Complex Conjugate
In short: A complex conjugate is the mirror partner of a number written with i: change the sign of the imaginary piece and leave the real piece alone.
Algebra 2Numbers and OperationsN-CN.A.2
What is Complex Conjugate?
The complex conjugate of \(a + bi\) is \(a – bi\): the real part stays, the sign of the imaginary part flips. Drawn on the complex plane, the two numbers are reflections of each other across the horizontal axis.
Its superpower shows up in a product. \((a + bi)(a – bi) = a^2 + b^2\), a plain non-negative number, because the imaginary terms cancel. That makes it the tool for dividing, and it explains why quadratics with real coefficients give imaginary solutions in matching pairs.
How to spot it
- An \(i\) appears in the denominator of a fraction and must be cleared away.
- A quadratic with a negative discriminant has two solutions that differ only by the sign in front of the \(i\) term.
- The task is to multiply \((a + bi)(a – bi)\), or to find the squared size of a complex number.
- A quotient must be rewritten in standard \(a + bi\) form, with a complex conjugate as the key move.
The method
- Write the partner of the denominator by flipping the sign between its two pieces.
- Multiply the top and the bottom of the fraction by that partner.
- Expand the numerator, swapping each \(i^2\) for \(-1\); the denominator becomes the real number \(a^2 + b^2\).
- Split the result into a real piece and an imaginary piece, and reduce each fraction.
Complex Conjugate: solved example
Problem
Write \(\frac{5 + 2i}{1 – 3i}\) in the form \(a + bi\).
Solution
The partner of the denominator is \(1 + 3i\). Multiply top and bottom by it.
Numerator: \((5 + 2i)(1 + 3i) = 5 + 15i + 2i + 6i^2 = 5 + 17i – 6 = -1 + 17i\). Denominator: \((1 – 3i)(1 + 3i) = 1 + 9 = 10\). The quotient is \(-\frac{1}{10} + \frac{17}{10}i\).
Try it
Question 1
Find the mirror partner of \(8 – 6i\) and multiply the two together.
Show the answer
The partner is \(8 + 6i\), and \((8 – 6i)(8 + 6i) = 64 + 36 = 100\).
Question 2
A quadratic with real coefficients has one solution \(2 + 5i\). Name the other solution and the sum of the two.
Show the answer
The other is the complex conjugate \(2 – 5i\), and the sum is \(4\).
Complex Conjugate vs. Additive Inverse
The additive inverse of \(a + bi\) is \(-a – bi\): both signs change, and adding the pair gives zero. A complex conjugate changes only the sign of the imaginary part, and multiplying the pair gives a positive real number. For \(3 + 4i\), the first is \(-3 – 4i\) and the second is \(3 – 4i\).
Keep going
Lessons
Worksheets and quizzes
- Complex Conjugates and Complex Solutions – Worksheet
- Complex Conjugates and Complex Solutions – Quiz
Related words
Whole dictionary: A–Z · Algebra 2 · Numbers and Operations