Introduction to Complex Numbers

Imaginary numbers were introduced to solve equations like x^2 + 1 = 0. The imaginary unit i is defined as i = sqrt(-1), so i^2 = -1.

A complex number is of the form a + ib, where a, b in R. Here a is the real part (Re(z)) and b is the imaginary part (Im(z)).

The set of complex numbers is denoted by C.

Powers of i

  • i^1 = i
  • i^2 = -1
  • i^3 = -i
  • i^4 = 1
  • In general, i^(4k + r) = i^r for integer k and r = 0, 1, 2, 3.

Algebra of Complex Numbers

Equality

a + ib = c + id if and only if a = c and b = d.

Addition

(a + ib) + (c + id) = (a + c) + i(b + d)

Subtraction

(a + ib) - (c + id) = (a - c) + i(b - d)

Multiplication

(a + ib)(c + id) = (ac - bd) + i(ad + bc)

Conjugate

Conjugate of z = a + ib is bar(z) = a - ib.

  • z bar(z) = a^2 + b^2
  • bar(z_1 + z_2) = bar(z_1) + bar(z_2)
  • bar(z_1 z_2) = bar(z_1) bar(z_2)

Division

(a + ib)/(c + id) = ((a + ib)(c - id))/(c^2 + d^2) (using rationalisation)

Modulus and Argument

Modulus (Absolute Value)

|z| = sqrt(a^2 + b^2) for z = a + ib.

Properties:

  • |z_1 z_2| = |z_1||z_2|
  • |z_1/z_2| = |z_1|/|z_2| (for z_2 != 0)
  • |z_1 + z_2| <= |z_1| + |z_2| (Triangle inequality)
  • |z_1 - z_2| >= ||z_1| - |z_2||

Argument (Amplitude)

arg(z) = theta where tan theta = b/a, measured in the appropriate quadrant.

  • theta = tan^(-1)(b/a) adjusted for quadrant.

Polar Form

z = r(cos theta + i sin theta), where r = |z| and theta = arg(z).

Euler's Form

z = r e^(i theta) where e^(i theta) = cos theta + i sin theta.

Square Roots of Complex Numbers

To find sqrt(a + ib), let sqrt(x + iy) = pm [sqrt((|z| + a)/2) + i sqrt((|z| - a)/2)] (with appropriate sign for y).

Quadratic Equations with Complex Roots

The quadratic equation ax^2 + bx + c = 0 with discriminant D = b^2 - 4ac:

  • If D > 0: Two distinct real roots.
  • If D = 0: Two equal real roots.
  • If D < 0: Two distinct complex conjugate roots.

Roots: x = (-b pm sqrt(D))/(2a)

For D < 0, roots are (-b)/(2a) pm i sqrt(|D|)/(2a).

Nature of Complex Roots

Complex roots always occur in conjugate pairs for quadratic equations with real coefficients.

Worked Examples

Example 1: Express (2 + 3i)/(1 - i) in a + ib form. Solution: Multiply numerator and denominator by (1 + i): ((2+3i)(1+i))/((1-i)(1+i)) = (2+2i+3i+3i^2)/(1+1) = (2+5i-3)/2 = (-1+5i)/2 = -1/2 + (5/2)i

Example 2: Find the modulus and argument of z = -1 + i*sqrt(3). Solution: r = sqrt((-1)^2 + 3) = sqrt(4) = 2. tan theta = sqrt(3)/(-1) = -sqrt(3). Since z is in QII, theta = pi - pi/3 = 2pi/3.

Example 3: Solve x^2 + 4x + 5 = 0. Solution: D = 16 - 20 = -4. x = (-4 pm sqrt(-4))/2 = (-4 pm 2i)/2 = -2 pm i.

Common Mistakes

  1. sqrt(a) sqrt(b) = sqrt(ab) fails for negative numbers: sqrt(-4) sqrt(-9) = 2i * 3i = -6, not sqrt(36) = 6.
  2. Argument quadrant: Always locate the quadrant before finding theta = tan^(-1)(b/a).
  3. Conjugate of sum: bar(z_1 + z_2) = bar(z_1) + bar(z_2), NOT bar(z_1) + z_2.
  4. Equality condition: a + ib = c + id requires both real and imaginary parts equal.

ISC Exam Focus

  • Theory (70%): Properties of conjugates, modulus, argument, polar form conversion.
  • Application (30%): Solving quadratic equations, finding square roots of complex numbers.
  • ISC frequently asks: express in a + ib form, find modulus and argument, solve quadratic with complex roots.
  • 4-6 mark questions involving these concepts appear regularly.

Self-Test Questions

Q1: Express (1 + i)/(1 - i) in a + ib form. Answer: (1 + i)/(1 - i) = ((1+i)^2)/(1+1) = (1 + 2i -1)/2 = i = 0 + 1i.

Q2: Find the modulus and argument of z = 1 + i*sqrt(3). Answer: |z| = sqrt(1 + 3) = 2. theta = tan^(-1)(sqrt(3)) = pi/3 (Q1).

Q3: Solve 2x^2 + x + 1 = 0. Answer: D = 1 - 8 = -7. x = (-1 pm i sqrt(7))/4.

Q4: If z = 2 + 3i, find z bar(z). Answer: z bar(z) = (2+3i)(2-3i) = 4 + 9 = 13.

Q5: Express z = sqrt(3) + i in polar form. Answer: r = sqrt(3 + 1) = 2, theta = tan^(-1)(1/sqrt(3)) = pi/6. Polar form: z = 2(cos(pi/6) + i sin(pi/6)).

Q6: Find the square root of 3 + 4i. Answer: Let sqrt(z) = a + ib. Then a^2 - b^2 = 3 and 2ab = 4. Also a^2 + b^2 = sqrt(3^2 + 4^2) = 5. Solving: a^2 = 4, b^2 = 1. So a = pm 2, b = pm 1. Since ab > 0, roots are 2 + i and -2 - i.

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