a complex number?
A complex number is a number of the form a + bi where a and b are real numbers and i is the imaginary unit with the property i^2 = -1. Here a is called the real part and b the imaginary part. The set of all complex numbers is denoted by C.
Notation: For z = a + bi, Re(z) = a, Im(z) = b, and the conjugate is z̄ = a - bi.
Why extend real numbers?
Complex numbers extend the real numbers so every polynomial equation (even with negative discriminant) has roots. For example, x^2 + 1 = 0 has no real solution, but in C it has solutions x = i and x = -i.
Algebraic operations (brief):
- Addition/subtraction: (a+bi) ± (c+di) = (a±c) + (b±d)i.
- Multiplication: (a+bi)(c+di) = (ac - bd) + (ad + bc)i.
- Division: (a+bi)/(c+di) = [(a+bi)(c-di)]/(c^2 + d^2) = [(ac+bd) + (bc-ad)i]/(c^2+d^2), provided c+di ≠ 0.
Geometric (Argand) representation:
A complex number a + bi corresponds to the point (a, b) in the plane (called the Argand plane). It can be treated as a vector from the origin to (a, b). The distance from the origin is the modulus |z| = sqrt(a^2 + b^2), and the angle with the positive real axis is the argument arg(z) = θ, where tan θ = b/a (take quadrant into account).
Polar form and Euler's formula:
Any nonzero z can be written as z = r(cos θ + i sin θ) where r = |z| and θ = arg(z). Using Euler's formula, z = r e^{iθ}. This makes multiplication and division geometric: multiplying multiplies moduli and adds arguments; dividing divides moduli and subtracts arguments.
Conjugate and properties:
The conjugate z̄ reflects z across the real axis. Useful identities: z z̄ = |z|^2, Re(z) = (z + z̄)/2, Im(z) = (z - z̄)/(2i).
Important consequences:
- Complex numbers form a field (you can add, subtract, multiply, divide except by zero).
- Non-real roots of real-coefficient polynomials occur in conjugate pairs.
- De Moivre's theorem: (r(cos θ + i sin θ))^n = r^n (cos nθ + i sin nθ) for integer n.
- Solving quadratics with negative discriminant: x^2 + 4 = 0 → x = ±2i.
- Alternating-current (AC) circuits: voltages and currents are represented as phasors (complex numbers) to encode amplitude and phase; impedance uses complex numbers (resistance + reactance).
- Signal processing and control systems: complex numbers describe sinusoidal signals and their phase relationships; poles and zeros of transfer functions lie in the complex plane.
- Fractals and complex dynamics: the Mandelbrot and Julia sets are defined using complex iterations.
- 2D rotations and scaling: multiply by a complex number re^{iθ} to rotate by θ and scale by r.
- \[i^2 = -1\]
- \[z = a + bi\]\[Re(z) = a\]\[Im(z) = b\]
- \[Equality: a+bi = c+di ⇔ a = c and b = d\]
- \[Addition: (a+bi)+(c+di) = (a+c) + (b+d)i\]
- \[Multiplication: (a+bi)(c+di) = (ac - bd) + (ad + bc)i\]
- \[Conjugate: z̄ = a - bi\]\[z z̄ = a^2 + b^2 = |z|^2\]
Algebra of Complex Numbers
Algebra of Complex Numbers
Key Point: z = x + iy, where x = Re(z), y = Im(z), and i^2 = -1
Introduction
A complex number is a number of the form z = x + iy where x and y are real numbers and i is the imaginary unit with i^2 = -1. Here x is called the real part, Re(z), and y is the imaginary part, Im(z). Complex numbers extend the real number system so equations like x^2 + 1 = 0 have solutions.
Geometric representation (Argand plane)
Complex numbers can be represented as points or vectors in a 2D plane: the horizontal axis is the real axis and the vertical axis is the imaginary axis. The point (x, y) corresponds to z = x + iy. The distance from the origin to the point is the modulus |z| = sqrt(x^2 + y^2) and the angle from the positive real axis to the vector is the argument arg(z) = theta.
Conjugate and modulus
The conjugate of z = x + iy is z̄ = x - iy. Conjugation reflects z about the real axis. Important facts: z z̄ = |z|^2 (a real nonnegative number) and |z̄| = |z|.
Algebraic operations
- Addition and subtraction: add/subtract corresponding real and imaginary parts: (x1 + iy1) ± (x2 + iy2) = (x1 ± x2) + i(y1 ± y2). Geometrically, addition is vector addition.
- Multiplication: use distributive law and i^2 = -1: (x1 + iy1)(x2 + iy2) = (x1x2 - y1y2) + i(x1y2 + x2y1). Geometrically multiplication scales and rotates vectors when expressed in polar form.
- Division: to divide use the conjugate: (x1 + iy1)/(x2 + iy2) = [(x1 + iy1)(x2 - iy2)]/(x2^2 + y2^2), provided the denominator is nonzero.
Polar (trigonometric) form and relation to Cartesian form
If z = x + iy and r = |z|, theta = arg(z), then x = r cos theta, y = r sin theta and z = r(cos theta + i sin theta). This form makes multiplication and powers simple: magnitudes multiply and arguments add.
Important properties
- |zw| = |z| |w|, arg(zw) = arg(z) + arg(w) (mod 2pi)
- |z/w| = |z|/|w| and arg(z/w) = arg(z) - arg(w)
- z + z̄ = 2 Re(z), z - z̄ = 2i Im(z)
de Moivre's formula (for integer n)
For any real theta and integer n, [r(cos theta + i sin theta)]^n = r^n (cos n theta + i sin n theta). This is useful for computing powers and roots (roots require more careful treatment).
Summary
Algebra of complex numbers combines algebraic manipulation (using i^2 = -1 and conjugates) with geometric intuition (vectors, rotation, scaling). This makes complex numbers a powerful tool in solving equations, modelling oscillations, waves and rotations, and in engineering applications.
- Addition: z1 = 3 + 2i, z2 = 1 - 4i. z1 + z2 = (3+1) + i(2-4) = 4 - 2i. Geometrically, place vectors for z1 and z2 and add head-to-tail.
- Multiplication: z1 = 2 + i, z2 = 1 + 2i. z1*z2 = (2*1 - 1*2) + i(2*2 + 1*1) = (2 - 2) + i(4 + 1) = 5i. In polar terms this multiplies magnitudes and adds angles.
- Division: z1 = 3 + 2i, z2 = 1 - i. z1/z2 = [(3+2i)(1+i)]/(1^2+(-1)^2) = [(3+3i+2i+2i^2)]/2 = [(3+5i-2)]/2 = (1+5i)/2 = 0.5 + 2.5i.
- Modulus and conjugate: For z = -1 + i*sqrt(3), |z| = sqrt[(-1)^2 + (sqrt(3))^2] = 2 and z̄ = -1 - i*sqrt(3).
