Multiplying and Dividing Complex Numbers

Expanding products and replacing i squared with minus one, what a conjugate is, why a number times its conjugate is always real, and how that clears i from a denominator.

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Multiplying complex numbers uses ordinary algebra with one extra move: whenever i^2 appears, replace it with -1 . Division needs a trick — the conjugate, which clears i out of the denominator entirely.

Example Multiplying step by step

Work out (2 + 4i)(9 - 3i) by expanding every pair.

(2)(9) = 18
(2)(-3i) = -6i
(4i)(9) = 36i
(4i)(-3i) = -12i^2
Replace i^2 with -1 : -12(-1) = 12
Real parts: 18 + 12 = 30
Imaginary parts: -6i + 36i = 30i
⟹ 30 + 30i

Notice that the i^2 term turned into a real number. That is what makes complex multiplication different from ordinary expansion.

Concept The conjugate
a + bi a − bi only the middle sign changes
The conjugate of a + bi is a - bi . The real part is untouched; only the sign of the imaginary part flips.
Theorem Why the conjugate helps
(a + bi)(a - bi) = a^2 + b^2

The middle terms cancel and -b^2i^2 becomes +b^2 . The answer is always a real number — which is exactly what we need in a denominator.

Example Dividing complex numbers

Work out \dfrac{2i}{3 + 6i} . The conjugate of the denominator is 3 - 6i .

Multiply top and bottom by that conjugate:
\frac{2i}{3 + 6i} \times \frac{3 - 6i}{3 - 6i}
Numerator: 2i(3 - 6i) = 6i - 12i^2 = 6i + 12
Denominator: 3^2 + 6^2 = 9 + 36 = 45
\frac{12 + 6i}{45}
Split and simplify each part:
\frac{12}{45} = \frac{4}{15} and \frac{6}{45} = \frac{2}{15}
⟹ 4/15 + (2/15)i

Multiplying by \frac{3-6i}{3-6i} is multiplying by 1, so the value never changes — only its appearance does.

Note Mistakes to avoid
Forgetting to replace i^2 with -1 , leaving the answer unfinished.
Changing the sign of the real part when writing the conjugate.
Multiplying only the denominator by the conjugate instead of both parts.
Leaving i in the denominator rather than reaching the form a + bi .
Not simplifying the final fractions.
Summary
  1. Multiply by expanding, then replace i² with −1 and collect parts.
  2. (2 + 4i)(9 − 3i) = 30 + 30i.
  3. The conjugate of a + bi is a − bi; only the imaginary sign changes.
  4. (a + bi)(a − bi) = a² + b², always a real number.
  5. To divide, multiply top and bottom by the denominator's conjugate, then simplify to a + bi.
231. Arithmetic Sequences and Series
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