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Mom's Basement University

Operations with Complex Numbers

What You’ll Learn

Section titled “What You’ll Learn”

In this lesson you’ll learn how to perform basic arithmetic operations (addition, subtraction, multiplication, and division) with complex numbers.

The Concept

Section titled “The Concept”

Complex numbers are added, subtracted, and multiplied similarly to binomials, with one important rule: i² = −1.

Addition and Subtraction - combine like terms (real parts with real parts, imaginary parts with imaginary parts).

The diagram shows complex addition as vectors on the complex plane. The blue arrow represents 3 + 4i and the orange arrow represents 5 − 2i. To add them, you combine the real parts (3 + 5 = 8) and the imaginary parts (4i − 2i = 2i). The green arrow is the result: 8 + 2i. The dashed lines form a parallelogram, just like adding vectors in physics.

Multiplication - use FOIL (First, Outer, Inner, Last) and replace i² with −1.

Example: (2 + 3i)(4 − i)

=2(4)+2(−i)+3i(4)+3i(−i)=8−2i+12i−3i2=8+10i−3(−1)=8+10i+3=11+10i\begin{aligned} &= 2(4) + 2(-i) + 3i(4) + 3i(-i) \\[1em] &= 8 - 2i + 12i - 3i^2 \\[1em] &= 8 + 10i - 3(-1) \\[1em] &= 8 + 10i + 3 = 11 + 10i \end{aligned}

Division - multiply numerator and denominator by the complex conjugate of the denominator. This eliminates the imaginary part from the denominator.

The complex conjugate of a + bi is a − bi. You just flip the sign of the imaginary part. For example:

  • The conjugate of 1 + 4i is 1 − 4i
  • The conjugate of 3 − 2i is 3 + 2i
  • The conjugate of 7i is −7i
  • The conjugate of 5 (a real number) is just 5

Why does this work? When you multiply a complex number by its conjugate, the imaginary parts cancel out and you get a real number:

(a+bi)(a−bi)=a2−(bi)2=a2−b2i2=a2+b2(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2

That’s always a real number (no i left). So multiplying top and bottom by the conjugate turns the denominator into a plain real number, which you can then divide normally.

Example: (3 + 2i) / (1 + 4i)

3+2i1+4i⋅1−4i1−4i=(3+2i)(1−4i)(1)2+(4)2=3−12i+2i−8i21+16=3−10i+817=11−10i17\frac{3 + 2i}{1 + 4i} \cdot \frac{1 - 4i}{1 - 4i} = \frac{(3 + 2i)(1 - 4i)}{(1)^2 + (4)^2} = \frac{3 - 12i + 2i - 8i^2}{1 + 16} = \frac{3 - 10i + 8}{17} = \frac{11 - 10i}{17}

Worked Example

Section titled “Worked Example”

1. Add: (7 − 3i) + (−2 + 5i)

(7−2)+(−3i+5i)=5+2i(7 - 2) + (-3i + 5i) = 5 + 2i

2. Multiply: (3 + 2i)(5 − i)

=3(5)+3(−i)+2i(5)+2i(−i)=15−3i+10i−2i2=15+7i−2(−1)=15+7i+2=17+7i\begin{aligned} &= 3(5) + 3(-i) + 2i(5) + 2i(-i) \\[1em] &= 15 - 3i + 10i - 2i^2 \\[1em] &= 15 + 7i - 2(-1) \\[1em] &= 15 + 7i + 2 = 17 + 7i \end{aligned}

3. Divide: (4 + 3i) / (2 − i)

Multiply by the conjugate (2 + i):

Numerator: (4+3i)(2+i)=8+4i+6i+3i2=8+10i−3=5+10iDenominator: (2)2+(1)2=4+1=5Result: 5+10i5=1+2i\begin{aligned} \text{Numerator: } &(4 + 3i)(2 + i) = 8 + 4i + 6i + 3i^2 = 8 + 10i - 3 = 5 + 10i \\[1em] \text{Denominator: } &(2)^2 + (1)^2 = 4 + 1 = 5 \\[1em] \text{Result: } &\frac{5 + 10i}{5} = 1 + 2i \end{aligned}

Real-World Application

Section titled “Real-World Application”

Operations with complex numbers are used in:

  • Electrical engineering (AC circuit analysis)
  • Signal processing and audio engineering
  • Computer graphics and 3D modeling
  • Quantum physics and advanced engineering

Even if you don’t work in these fields, mastering these operations builds confidence with algebra and prepares you for more advanced math.

Quiz

Section titled “Quiz”
Add: $(5 + 3i) + (2 - 7i)$
Multiply: $(3 + 2i)(4 - i)$
What is $i^2$?
To divide complex numbers, multiply numerator and denominator by:
Subtract: $(8 - 2i) - (3 + 5i)$