- \[z = x + iy\]\[where x = Re(z)\]\[y = Im(z)\]\[and i^2 = -1\]
- \[Conjugate: z̄ = x - iy\]
- \[Modulus: |z| = sqrt(x^2 + y^2)\]\[|z|^2 = z z̄\]
- \[Argument: arg(z) = atan2(y\]\[x) (principal value usually in (-pi\]\[pi] or [0, 2pi))\]
- \[Addition/Subtraction: (x1 + iy1) ± (x2 + iy2) = (x1 ± x2) + i(y1 ± y2)\]
- \[Multiplication: (x1 + iy1)(x2 + iy2) = (x1x2 - y1y2) + i(x1y2 + x2y1)\]
Complex Conjugate
Complex Conjugate
Key Point: If z = a + ib then z̄ = a − ib
Definition: For a complex number z = a + ib (where a and b are real and i = √-1), the complex conjugate of z is z̄ = a − ib. Geometrically, z̄ is the reflection of z across the real axis in the Argand plane.
Key ideas and interpretation:
- Real part: Re(z) = a. Imaginary part: Im(z) = b. Then z̄ = Re(z) − i·Im(z).
- If z = r(cos θ + i sin θ) = r e^{iθ} in polar form, then z̄ = r(cos(−θ) + i sin(−θ)) = r e^{-iθ}.
- Conjugation is a reflection: point (a, b) maps to (a, −b) on the plane.
- When coefficients of a polynomial are real, non-real roots occur in conjugate pairs: if z is a root, so is z̄.
Important properties (useful in algebra and solving equations):
- (z̄)̄ = z
- z + z̄ = 2·Re(z) (a real number)
- z − z̄ = 2i·Im(z) (a purely imaginary number)
- z·z̄ = a^2 + b^2 = |z|^2 (a nonnegative real number)
- (z + w)̄ = z̄ + w̄, and (z − w)̄ = z̄ − w̄
- (z·w)̄ = z̄·w̄
- (z/w)̄ = z̄ / w̄ for w ≠ 0
- If z ≠ 0 then 1/z̄ = (1/z)̄
Why conjugates are useful: Multiply numerator and denominator by the conjugate to rationalize complex denominators: (a + ib)/(c + id) = [(a + ib)(c − id)]/(c^2 + d^2). Conjugates appear in modulus computations, simplifying expressions and proving results about polynomials with real coefficients.
- If z = 3 + 4i then z̄ = 3 − 4i. Also z·z̄ = (3 + 4i)(3 − 4i) = 9 + 16 = 25 = |z|^2.
- For z = 5e^{iπ/6}, the conjugate is z̄ = 5e^{-iπ/6} (angle reversed).
- Solve quadratic x^2 − 2x + 5 = 0. Roots: x = [2 ± √(4 − 20)]/2 = 1 ± 2i. These roots are conjugates: 1 + 2i and 1 − 2i.
- Rationalize denominator: (2 + i)/(1 − 3i) = [(2 + i)(1 + 3i)]/(1 + 9) = (2 + 6i + i + 3i^2)/10 = (−1 + 7i)/10.
- Reflection example: point P(1, 2) representing z = 1 + 2i has conjugate P' = (1, −2) representing z̄ = 1 − 2i (reflection across real axis).
- \[If z = a + ib then z̄ = a − ib\]
- \[z + z̄ = 2·Re(z)\]
- \[z − z̄ = 2i·Im(z)\]
- \[z·z̄ = |z|^2 = a^2 + b^2\]
- \[(z̄)̄ = z\]
- \[(z + w)̄ = z̄ + w̄\]\[(z·w)̄ = z̄·w̄\]\[(z/w)̄ = z̄/w̄ (w ≠ 0)\]
Modulus and Argument
Modulus and Argument
Key Point: |z| = sqrt(x^2 + y^2)
Definition (Geometric view)
A complex number z = x + iy corresponds to the point (x,y) in the Argand plane. The modulus |z| is the distance from the origin to (x,y). The argument arg(z) is the directed angle θ measured from the positive real axis to the line joining the origin to (x,y).
Formulas and interpretation
- Modulus: |z| = sqrt(x² + y²). Equivalently |z| = sqrt(z·z̄) where z̄ is the conjugate.
- Argument: arg(z) = θ where tan θ = y/x, but θ must be chosen according to the quadrant of (x,y). The argument is multi-valued: arg(z) = θ + 2kπ (k ∈ ℤ). The principal value is Arg(z) ∈ (−π, π].
- Polar (trigonometric) form: z = r(cos θ + i sin θ) = r·cis θ, where r = |z| and θ = arg(z). Conversion to rectangular: x = r cos θ, y = r sin θ.
Important properties
- |zw| = |z|·|w| and arg(zw) = arg(z) + arg(w) (angles add, moduli multiply).
- |z/w| = |z|/|w| and arg(z/w) = arg(z) − arg(w) (division: subtract angles).
- Conjugation: |z̄| = |z| and arg(z̄) = −arg(z) (reflected across real axis).
- Triangle inequality: |z1 + z2| ≤ |z1| + |z2|.
How to compute argument in practice
- Compute θ0 = arctan(|y/x|).
- Decide the correct quadrant using signs of x and y and adjust θ0 to get θ in (−π, π] (use atan2(y,x) if available).
Why it matters
Modulus gives magnitude; argument gives direction. This polar viewpoint makes multiplication/division of complex numbers simple (scale + rotate), and is fundamental in trigonometry, roots of unity, phasors in physics and engineering.
- z = 3 + 4i. Modulus |z| = sqrt(3^2 + 4^2) = 5. Argument arg(z) = arctan(4/3) ≈ 53.13° (first quadrant). Polar form: 5(cos 53.13° + i sin 53.13°).
- z = -1 + √3 i. x = -1, y = √3 ≈ 1.732 so |z| = 2. tan θ = (√3)/(-1) → reference angle 60°, point is in quadrant II, so Arg(z) = 180° − 60° = 120°. Polar: 2 cis 120°.
- Multiply z1 = 3 + 4i (r1=5, θ1≈53.13°) and z2 = -1 + √3 i (r2=2, θ2=120°). Modulus of product = 5·2 = 10. Argument of product = 53.13° + 120° = 173.13°. So z1·z2 has modulus 10 and argument ≈173.13° (convert to rectangular if needed).
- Using conjugation: if z = a + ib has arg θ, then z̄ = a − ib has arg −θ (reflection across real axis). For example z = 1 + i (Arg 45°) ⇒ z̄ = 1 − i (Arg −45°).
- \[|z| = sqrt(x^2 + y^2)\]
- \[|z| = sqrt(z · z̄)\]
- \[arg(z) = θ where tan θ = y/x (choose θ by quadrant)\]\[principal Arg(z) ∈ (−π, π]\]
- \[z = r(cos θ + i sin θ) = r·cis θ where r = |z|, θ = arg(z)\]
- \[x = r cos θ\]\[y = r sin θ\]
- \[|zw| = |z|·|w|\]\[arg(zw) = arg(z) + arg(w)\]
Geometric Representation (Argand Plane)
Geometric Representation (Argand Plane)
Key Point: z = x + iy, where x = Re(z), y = Im(z)
Introduction: A complex number z = x + iy (x, y real; i^2 = −1) can be represented as a point (x, y) or as a position vector from the origin in a two‑dimensional plane called the Argand plane. The horizontal axis (real axis) represents the real part Re(z) and the vertical axis (imaginary axis) represents the imaginary part Im(z).
