Complex Numbers and Operations

Understand complex numbers, perform operations with complex numbers, and solve equations involving complex solutions.

Complex Numbers and Operations

Introduction to Complex Numbers

A complex number has the form: z=a+biz = a + bi

Where:

  • aa is the real part: Re(z)=a\text{Re}(z) = a
  • bb is the imaginary part: Im(z)=b\text{Im}(z) = b
  • ii is the imaginary unit: i=1i = \sqrt{-1}, so i2=1i^2 = -1

Key Properties of ii

  • i1=ii^1 = i
  • i2=1i^2 = -1
  • i3=i2i=ii^3 = i^2 \cdot i = -i
  • i4=i2i2=1i^4 = i^2 \cdot i^2 = 1
  • Pattern repeats: i5=ii^5 = i, i6=1i^6 = -1, etc.

To find ini^n: divide nn by 4 and use the remainder.

Complex Number Operations

Addition and Subtraction

Combine like terms (real with real, imaginary with imaginary): (a+bi)+(c+di)=(a+c)+(b+d)i(a + bi) + (c + di) = (a + c) + (b + d)i (a+bi)(c+di)=(ac)+(bd)i(a + bi) - (c + di) = (a - c) + (b - d)i

Multiplication

Use the distributive property and i2=1i^2 = -1: (a+bi)(c+di)=ac+adi+bci+bdi2(a + bi)(c + di) = ac + adi + bci + bdi^2 =ac+adi+bcibd= ac + adi + bci - bd =(acbd)+(ad+bc)i= (ac - bd) + (ad + bc)i

FOIL method works:

  • First: acac
  • Outer: adiadi
  • Inner: bcibci
  • Last: bdi2=bdbdi^2 = -bd

Division

To divide complex numbers, multiply by the conjugate of the denominator: a+bic+di=a+bic+dicdicdi\frac{a + bi}{c + di} = \frac{a + bi}{c + di} \cdot \frac{c - di}{c - di}

The denominator becomes real: (a+bi)(cdi)c2+d2\frac{(a + bi)(c - di)}{c^2 + d^2}

Complex Conjugate

The conjugate of z=a+biz = a + bi is: zˉ=abi\bar{z} = a - bi

Properties:

  • zzˉ=(a+bi)(abi)=a2+b2z \cdot \bar{z} = (a + bi)(a - bi) = a^2 + b^2 (always real and non-negative)
  • (z+w)=zˉ+wˉ(z + w)^* = \bar{z} + \bar{w}
  • (zw)=zˉwˉ(zw)^* = \bar{z} \cdot \bar{w}
  • zˉ=z\overline{\bar{z}} = z

Absolute Value (Modulus)

The absolute value or modulus of z=a+biz = a + bi is: z=a2+b2|z| = \sqrt{a^2 + b^2}

This represents the distance from the origin in the complex plane.

Properties:

  • z0|z| \geq 0
  • z=0|z| = 0 if and only if z=0z = 0
  • zw=zw|zw| = |z| \cdot |w|
  • zw=zw|\frac{z}{w}| = \frac{|z|}{|w|} (if w0w \neq 0)
  • z2=zzˉ|z|^2 = z \cdot \bar{z}

Complex Plane (Argand Diagram)

Complex numbers can be plotted on a coordinate plane:

  • Horizontal axis: Real part
  • Vertical axis: Imaginary part
  • z=a+biz = a + bi corresponds to point (a,b)(a, b)

Polar Form

A complex number can also be written in polar form: z=r(cosθ+isinθ)=rcisθz = r(\cos\theta + i\sin\theta) = r\text{cis}\theta

Or using Euler's formula: z=reiθz = re^{i\theta}

Where:

  • r=z=a2+b2r = |z| = \sqrt{a^2 + b^2} (modulus)
  • θ=arg(z)=arctan(ba)\theta = \arg(z) = \arctan(\frac{b}{a}) (argument/angle, adjusted for quadrant)

Solving Equations with Complex Numbers

Quadratic Equations

For ax2+bx+c=0ax^2 + bx + c = 0 with discriminant b24ac<0b^2 - 4ac < 0:

x=b±b24ac2a=b±ib24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm i\sqrt{|b^2 - 4ac|}}{2a}

Complex Conjugate Root Theorem: If a+bia + bi is a root of a polynomial with real coefficients, then abia - bi is also a root.

Higher Degree Equations

Complex solutions come in conjugate pairs for polynomials with real coefficients.

