Complex Numbers And Quadratic Equations Class 11 Chapter Notes

Key Definitions

Complex Number

A number of the form z = a + ib, where a and b are real numbers, and i = √(-1) is the imaginary unit. a is the real part (Re(z)) and b is the imaginary part (Im(z)).

Imaginary Unit (i)

Defined as i = √(-1), implying i² = -1. It allows the square root of negative numbers to be expressed.

Conjugate of a Complex Number

For z = a + ib, its conjugate, denoted as z̄ or z*, is a - ib. Geometrically, it's the reflection of z across the real axis.

Modulus of a Complex Number

For z = a + ib, the modulus, denoted |z|, is √(a² + b²). It represents the distance of the point (a, b) from the origin in the Argand plane.

Argument (or Amplitude) of a Complex Number

For z = a + ib, it's the angle θ that the line segment from the origin to (a, b) makes with the positive real axis. The principal argument Arg(z) lies in the interval (-π, π].

Argand Plane (Complex Plane)

A coordinate plane used to represent complex numbers graphically. The horizontal axis represents the real part (Re(z)), and the vertical axis represents the imaginary part (Im(z)).

Purely Real/Imaginary

A complex number z = a + ib is purely real if b = 0 (i.e., z = a). It is purely imaginary if a = 0 (i.e., z = ib, b ≠ 0).

Understanding Complex Numbers: The Extension of Real Numbers

Historically, mathematicians encountered equations like x² + 1 = 0 which had no solutions within the realm of real numbers. To address this, the concept of an imaginary unit, denoted by 'i', was introduced. By definition, i = √(-1), which directly implies that i² = -1. This seemingly simple definition opened up an entirely new system of numbers.

A complex number z is then defined as any number that can be expressed in the form a + ib, where a and b are real numbers, and i is the imaginary unit. Here, a is called the real part of z (written as Re(z) = a), and b is called the imaginary part of z (written as Im(z) = b). Every real number can be considered a complex number with its imaginary part equal to zero (e.g., 5 = 5 + 0i).

The powers of i follow a cyclical pattern, which is crucial for simplifying expressions:

• i¹ = i
• i² = -1
• i³ = i² i = -1 i = -i
• i⁴ = i² i² = (-1) (-1) = 1

This cycle repeats every four powers. So, i⁵ = i, i⁶ = -1, and so on. In general, for any integer k, i^(4k) = 1, i^(4k+1) = i, i^(4k+2) = -1, and i^(4k+3) = -i. Understanding this cycle is vital for simplifying higher powers of i quickly and accurately. The introduction of complex numbers allows us to find solutions for all quadratic equations, even when the discriminant is negative.

Key Formulas and Properties (Must Remember)

• Powers of i: i¹=i, i²=-1, i³=-i, i⁴=1. Generally, i^(4n+k) = i^k.
• Algebra of Complex Numbers: If z₁ = a + ib and z₂ = c + id: Addition: z₁ + z₂ = (a+c) + i(b+d) Subtraction: z₁ - z₂ = (a-c) + i(b-d) Multiplication: z₁z₂ = (ac-bd) + i(ad+bc) Division: z₁/z₂ = (z₁z₂̄) / |z₂|² = [(a+ib)(c-id)] / (c²+d²) = [(ac+bd) + i(bc-ad)] / (c²+d²)
• Modulus Properties: |z| = √(Re(z)² + Im(z)²). Also, |z₁z₂| = |z₁||z₂|, |z₁/z₂| = |z₁|/|z₂|, |zⁿ| = |z|ⁿ, |z̄| = |z|, z z̄ = |z|².
• Conjugate Properties: z̄ = Re(z) - i Im(z). Also, (z₁+z₂)̄ = z̄₁+z̄₂, (z₁-z₂)̄ = z̄₁-z̄₂, (z₁z₂)̄ = z̄₁z̄₂, (z₁/z₂)̄ = z̄₁/z̄₂.
• Multiplicative Inverse: For z ≠ 0, z⁻¹ = 1/z = z̄ / |z|².
• Polar Form (Trigonometric Form): A complex number z = x + iy can be written as z = r(cosθ + i sinθ), where r = |z| = √(x² + y²) and θ is the argument such that cosθ = x/r and sinθ = y/r.
• Principal Argument: The value of θ such that -π < θ ≤ π. Quadrant rules: Q1 (x>0, y>0): θ = α Q2 (x<0, y>0): θ = π - α Q3 (x<0, y<0): θ = - (π - α) or α - π Q4 (x>0, y<0): θ = -α where α = tan⁻¹(|y/x|) is the reference angle.
• Quadratic Equations: For ax² + bx + c = 0 (a≠0), if the discriminant D = b² - 4ac < 0, then the roots are complex conjugates given by x = [-b ± i√(4ac - b²)] / 2a.