Plotting: To plot z = x + iy, mark the point with coordinates (x, y) and draw an arrow from the origin to this point. This arrow is often called the vector of z.
Modulus and Argument: The modulus (or absolute value) of z is the distance of the point from the origin:
|z| = sqrt(x^2 + y^2).
The argument arg(z) is the angle θ made by the vector with the positive real axis (measured counterclockwise). In Cartesian terms, θ = arctan(y/x), adjusted for the correct quadrant.
Polar (Trigonometric) Form: Using r = |z| and θ = arg(z),
z = x + iy = r(cos θ + i sin θ) = r e^{iθ}.
Geometric meaning of basic operations:
- Addition/Subtraction: z1 + z2 is the vector sum — place the tail of z2 at the head of z1 (or use the parallelogram rule).
- Conjugation: z̄ = x − iy is reflection of z across the real axis.
- Multiplication: If z1 = r1 e^{iθ1} and z2 = r2 e^{iθ2}, then z1 z2 = r1 r2 e^{i(θ1+θ2)}. Geometrically this scales by r2 and rotates by θ2 (or vice versa).
- Division: z1 / z2 = (r1 / r2) e^{i(θ1−θ2)} — scale and rotate in the opposite direction.
- Distance between points: |z1 − z2| gives the length of the line segment joining the points representing z1 and z2.
Loci: Many plane loci can be expressed using complex numbers. Examples useful in Class 11:
- |z − a| = r represents a circle centered at a with radius r.
- |z − a| = |z − b| represents the perpendicular bisector of segment AB (a and b are complex numbers as points).
- arg(z − a) = constant represents a half‑line (ray) from a making that constant angle with the positive real axis.
Why this is useful: The Argand plane connects algebraic operations on complex numbers with geometric transformations (translations, rotations, scalings, reflections). This is used in geometry, physics (phasors in AC circuits), signal/wave representation, and transformations in engineering.
- 1) Plot z = 3 + 4i. Find |z| and arg(z). Solution: Point (3,4). |z| = sqrt(3^2+4^2)=5. arg(z) = arctan(4/3) ≈ 53.13°.
- 2) Add z1 = 2 + i and z2 = −1 + 3i geometrically. Algebraically z1 + z2 = (2−1) + (1+3)i = 1 + 4i. Plotting shows the parallelogram result (point (1,4)).
- 3) Multiply z1 = 2(cos 30° + i sin 30°) and z2 = 3(cos 45° + i sin 45°). Then z1 z2 = 6[cos(75°) + i sin(75°)]: magnitude multiplied (2·3=6) and angles added (30°+45°=75°).
- 4) Locus: |z − (1 + i)| = 2 is the circle center (1,1) radius 2. All complex numbers z satisfying this lie on that circle.
- \[z = x + iy\]\[where x = Re(z)\]\[y = Im(z)\]
- \[|z| = sqrt(x^2 + y^2)\]
- \[arg(z) = θ\]\[where θ = arctan(y/x) (take care of quadrant)\]
- \[Polar form: z = r(cos θ + i sin θ) = r e^{iθ}\]\[with r = |z|, θ = arg(z)\]
- \[Conjugate: z̄ = x − iy (reflection about real axis)\]
- \[Product: z1 z2 = r1 r2 e^{i(θ1 + θ2)} (multiply moduli\]\[add arguments)\]
Polar Form and Exponential Form
Polar Form and Exponential Form
Key Point: z = x + iy
Overview
A complex number z = x + iy can be represented as a point or vector in the complex (Argand) plane. Polar form expresses z using its distance from the origin (modulus) and the angle it makes with the positive real axis (argument).
Conversion to polar form
- Modulus: r = |z| = sqrt(x^2 + y^2).
- Argument: θ = arg(z) = atan2(y, x). (Choose principal value Arg z typically in (−π, π] or [0, 2π).)
- Polar form: z = r(cos θ + i sin θ). Often written z = r cis θ.
Exponential form (Euler's formula)
Euler's identity: e^{iθ} = cos θ + i sin θ. Using this,
z = r(cos θ + i sin θ) = r e^{iθ}.
This is the exponential form and is very convenient for multiplication, division, powers and roots.
Key properties
- Multiplication: if z1 = r1 e^{iθ1} and z2 = r2 e^{iθ2}, then z1 z2 = r1 r2 e^{i(θ1+θ2)}. Geometrically: multiply lengths and add angles (scale + rotate).
- Division: z1 / z2 = (r1 / r2) e^{i(θ1−θ2)} (r2 ≠ 0).
- Conjugate: if z = r e^{iθ}, then z̄ = r e^{−iθ}.
- De Moivre's theorem (powers): (r cis θ)^n = r^n cis(nθ) or (r e^{iθ})^n = r^n e^{inθ} for integer n.
- n-th roots: Solutions of w^n = z (z = R e^{iΦ}) are w_k = R^{1/n} e^{i( (Φ + 2kπ)/n )}, k = 0,1,...,n−1. They are placed uniformly on a circle of radius R^{1/n}.
Multi-valued argument and principal value
Argument is multi-valued: arg(z) = θ + 2kπ. The principal value Arg z is the single value chosen in a standard interval (e.g. (−π, π]).
Why this matters (intuition)
Polar/exponential forms separate magnitude and direction. This makes repeated multiplication (rotations & scalings), solving equations like z^n = a, and understanding oscillatory behavior (waves, phasors) simple and geometric.
- Example 1 — Convert z = 1 + i to polar & exponential forms: r = sqrt(1^2 + 1^2) = sqrt(2). θ = atan2(1,1) = π/4. So z = sqrt(2)(cos π/4 + i sin π/4) = sqrt(2) e^{iπ/4}.
- Example 2 — Convert z = −1 + √3 i: r = sqrt(1 + 3) = 2. θ = atan2(√3, −1) = 120° = 2π/3. So z = 2 cis(2π/3) = 2 e^{i2π/3}.
- Example 3 — Multiply (2 + 2i) and (1 − i) using polar forms: 2+2i has r1 = 2√2, θ1 = π/4. 1−i has r2 = √2, θ2 = −π/4. Product modulus = (2√2)(√2) = 4. Angle = π/4 + (−π/4) = 0. So product = 4 e^{i0} = 4 (real). Check by rectangular multiplication: (2+2i)(1−i) = 4.
- Example 4 — Cube roots of 8: write 8 = 8 e^{i0}. Cube roots are 8^{1/3} e^{i(0 + 2kπ)/3} = 2 e^{i 2kπ/3}, k = 0,1,2. These are 2, 2 e^{i 2π/3} and 2 e^{i 4π/3} (vertices of an equilateral triangle on circle radius 2).
- Example 5 — n-th roots of unity (z^n = 1): roots are e^{i 2kπ/n}, k = 0,...,n−1. They lie on unit circle at equal angles and form a regular n-gon (useful in Fourier analysis and signal processing).
