Math · Precalculus · Worked example

Multiply and divide complex numbers

For z₁ = 3 + 2i and z₂ = 1 − 4i, find z₁z₂ and z₁/z₂ in the form a + bi.

Multiply like binomials

Distribute each term of the first factor over the second.

(3 + 2 i) (1 - 4 i) = 3 - 12 i + 2 i - 8 i^{2}

\begin{array}{c} (3 + 2 i) (1 - 4 i) \\ = 3 - 12 i + 2 i - 8 i^{2} \end{array}

−8i² = +8, so the real part is 3 + 8 = 11 and the imaginary part is −12 + 2 = −10.

3 - 10 i + 8 = 11 - 10 i

Multiply the numerator and the denominator by 1 + 4i, the conjugate of the denominator. The denominator becomes 1² + 4² = 17.

\frac{3 + 2 i}{1 - 4 i} \cdot \frac{1 + 4 i}{1 + 4 i} = \frac{(3 + 2 i) (1 + 4 i)}{17}

\begin{array}{c} \frac{3 + 2 i}{1 - 4 i} \cdot \frac{1 + 4 i}{1 + 4 i} \\ = \frac{(3 + 2 i) (1 + 4 i)}{17} \end{array}

(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i² = −5 + 14i.

\frac{- 5 + 14 i}{17} = - \frac{5}{17} + \frac{14}{17} i

(−5/17 + (14/17)i)(1 − 4i) must give 3 + 2i. Its real part is −5/17 + 56/17 = 3, and its imaginary part is 20/17 + 14/17 = 2.

\begin{array}{l} - \frac{5}{17} + \frac{56}{17} = 3 \\ \frac{20}{17} + \frac{14}{17} = 2 \end{array}

z₁z₂ = 11 − 10i and z₁/z₂ = −5/17 + (14/17)i.

Write (2 − i)/(3 + 4i) in the form a + bi.

Show the answer and explanation

Multiply by (3 − 4i)/(3 − 4i). The numerator is (2 − i)(3 − 4i) = 6 − 8i − 3i + 4i² = 2 − 11i, and the denominator is 9 + 16 = 25.

\frac{2 - i}{3 + 4 i} = \frac{2}{25} - \frac{11}{25} i

In Complex plane & roots, switch the task to division: the exact parts read −5/17 and 14/17. The modulus of the product, about 14.87, is √13 · √17.

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