Converting a Complex Number to Polar Form

1. — Find r = |z| = √(x² + y²). This is the distance of the complex number from the origin on the Argand plane.
2. — Calculate α = tan⁻¹(|y/x|). This is an acute angle representing the basic angular magnitude, usually in [0, π/2].
3. — Based on the quadrant where the point (x, y) lies, find θ such that -π < θ ≤ π: If x > 0, y > 0 (Quadrant I): θ = α If x < 0, y > 0 (Quadrant II): θ = π - α If x < 0, y < 0 (Quadrant III): θ = -(π - α) or α - π If x > 0, y < 0 (Quadrant IV): θ = -α Special cases: If x=0, θ = π/2 (for y>0) or θ = -π/2 (for y<0). If y=0, θ = 0 (for x>0) or θ = π (for x<0).
4. — Substitute the calculated r and θ into the polar form: z = r(cosθ + i sinθ).

Worked Examples

• Example 1: Express in a + ib form Simplify (2 + 3i) / (1 - i). Solution: Multiply numerator and denominator by the conjugate of the denominator: (2 + 3i) / (1 - i) (1 + i) / (1 + i) = (21 + 2i + 3i1 + 3i*i) / (1² - i²) = (2 + 2i + 3i - 3) / (1 - (-1)) = (-1 + 5i) / 2 = -1/2 + 5/2 i
• Example 2: Find Modulus and Principal Argument Find the modulus and principal argument of z = 1 - i√3. Solution: Here, x = 1, y = -√3. Modulus r = |z| = √(1² + (-√3)²) = √(1 + 3) = √4 = 2. Since x > 0 and y < 0, z lies in the IV Quadrant. Reference angle α = tan⁻¹(|-√3/1|) = tan⁻¹(√3) = π/3. Principal Argument θ = -α = -π/3. So, |z| = 2 and Arg(z) = -π/3.
• Example 3: Solve Quadratic Equation with Complex Roots Solve x² + x + 1 = 0. Solution: Comparing with ax² + bx + c = 0, we have a=1, b=1, c=1. Discriminant D = b² - 4ac = 1² - 4(1)(1) = 1 - 4 = -3. Since D < 0, the roots are complex: x = [-b ± √D] / 2a = [-1 ± √(-3)] / (2*1) x = [-1 ± i√3] / 2 Thus, the roots are (-1 + i√3)/2 and (-1 - i√3)/2.

Exam Tip: Avoiding Common Traps

When calculating the principal argument, always determine the correct quadrant of the complex number first. A common mistake is to only use tan⁻¹(y/x) directly, which only gives the angle in the first or fourth quadrant. Remember the transformations for Quadrant II (π - α) and Quadrant III (- (π - α) or α - π). For powers of i, always divide the exponent by 4 and use the remainder to determine the equivalent power (i^0 = 1, i^1 = i, i^2 = -1, i^3 = -i). Finally, when performing division, ensure you multiply by the conjugate of the denominator and simplify correctly to the a + ib form.

Practice Questions with Solutions

• Q1: What is the conjugate of (3 - 2i)(1 + i)? A1: First, multiply: (3 - 2i)(1 + i) = 3 + 3i - 2i - 2i² = 3 + i + 2 = 5 + i. The conjugate is 5 - i.
• Q2: Find the modulus of (1 + i√3) / (1 - i√3). A2: Using |z₁/z₂| = |z₁|/|z₂|, we get |1 + i√3| / |1 - i√3| = √(1²+ (√3)²) / √(1² + (-√3)²) = √4 / √4 = 2/2 = 1.
• Q3: Express (i²⁵)³ in the form a + ib. A3: i²⁵ = i^(4*6 + 1) = i¹ = i. So, (i²⁵)³ = i³ = -i. In a+ib form, it is 0 - 1i.
• Q4: What are the roots of x² + 4 = 0? A4: x² = -4, so x = ±√(-4) = ±√(4 * -1) = ±2i. The roots are 2i and -2i.

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