- \[z = x + iy\]
- \[r = |z| = sqrt(x^2 + y^2)\]
- \[arg(z) = θ = atan2(y\]\[x) (multi-valued: θ + 2kπ)\]
- \[Polar form: z = r(cos θ + i sin θ) = r cis θ\]
- \[Euler: e^{iθ} = cos θ + i sin θ\]
- \[Exponential form: z = r e^{iθ}\]
De Moivre's Theorem
De Moivre's Theorem
Key Point: (cos θ + i sin θ)^n = cos(nθ) + i sin(nθ), for integer n
Statement. For any real angle θ and any integer n,
(cos θ + i sin θ)n = cos(nθ) + i sin(nθ).
This is De Moivre's theorem. It gives a direct way to compute powers of complex numbers when they are written in trigonometric (polar) form. More generally, if z = r(cos θ + i sin θ) with r ≥ 0, then
zn = rn(cos(nθ) + i sin(nθ)).
Short proof (by induction for integer n ≥ 1). The base n=1 is trivial. Assume the formula holds for n. Then
(cos θ + i sin θ)n+1 = (cos nθ + i sin nθ)(cos θ + i sin θ)
= cos nθ cos θ − sin nθ sin θ + i(cos nθ sin θ + sin nθ cos θ) = cos((n+1)θ) + i sin((n+1)θ).
So it holds for n+1. For negative integers, apply the theorem to 1/z or use n → −n. Thus the theorem is true for all integers n.
Related results.
- Multiple-angle formulas: cos(nθ) = Re((cos θ + i sin θ)n), sin(nθ) = Im((cos θ + i sin θ)n).
- Power of a general complex number: z = r(cos θ + i sin θ) ⇒ zn = rn(cos nθ + i sin nθ).
- Formula for nth roots: The n distinct nth roots of z = r(cos θ + i sin θ) are r1/n(cos((θ+2kπ)/n) + i sin((θ+2kπ)/n)), k = 0,1,...,n−1.
Notes and limitations. In Class 11 we apply the theorem for integer n (positive, zero and negative). Extensions to rational or real exponents require complex logarithms and branch choices, which are beyond the basic scope.
- Example 1: Compute (1 + i)^5. Write 1 + i = √2(cos π/4 + i sin π/4). By De Moivre: (1 + i)^5 = (√2)^5(cos(5π/4) + i sin(5π/4)) = 4√2(−√2/2 − i√2/2) = −4 − 4i.
- Example 2: Using De Moivre to get triple-angle formulas. (cos θ + i sin θ)^3 = cos 3θ + i sin 3θ. Equating real and imaginary parts gives cos 3θ = 4 cos^3 θ − 3 cos θ and sin 3θ = 3 sin θ − 4 sin^3 θ.
- Example 3: Find the cube roots of 8. Write 8 = 8(cos 0 + i sin 0) so its cube roots are 8^(1/3)(cos((0 + 2kπ)/3) + i sin((0 + 2kπ)/3)) = 2(cos(2kπ/3) + i sin(2kπ/3)), k = 0,1,2. These are 2, −1 + i√3, −1 − i√3.
- \[(cos θ + i sin θ)^n = cos(nθ) + i sin(nθ)\]\[for integer n\]
- \[If z = r(cos θ + i sin θ) then z^n = r^n(cos(nθ) + i sin(nθ))\]
- \[nth roots of z = r(cos θ + i sin θ): z_k = r^{1/n}[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]\]\[k = 0,1,...,n−1\]
- \[cos(nθ) = Re((cos θ + i sin θ)^n)\]\[sin(nθ) = Im((cos θ + i sin θ)^n)\]
- \[For r = 1\]\[the nth roots of unity are cos(2kπ/n) + i sin(2kπ/n)\]\[k = 0,1,...,n−1\]
Roots of Complex Numbers and nth Roots of Unity
Roots of Complex Numbers and nth Roots of Unity
Key Point: Polar form: z = r(cos θ + i sin θ), r = |z|, θ = arg(z)
Overview
Every nonzero complex number z can be written in polar form z = r(cos θ + i sin θ) where r = |z| > 0 and θ = arg(z). An nth root of z is any complex number w such that w^n = z. By De Moivre's theorem the n distinct nth roots of z are obtained by taking the nth root of the modulus and dividing the argument into n equal parts (adding multiples of 2π to get all distinct solutions).
General formula
If z = r(cos θ + i sin θ) with r > 0, then the n nth roots of z are
w_k = r^{1/n} [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)], k = 0, 1, ..., n-1.
Equivalently using exponentials: w_k = r^{1/n} exp(i(θ + 2πk)/n).
Special case — nth roots of unity
When r = 1 and z = 1, the nth roots of z are the nth roots of unity:
ω_k = cos(2πk/n) + i sin(2πk/n) = exp(2πik/n), k = 0, 1, ..., n-1.
These are n distinct points on the unit circle, equally spaced by angle 2π/n and forming the vertices of a regular n-gon. Important properties include:
- Sum of all nth roots of unity = 0 (for n > 1).
- They are the roots of the polynomial x^n − 1 = 0.
- Exactly φ(n) of them are primitive (where φ is Euler's totient function): a primitive nth root is one with multiplicative order n (not some smaller divisor).
Geometric interpretation
Each root w_k has modulus r^{1/n} and argument (θ + 2πk)/n. Geometrically the n roots of z are points lying on the circle centered at origin with radius r^{1/n}, placed at equal angular separations of 2π/n. For nth roots of unity the radius is 1, so the points are on the unit circle and form a regular polygon.
Notes and special cases
- If z = 0 then the only nth root is 0.
- De Moivre's theorem: [r(cos θ + i sin θ)]^n = r^n(cos nθ + i sin nθ) — used to derive the root formula.
- Principal nth root: one typically takes k = 0 (argument divided by n and principal branch of argument) to get the principal root; the other roots differ by multiplication by nth roots of unity.
- Example 1 — Cube roots of 8: Write 8 = 8(cos 0 + i sin 0). Here r = 8, r^{1/3} = 2, θ = 0. The cube roots are 2[cos(2πk/3) + i sin(2πk/3)] for k = 0,1,2. So the roots are: 2, -1 + i√3, -1 - i√3.
- Example 2 — Fourth roots of -16: Write -16 = 16(cos π + i sin π). r^{1/4} = 2. Roots are 2[cos((π+2πk)/4) + i sin((π+2πk)/4)], k = 0..3. Explicitly: √2 + i√2, -√2 + i√2, -√2 - i√2, √2 - i√2.
- Example 3 — Cube roots of unity: Solve w^3 = 1. Roots are 1, cos(2π/3)+i sin(2π/3) = -1/2 + i√3/2, and cos(4π/3)+i sin(4π/3) = -1/2 - i√3/2. These are vertices of an equilateral triangle on the unit circle.
- \[Polar form: z = r(cos θ + i sin θ)\]\[r = |z|, θ = arg(z)\]
- \[De Moivre's theorem: (r(cos θ + i sin θ))^n = r^n(cos nθ + i sin nθ)\]
- \[General nth roots: w_k = r^{1/n}[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)]\]\[k = 0,1,...,n-1\]
- \[Exponential form: w_k = r^{1/n} exp(i(θ + 2πk)/n)\]
- \[nth roots of unity: ω_k = cos(2πk/n) + i sin(2πk/n) = exp(2πik/n)\]
- \[Sum of nth roots of unity (n>1): Σ_{k=0}^{n-1} ω_k = 0\]
Quadratic Equations — Basics
Quadratic Equations — Basics
Key Point: General form: ax^2 + bx + c = 0, a ≠ 0
What is a quadratic equation?