Key Theorems

  1. Fundamental Theorem of Algebra: Every polynomial of degree n1n \geq 1 has exactly nn complex roots (counting multiplicity).

  2. Complex Conjugate Root Theorem: If a polynomial has real coefficients and a+bia + bi is a root, then abia - bi is also a root.

📚 Practice Problems

1Problem 1easy

Question:

Simplify (3+2i)+(54i)(3 + 2i) + (5 - 4i) and (3+2i)(54i)(3 + 2i) - (5 - 4i).

💡 Show Solution

Solution:

Part 1: Addition (3+2i)+(54i)(3 + 2i) + (5 - 4i)

Combine real parts and imaginary parts separately:

  • Real parts: 3+5=83 + 5 = 8
  • Imaginary parts: 2+(4)=22 + (-4) = -2

Answer: 82i8 - 2i

Part 2: Subtraction (3+2i)(54i)(3 + 2i) - (5 - 4i)

Distribute the negative sign and combine: =3+2i5+4i= 3 + 2i - 5 + 4i

  • Real parts: 35=23 - 5 = -2
  • Imaginary parts: 2+4=62 + 4 = 6

Answer: 2+6i-2 + 6i

Verification:

  • Addition: Real = 8, Imaginary = -2 ✓
  • Subtraction: Real = -2, Imaginary = 6 ✓

2Problem 2medium

Question:

Multiply (2+3i)(4i)(2 + 3i)(4 - i) and express in standard form a+bia + bi.

💡 Show Solution

Solution:

Given: (2+3i)(4i)(2 + 3i)(4 - i)

Use FOIL method:

  • First: (2)(4)=8(2)(4) = 8
  • Outer: (2)(i)=2i(2)(-i) = -2i
  • Inner: (3i)(4)=12i(3i)(4) = 12i
  • Last: (3i)(i)=3i2=3(1)=3(3i)(-i) = -3i^2 = -3(-1) = 3

Combine: 82i+12i+38 - 2i + 12i + 3 =(8+3)+(2+12)i= (8 + 3) + (-2 + 12)i =11+10i= 11 + 10i

Answer: 11+10i11 + 10i

Verification using the formula: (a+bi)(c+di)=(acbd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i

Where a=2,b=3,c=4,d=1a = 2, b = 3, c = 4, d = -1:

  • Real part: (2)(4)(3)(1)=8+3=11(2)(4) - (3)(-1) = 8 + 3 = 11
  • Imaginary part: (2)(1)+(3)(4)=2+12=10(2)(-1) + (3)(4) = -2 + 12 = 10

3Problem 3hard

Question:

Divide 3+2i12i\frac{3 + 2i}{1 - 2i} and express in standard form a+bia + bi.

💡 Show Solution

Solution:

Given: 3+2i12i\frac{3 + 2i}{1 - 2i}

Step 1: Multiply by the conjugate of the denominator

The conjugate of 12i1 - 2i is 1+2i1 + 2i:

3+2i12i1+2i1+2i\frac{3 + 2i}{1 - 2i} \cdot \frac{1 + 2i}{1 + 2i}

Step 2: Multiply the numerators (3+2i)(1+2i)(3 + 2i)(1 + 2i)

  • F: (3)(1)=3(3)(1) = 3
  • O: (3)(2i)=6i(3)(2i) = 6i
  • I: (2i)(1)=2i(2i)(1) = 2i
  • L: (2i)(2i)=4i2=4(2i)(2i) = 4i^2 = -4

=3+6i+2i4=1+8i= 3 + 6i + 2i - 4 = -1 + 8i

Step 3: Multiply the denominators (12i)(1+2i)=12(2i)2=14i2=1+4=5(1 - 2i)(1 + 2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 + 4 = 5

Step 4: Combine 1+8i5=15+85i\frac{-1 + 8i}{5} = -\frac{1}{5} + \frac{8}{5}i

Answer: 15+85i-\frac{1}{5} + \frac{8}{5}i

Verification: Check by multiplying: (12i)(15+85i)(1 - 2i)(-\frac{1}{5} + \frac{8}{5}i) =15+85i+25i165i2= -\frac{1}{5} + \frac{8}{5}i + \frac{2}{5}i - \frac{16}{5}i^2 =15+105i+165= -\frac{1}{5} + \frac{10}{5}i + \frac{16}{5} =155+105i=3+2i= \frac{15}{5} + \frac{10}{5}i = 3 + 2i