Topic
Multiplying and dividing complex numbersMy Progress
Pug Score
0%
Getting Started
Let's build your foundation!
Best Streak
0 in a row
Study Points
+0
Overview
Practice
Watch
Read
Quiz
Next Steps
Back to Menu
Topic Progress
Pug Score
0%
Getting Started
"Let's build your foundation!"
Videos Watched
0/0
Best Practice
No score
Read
Not viewed
Best Quiz
No attempts
Best Streak
0 in a row
Study Points
+0
Overview
Practice
Watch
Read
Quiz
Next Steps
Read
Multiplying and Dividing Complex Numbers
This lesson explains how to multiply complex numbers using the FOIL method and how to divide them by multiplying the numerator and denominator by the complex conjugate. It covers the key formula, why i squared equals negative 1, and works through full examples of each operation.
Introduction
Multiplying and dividing complex numbers builds directly on what you already know about multiplying binomials and rationalizing denominators. The only new rule you need is that \(i^2 = -1\). If you haven't seen how complex numbers are built and plotted yet, it helps to review complex numbers and complex planes and introduction to imaginary numbers before working through this lesson.
Multiplying Complex Numbers
A complex number has the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. To multiply two complex numbers, treat \(i\) like a variable and expand using the same distributive process you use for binomials (FOIL: First, Outer, Inner, Last), then simplify \(i^2\) to \(-1\).
This gives the complex number multiplication formula:
\((a + bi)(c + di) = (ac - bd) + (ad + bc)i\)
You don't need to memorize this formula word for word — if you can expand the product term by term and simplify \(i^2\) to \(-1\), you'll always get the correct answer.
Worked Example: Multiplying Complex Numbers
Multiply \((3 + 2i)(4 - 5i)\).
\((3 + 2i)(4 - 5i) = 3(4) + 3(-5i) + 2i(4) + 2i(-5i)\)
\(= 12 - 15i + 8i - 10i^2\)
Since \(i^2 = -1\), the last term becomes \(-10(-1) = 10\):
\(= 12 - 15i + 8i + 10 = 22 - 7i\)
Dividing Complex Numbers
Dividing complex numbers is trickier because a fraction with \(i\) in the denominator isn't considered simplified. The fix is the same idea used when you rationalize the denominator of a square root: multiply the top and bottom by a matching expression that clears the unwanted term.
For complex numbers, that matching expression is the complex conjugate. The conjugate of \(c + di\) is \(c - di\) (just flip the sign of the imaginary part). Multiplying a complex number by its conjugate always produces a real number, because:
\((c + di)(c - di) = c^2 - (di)^2 = c^2 + d^2\)
So to divide, multiply the number by its complex conjugate on both the top and bottom of the fraction:
\(\dfrac{a + bi}{c + di} = \dfrac{(a + bi)(c - di)}{(c + di)(c - di)} = \dfrac{(ac + bd) + (bc - ad)i}{c^2 + d^2}\)
Worked Example: Dividing Complex Numbers
Divide \(\dfrac{5 + 2i}{3 - i}\).
The conjugate of \(3 - i\) is \(3 + i\). Multiply top and bottom by it:
\(\dfrac{5 + 2i}{3 - i} \times \dfrac{3 + i}{3 + i} = \dfrac{(5 + 2i)(3 + i)}{(3 - i)(3 + i)}\)
Numerator: \((5 + 2i)(3 + i) = 15 + 5i + 6i + 2i^2 = 15 + 11i - 2 = 13 + 11i\)
Denominator: \((3 - i)(3 + i) = 9 - i^2 = 9 + 1 = 10\)
\(\dfrac{13 + 11i}{10} = \dfrac{13}{10} + \dfrac{11}{10}i\)
Tips for Staying Accurate
Watch for two common slip-ups: forgetting to distribute the negative sign when writing a conjugate, and stopping before simplifying \(i^2\) to \(-1\). Every term with \(i^2\) must be rewritten as a real number before you combine like terms.
Once you're comfortable multiplying and dividing complex numbers in \(a + bi\) form, you can compare this to working with adding and subtracting complex numbers, which uses simpler like-term rules instead of FOIL and conjugates.