A quadratic equation in one variable x is a polynomial equation of degree two. Its general form is ax^2 + bx + c = 0, where a, b, c are real (or complex) numbers and a ≠ 0. The graph of y = ax^2 + bx + c is a parabola.
Key features
- Coefficient a determines opening and width: if a > 0 the parabola opens upward; if a < 0 it opens downward. Larger |a| makes it narrower.
- Vertex: the highest or lowest point of the parabola. Its x-coordinate is x = -b/(2a). The vertex coordinates are ( -b/(2a), -D/(4a) ), where D is the discriminant.
- Axis of symmetry: the vertical line x = -b/(2a) that passes through the vertex.
- Roots (zeros or solutions): values of x for which ax^2 + bx + c = 0. They correspond to x-intercepts of the parabola.
How to solve
- Factorization: write ax^2 + bx + c = (px + q)(rx + s) and solve for x when factors equal zero (works when factors exist with rational coefficients).
- Completing the square: rewrite ax^2 + bx + c in vertex form a(x + b/(2a))^2 + constant and solve.
- Quadratic formula: derived by completing the square, gives roots directly: x = [-b ± sqrt(b^2 - 4ac)]/(2a).
Discriminant and nature of roots
Discriminant D = b^2 - 4ac determines the nature of roots:
- D > 0: two distinct real roots (parabola cuts x-axis at two points).
- D = 0: one real repeated root (parabola is tangent to x-axis at vertex).
- D < 0: two complex conjugate roots (parabola does not meet the real x-axis).
Relations between roots and coefficients
If the roots are α and β, then α + β = -b/a and αβ = c/a. These are useful for writing quadratic equations with given sum and product of roots and for many problem-solving techniques.
Behaviour at infinity
As x → ±∞, the ax^2 term dominates so y → +∞ if a > 0 and y → -∞ if a < 0.
Connections to complex numbers
When D < 0 the solutions are complex and appear as conjugate pairs α ± iβ. This ties quadratic equations to the study of complex numbers in the same chapter.
- Solve x^2 - 5x + 6 = 0 by factorization. Factors: (x - 2)(x - 3) = 0 → roots x = 2, 3.
- Solve 2x^2 + 3x - 2 = 0 using quadratic formula. D = 3^2 - 4·2·(-2) = 9 + 16 = 25. Roots: x = [-3 ± 5]/(4) → x = 1/2 or x = -2.
- Projectile motion: height h(t) = -5t^2 + 20t + 1 (t in seconds). Vertex at t = -b/(2a) = -20/(2·-5) = 2 s gives time of maximum height. Roots of h(t) = 0 give times when projectile hits the ground.
- Optimization / area: Given a fixed perimeter, area A of a rectangle of sides x and (P/2 - x) is A(x) = x(P/2 - x) = -x^2 + (P/2)x, a quadratic in x. Vertex gives maximum area.
- Kinetic energy relation: KE = 1/2 m v^2 is quadratic in speed v. For m = 2 kg, KE(v) = v^2 J; graph of KE vs v is a parabola opening upward.
- \[General form: ax^2 + bx + c = 0\]\[a ≠ 0\]
- \[Quadratic formula (roots): x = [-b ± sqrt(b^2 - 4ac)]/(2a)\]
- \[Discriminant: D = b^2 - 4ac\]\[Nature of roots depends on D (D>\]\[0 two real\]\[D=0 one real repeated\]\[D<\]\[0 two complex)\]
- \[Sum of roots: α + β = -b/a\]
- \[Product of roots: αβ = c/a\]
- \[Axis of symmetry: x = -b/(2a)\]
Discriminant and Nature of Roots
Discriminant and Nature of Roots
Key Point: Discriminant: Δ = b² − 4ac
What is the discriminant?
For a quadratic equation ax² + bx + c = 0 (a ≠ 0), the discriminant Δ is defined as Δ = b² − 4ac. The discriminant determines the nature (type and number) of the roots without solving the equation.
Rules relating Δ to the nature of roots
- Δ > 0: two distinct real roots. If Δ is a perfect square (and a, b, c rational/integer), the roots are rational; otherwise they are irrational.
- Δ = 0: exactly one real root (a repeated or double root). Both roots are equal: −b/(2a).
- Δ < 0: two complex (non-real) conjugate roots. They are of the form (−b ± i√|Δ|)/(2a).
Geometric interpretation
Consider the graph y = ax² + bx + c, a parabola.
- Δ > 0: the parabola intersects the x-axis at two distinct points (two real zeros).
- Δ = 0: the parabola is tangent to the x-axis at the vertex (one real, repeated root).
- Δ < 0: the parabola does not meet the x-axis (no real zeros), it lies entirely above or below the x-axis depending on the sign of a.
Useful related results
- Roots (quadratic formula): x = [−b ± √Δ]/(2a).
- Sum and product of roots (if roots are α and β): α + β = −b/a, αβ = c/a.
- Vertex coordinates: x_vertex = −b/(2a), y_vertex = f(−b/(2a)) = −Δ/(4a).
How to decide quickly
- Compute Δ = b² − 4ac.
- If Δ > 0 → two real and distinct roots. Check if Δ is a perfect square for rational roots.
- If Δ = 0 → one real repeated root at x = −b/(2a).
- If Δ < 0 → two complex conjugate roots; no real x-intercepts.
- Example 1 (distinct real roots): x² − 3x + 2 = 0. Δ = (−3)² − 4·1·2 = 9 − 8 = 1 > 0. Roots: (3 ± 1)/2 ⇒ 2 and 1. Parabola crosses x-axis at x = 1 and x = 2.
- Example 2 (repeated root): x² − 2x + 1 = 0. Δ = (−2)² − 4·1·1 = 4 − 4 = 0. Single repeated root: x = 1 (parabola tangent to x-axis at vertex).
- Example 3 (complex roots): x² + x + 1 = 0. Δ = 1 − 4 = −3 < 0. Roots: (−1 ± i√3)/2 (complex conjugates). No real intersections with x-axis.
- Real-life example — projectile height: If height s(t) = −5t² + 20t + 1 (m) and you ask when s(t) = h0 (a given height), you solve −5t² + 20t + (1 − h0) = 0. The discriminant shows whether the projectile reaches h0 twice (Δ > 0), exactly once at peak (Δ = 0), or never (Δ < 0).
- Real-life example — break-even analysis: Profit P(x) modeled as ax² + bx + c. Solving P(x) = 0 gives break-even quantities. Δ > 0 means two break-even points (two production levels with zero profit), Δ = 0 one break-even level, Δ < 0 no real break-even.
- \[Discriminant: Δ = b² − 4ac\]
- \[Quadratic formula: x = [−b ± √Δ] / (2a)\]
- \[Sum of roots: α + β = −b/a\]
- \[Product of roots: αβ = c/a\]
- \[Vertex: x_v = −b/(2a)\]\[y_v = −Δ/(4a)\]
- \[Conditions: Δ >\]\[0 ⇒ two distinct real roots\]\[Δ = 0 ⇒ one real (double) root\]\[Δ <\]\[0 ⇒ two complex conjugate roots\]
Relations Between Roots and Coefficients
Relations Between Roots and Coefficients
Key Point: For ax² + bx + c = 0 with roots α, β: α + β = -b/a, αβ = c/a.
Overview. For a quadratic equation ax² + bx + c = 0 (a ≠ 0) with roots α and β, the coefficients are directly related to the sum and product of the roots. These relations are known as Viète's formulas for quadratics and follow from factoring the polynomial:
ax² + bx + c = a[x - α][x - β] = ax² - a(α + β)x + aαβ.
Comparing coefficients gives the fundamental relations:
- Sum of roots: α + β = -b/a
- Product of roots: αβ = c/a
Discriminant and nature of roots. The discriminant Δ = b² - 4ac determines the nature of the roots:
- Δ > 0: two distinct real roots
- Δ = 0: one repeated real root (α = β)
- Δ < 0: two complex conjugate roots (if a, b, c are real): α = p + iq, β = p - iq
For complex conjugate roots p ± iq, the sum and product give p and q: p = (α + β)/2 = -b/(2a) and p² + q² = αβ = c/a, so q = sqrt(p² - c/a) (with sign depending on orientation).
Useful transformations. If α and β are roots of ax² + bx + c = 0, then:
- Quadratic with roots 1/α and 1/β: c x² + b x + a = 0 (reverse coefficients).
- Quadratic with roots -α and -β: a x² - b x + c = 0 (change the sign of b).
- Quadratic with roots α + k and β + k: x² - (α + β + 2k)x + (αβ + k(α + β) + k²) = 0. Using α + β = -b/a and αβ = c/a, substitute to form the new equation.
Vertex and axis of symmetry. The axis of symmetry is x = -b/(2a) = (α + β)/2. The y-coordinate of the vertex is f( -b/(2a) ) = (4ac - b²) / (4a) = -Δ/(4a).
How to form a quadratic from given roots. If roots are r1 and r2, the monic quadratic is x² - (r1 + r2)x + r1 r2 = 0. Multiply by any nonzero constant to get a non-monic quadratic.
- Example 1: For 2x² - 3x + 5 = 0, sum = -b/a = 3/2, product = c/a = 5/2. Δ = (-3)² - 4·2·5 = 9 - 40 = -31 < 0 so roots are complex: 3/4 ± (√31 / 4) i.
- Example 2: Given roots 2 + i and 2 - i. Sum = 4, product = (2 + i)(2 - i) = 4 + 1 = 5. Quadratic: x² - 4x + 5 = 0.
- Example 3 (reciprocal roots): If 3x² + 5x - 2 = 0 has roots α, β, then the quadratic with roots 1/α and 1/β is -2 x² + 5x + 3 = 0 (or multiplied by -1 → 2x² - 5x - 3 = 0).
- Example 4 (shifted roots): If α, β are roots of x² - 7x + 10 = 0 (so α + β = 7, αβ = 10), then quadratic with roots α + 2 and β + 2 has sum 7 + 4 = 11 and product 10 + 2·7 + 4 = 28, so x² - 11x + 28 = 0.
- Real-life example: In projectile motion modeled by h(t) = -4.9t² + v₀t + h₀, the times when the projectile is at ground level are the roots t1, t2. Their sum t1 + t2 = -b/a = -v₀/(-4.9) = v₀/4.9 and product t1 t2 = c/a = h₀/(-4.9) (useful to relate launch speed, flight time and initial height).
- Applied example (economics): If profit P(x)=ax²+bx+c is zero at months x1 and x2 (break-even points), then x1 + x2 = -b/a and x1 x2 = c/a. Knowledge of one root and coefficients gives the other root directly.
- \[For ax² + bx + c = 0 with roots α, β: α + β = -b/a, αβ = c/a.\]
- \[Discriminant: Δ = b² - 4ac\]\[Nature: Δ>0 (two real), Δ=0 (equal), Δ<0 (complex conjugates).\]
- \[Vertex x-coordinate: x_v = -b/(2a) = (α + β)/2\]\[Vertex y-coordinate: y_v = -Δ/(4a) = (4ac - b²)/(4a).\]
- \[Quadratic with roots r1\]\[r2: k(x - r1)(x - r2) = 0 → kx² - k(r1 + r2)x + k r1 r2 = 0.\]
- \[Reciprocal roots: if α, β roots of ax² + bx + c = 0\]\[then roots 1/α, 1/β satisfy c x² + b x + a = 0.\]
- \[Shifted roots α + k, β + k: new sum = (α + β) + 2k = -b/a + 2k\]\[new product = αβ + k(α + β) + k² = c/a + k(-b/a) + k².\]
Quadratic Equations with Complex Roots and Conjugate Root Theorem
Quadratic Equations with Complex Roots and Conjugate Root Theorem
Key Point: Quadratic formula: x = (−b ± √(b² − 4ac)) / (2a)
Overview
A quadratic equation ax² + bx + c = 0 (a ≠ 0) with real coefficients can have three types of roots depending on the discriminant Δ = b² − 4ac:
- Δ > 0: two distinct real roots
- Δ = 0: one repeated real root
- Δ < 0: two complex (non-real) roots which are conjugates of each other
Quadratic formula and complex roots
The roots are given by the quadratic formula
x = (−b ± √(b² − 4ac)) / (2a).
When Δ < 0, write √(b² − 4ac) = i√(4ac − b²). Thus the roots become
x = −b/(2a) ± i·(√(4ac − b²) / (2a)).
Conjugate Root Theorem (for real-coefficient polynomials)
If a polynomial has all real coefficients and a complex number z = p + iq (q ≠ 0) is a root, then its complex conjugate z̄ = p − iq is also a root. Proof sketch: if P(z) = 0 and coefficients are real, taking complex conjugates of both sides yields P(z̄) = 0.
Properties for a quadratic with complex roots
- Sum of roots = α + β = −b/a (real)
- Product of roots = αβ = c/a (real)
- If α = p + iq and β = p − iq, then p = −b/(2a) and q = √(|Δ|)/(2a)
- Modulus of each root: |α| = √(p² + q²)
Interpretation
For Δ < 0 the parabola y = ax² + bx + c does not intersect the real x-axis. The real part of the complex roots equals the x-coordinate of the axis of symmetry and vertex (−b/(2a)). The imaginary parts measure how far 'off' the real axis the complex roots lie (± value).
Connection to differential equations and physical systems
Quadratic equations appear as characteristic equations of second-order linear systems (e.g., mechanical oscillators, electrical RLC circuits). Complex conjugate roots lead to oscillatory responses whose amplitude may decay or grow depending on the real part (damped/unstable oscillations).
Summary steps to handle a quadratic with complex roots
- Compute Δ = b² − 4ac. If Δ < 0, proceed.
- Compute real part: Re(x) = −b/(2a).
- Compute imaginary magnitude: Im(x) = √(|Δ|)/(2a). Roots = Re(x) ± i·Im(x).
- Use conjugate root theorem: roots come in pairs p ± iq.
- Example 1 — Pure algebraic: Solve x² + 2x + 5 = 0. Here a=1, b=2, c=5. Discriminant Δ = 4 − 20 = −16 < 0. Roots: x = (−2 ± √(−16))/2 = −1 ± 2i. Sum = (−1+2i) + (−1−2i) = −2 = −b/a. Product = (−1+2i)(−1−2i) = 1 + 4 = 5 = c/a.
- Example 2 — RLC circuit / damped oscillator: Characteristic equation s² + 4s + 13 = 0 (a=1,b=4,c=13). Δ = 16 − 52 = −36. Roots: s = (−4 ± √(−36))/2 = −2 ± 3i. Time-domain homogeneous response (for e.g. current or displacement) is C·e^{(−2+3i)t} + D·e^{(−2−3i)t} = e^{−2t}[A cos(3t) + B sin(3t)], a damped oscillation with decay rate 2 and angular frequency 3.
- Example 3 — Using conjugate root theorem: If a quadratic with real coefficients has one root 3 + 4i, the other must be 3 − 4i. The quadratic is (x − (3+4i))(x − (3−4i)) = x² − 6x + 25 (coefficients real: a=1, b=−6, c=25).
- \[Quadratic formula: x = (−b ± √(b² − 4ac)) / (2a)\]
- \[Discriminant: Δ = b² − 4ac (nature of roots depends on sign of Δ)\]
- \[If Δ <\]\[complex roots: x = −b/(2a) ± i·(√(|Δ|) / (2a))\]
- \[Sum of roots: α + β = −b/a (real for real coefficients)\]
- \[Product of roots: αβ = c/a (real for real coefficients)\]
- \[Conjugate Root Theorem: If coefficients are real and z is a non-real root\]\[then z̄ is also a root\]
Applications and Problem Types
Applications and Problem Types
Key Point: Quadratic formula: x = [−b ± sqrt(b^2 − 4ac)]/(2a)
Overview: This topic connects quadratic equations and complex numbers. It covers how complex roots arise for quadratic equations, how to work with them using algebraic and geometric (Argand plane) methods, and typical problem types: finding/classifying roots, forming quadratics from given roots, solving parameter problems, and locus/geometric problems in the complex plane.
Core ideas and strategies:
- Discriminant and nature of roots: For ax² + bx + c = 0, Δ = b² − 4ac. If Δ > 0: two distinct real roots; Δ = 0: real and equal roots; Δ < 0: two complex conjugate roots (when coefficients are real).
- Complex conjugate root theorem: If coefficients are real and α = p + iq is a root, then ᾱ = p − iq is also a root. Use this to form quadratics with given complex roots.
- Vieta’s relations: If roots are α and β, then α + β = −b/a and αβ = c/a. These let you form or identify quadratics from sums/products of roots, including complex roots.
- Quadratic formula (works for complex roots): x = [−b ± sqrt(Δ)]/(2a). For Δ < 0, write sqrt(Δ) = i·sqrt(4ac − b²).
- Algebraic method for a complex root: write z = x + iy, separate real and imaginary parts, and solve the resulting system. Useful when equation or condition is given in complex form.
- Polar/Exponential form: Express z = r(cosθ + i sinθ) = re^{iθ} for multiplicative or argument problems (roots of unity, powers, finding modulus/argument).
- Locus problems: Translate geometric conditions (distance, angle, perpendicular bisector) into equations in x and y, or complex form |z − z0| = r, |z − z1| = |z − z2|, arg((z − a)/(z − b)) = constant, etc.
Typical problem types and short solution outlines:
- Find roots of a quadratic: compute Δ, use quadratic formula; if Δ < 0, express roots as conjugates.
- Form quadratic from given roots: Use Vieta. For given α and β, polynomial k[x² − (α + β)x + αβ]. If α is complex, include its conjugate if coefficients required to be real.
- Parameter problems: find parameter values so roots are real/equal/complex (use sign of Δ) or satisfy additional conditions (e.g., product = 1 → set αβ = 1).
- Geometric interpretation / locus: Convert conditions to |z − z0|, equate squares to avoid roots, or convert to x,y and get circle/line/perpendicular bisector.
- Transformations of roots: Given a quadratic with roots α, β, find polynomial with roots f(α), f(β) by expressing symmetric sums in terms of α+β and αβ.
Tips for solving exam problems:
- Always check whether coefficients are real — then expect conjugate pairs when roots are non-real.
- Use Vieta’s relations to avoid solving for roots explicitly when only symmetric expressions are needed.
- For locus problems, square modulus expressions to convert to algebraic equations; watch for extraneous solutions when squaring.
- Sketch Argand diagrams for geometric insight: conjugates are symmetric about the real axis; equal-modulus loci are circles centered at the origin; equal-distance loci are perpendicular bisectors.
- 1) Find roots of x^2 + 2x + 5 = 0. Solution: Δ = 4 − 20 = −16 → roots = (−2 ± 4i)/2 = −1 ± 2i.
- 2) Form a quadratic with real coefficients whose roots are 2 + 3i and 2 − 3i. Solution: Sum = 4, Product = (2+3i)(2−3i) = 4+9 = 13 → x^2 − 4x + 13 = 0.
- 3) If α and β are roots of x^2 + px + q = 0 and αβ = 1, find relation between p and q. Using Vieta: q = αβ = 1, so q = 1 (and p = −(α+β) can be any value consistent with roots).
- 4) Parameter problem: For which m does x^2 − 4x + m = 0 have real roots? Δ = 16 − 4m ≥ 0 → m ≤ 4.
- 5) Locus problem: Describe points z satisfying |z − (1 + i)| = 2. Answer: Circle in Argand plane centered at (1,1) with radius 2. Algebraic form: (x − 1)^2 + (y − 1)^2 = 4.
- 6) Transform roots: Given roots α, β of x^2 − 3x + 2 = 0 (so α + β = 3, αβ = 2), find quadratic with roots 1/α and 1/β. Sum = (α+β)/αβ = 3/2, product = 1/2 → x^2 − (3/2)x + 1/2 = 0 (or 2x^2 − 3x + 1 = 0).
- \[Quadratic formula: x = [−b ± sqrt(b^2 − 4ac)]/(2a)\]
- \[Discriminant: Δ = b^2 − 4ac (sign determines nature of roots)\]
- \[Vieta: If roots are α, β of ax^2 + bx + c = 0 then α + β = −b/a, αβ = c/a\]
- \[Conjugate roots: For real coefficients\]\[nonreal root α implies ᾱ is also a root\]
- \[Complex number (rectangular): z = x + iy, |z| = sqrt(x^2 + y^2)\]\[arg(z) = arctan(y/x)\]
- \[Polar/exponential form: z = r(cosθ + i sinθ) = re^{iθ}\]
Key Concepts
- Complex number
- A number of the form a + bi where a and b are real numbers and i is the imaginary unit with i^2 = -1.
- Imaginary unit
- Denoted by i, it satisfies i^2 = -1 and generates imaginary numbers.
- Real part
- For z = a + ib, the real part Re(z) is the real number a.
- Imaginary part
- For z = a + ib, the imaginary part Im(z) is the real number b.
- Conjugate
- For z = a + ib, the conjugate is ̅z = a - ib; z̅z = |z|^2 (a^2 + b^2).
- Modulus
- The modulus (or absolute value) of z = a + ib is |z| = sqrt(a^2 + b^2), the distance from origin.
- Argument (arg)
- The angle θ (in radians) between the positive real axis and the line representing z; principal argument Arg(z) is usually taken in (-π, π].
- Polar (trigonometric) form
- Expressing z as z = r(cos θ + i sin θ) where r = |z| and θ = arg z.
- Euler's formula
- Relates exponential and trigonometric forms: e^{iθ} = cos θ + i sin θ, useful for complex exponentials.
- cis notation
- Short form cis θ = cos θ + i sin θ, so z = r cis θ is the polar form.
- Equality of complex numbers
- Two complex numbers a + ib and c + id are equal iff a = c and b = d (equality of real and imaginary parts).
- Algebra of complex numbers
- Addition/subtraction performed component-wise; multiplication uses distributive law with i^2 = -1; division uses conjugate to rationalize denominator.
- Division of complex numbers
- To divide z1 by z2, multiply numerator and denominator by the conjugate of z2: z1/z2 = (z1̅z2)/|z2|^2.
- Triangle inequality
- For any complex z and w: |z + w| ≤ |z| + |w| and ||z| - |w|| ≤ |z - w|.
- Quadratic equation
- A polynomial equation of degree two: ax^2 + bx + c = 0 with a ≠ 0; has two roots (possibly equal or complex).
- Discriminant
- Δ = b^2 - 4ac; it determines the nature of roots of ax^2 + bx + c = 0.
- Nature of roots
- If Δ > 0 → two distinct real roots; Δ = 0 → real and equal roots; Δ < 0 → two non-real complex conjugate roots.
- Sum and product of roots (Vieta's relations)
- If α and β are roots of ax^2 + bx + c = 0 then α + β = -b/a and αβ = c/a.
- Quadratic formula
- Roots of ax^2 + bx + c = 0 are x = (-b ± √Δ) / (2a), where Δ = b^2 - 4ac.
- Conjugate roots theorem
- If a polynomial with real coefficients has a non-real complex root z, then its complex conjugate ̅z is also a root; applies to quadratic equations with real coefficients.
Practice Questions
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Define a complex number and state what is meant by its real and imaginary parts. / सम्मिश्र संख्या को परिभाषित कीजिए और इसके वास्तविक एवं काल्पनिक भागों से क्या तात्पर्य है, बताइए।
Show answer
A complex number is a number of the form z = a + bi, where a, b are real and i satisfies i² = −1; here a = Re(z) is the real part and b = Im(z) is the imaginary part. / सम्मिश्र संख्या z = a + bi रूप की होती है, जहाँ a, b वास्तविक हैं और i² = −1 है; यहाँ a = Re(z) वास्तविक भाग और b = Im(z) काल्पनिक भाग है।
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Express (3 + 2i)/(1 − i) in the form a + bi. / (3 + 2i)/(1 − i) को a + bi रूप में व्यक्त कीजिए।
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Multiply numerator and denominator by the conjugate (1 + i): [(3+2i)(1+i)]/(1²+1²) = (3+3i+2i+2i²)/2 = (1+5i)/2 = 0.5 + 2.5i. / अंश और हर को संयुग्मी (1 + i) से गुणा करें: [(3+2i)(1+i)]/(1²+1²) = (3+3i+2i+2i²)/2 = (1+5i)/2 = 0.5 + 2.5i।
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Find the modulus and principal argument of z = −1 + √3 i, and write it in polar form. / z = −1 + √3 i का मापांक और मुख्य कोणांक ज्ञात कीजिए, तथा इसे ध्रुवीय रूप में लिखिए।
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|z| = √((−1)² + (√3)²) = √4 = 2; the point lies in quadrant II so Arg(z) = 180° − 60° = 120° (2π/3); polar form z = 2(cos 120° + i sin 120°). / |z| = √((−1)² + (√3)²) = √4 = 2; बिंदु द्वितीय चतुर्थांश में है इसलिए Arg(z) = 180° − 60° = 120° (2π/3); ध्रुवीय रूप z = 2(cos 120° + i sin 120°)।
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Prove that for any complex number z, z·z̄ = |z|² and hence state why z·z̄ is always real and non-negative. / सिद्ध कीजिए कि किसी भी सम्मिश्र संख्या z के लिए z·z̄ = |z|² है और बताइए कि z·z̄ सदैव वास्तविक एवं अऋणात्मक क्यों होता है।
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If z = a + bi then z̄ = a − bi, so z·z̄ = (a+bi)(a−bi) = a² − (bi)² = a² + b² = |z|²; since a² + b² is a sum of real squares, it is real and ≥ 0. / यदि z = a + bi तो z̄ = a − bi, अतः z·z̄ = (a+bi)(a−bi) = a² − (bi)² = a² + b² = |z|²; क्योंकि a² + b² वास्तविक वर्गों का योग है, यह वास्तविक एवं ≥ 0 होता है।
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Using the discriminant, determine the nature of the roots of x² + x + 1 = 0 and find them. / विविक्तकर का प्रयोग करके x² + x + 1 = 0 के मूलों की प्रकृति ज्ञात कीजिए और मूल निकालिए।
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Δ = b² − 4ac = 1 − 4 = −3 < 0, so the roots are complex conjugates; x = (−1 ± i√3)/2. / Δ = b² − 4ac = 1 − 4 = −3 < 0, अतः मूल सम्मिश्र संयुग्मी हैं; x = (−1 ± i√3)/2।
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If a quadratic equation with real coefficients has 2 + i as one root, write the equation. / यदि वास्तविक गुणांकों वाले द्विघात समीकरण का एक मूल 2 + i है, तो समीकरण लिखिए।
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Real coefficients force the other root to be the conjugate 2 − i; sum = 4, product = (2+i)(2−i) = 5, so x² − 4x + 5 = 0. / वास्तविक गुणांक होने से दूसरा मूल संयुग्मी 2 − i होगा; योग = 4, गुणनफल = (2+i)(2−i) = 5, अतः x² − 4x + 5 = 0।
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Apply De Moivre's theorem to compute (1 + i)⁵. / De Moivre प्रमेय का प्रयोग करके (1 + i)⁵ की गणना कीजिए।
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Write 1 + i = √2(cos 45° + i sin 45°); then (1+i)⁵ = (√2)⁵(cos 225° + i sin 225°) = 4√2(−√2/2 − i√2/2) = −4 − 4i. / 1 + i = √2(cos 45° + i sin 45°) लिखें; तब (1+i)⁵ = (√2)⁵(cos 225° + i sin 225°) = 4√2(−√2/2 − i√2/2) = −4 − 4i।
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If α and β are roots of 2x² − 3x + 5 = 0, find α + β and αβ without solving the equation. / यदि α और β समीकरण 2x² − 3x + 5 = 0 के मूल हैं, तो समीकरण हल किए बिना α + β और αβ ज्ञात कीजिए।
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By relations between roots and coefficients, α + β = −b/a = 3/2 and αβ = c/a = 5/2. / मूल और गुणांकों के संबंधों द्वारा, α + β = −b/a = 3/2 और αβ = c/a = 5/2।
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