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Mathematics · Ch 10 — Complex Numbers

Division of Complex Number

10.2.8

Division of Complex Number

Division of Complex Number

Let z1=a+ibz_1=a+ib and z2=c+idz_2=c+id be complex numbers with z2≠0z_2\neq0 (i.e. c+id≠0c+id\neq0). To find z1z2=a+ibc+id\dfrac{z_1}{z_2}=\dfrac{a+ib}{c+id}, multiply numerator and denominator by the conjugate of the denominator, z2‾=c−id\overline{z_2}=c-id:

z1z2=a+ibc+id×c−idc−id=(ac+bd)+(bc−ad)ic2+d2\frac{z_1}{z_2}=\frac{a+ib}{c+id}\times\frac{c-id}{c-id}=\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}

so that

z1z2=ac+bdc2+d2+bc−adc2+d2 i,where ac+bdc2+d2∈R and bc−adc2+d2∈R\frac{z_1}{z_2}=\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}\,i,\qquad\text{where }\frac{ac+bd}{c^2+d^2}\in\mathbb{R}\text{ and }\frac{bc-ad}{c^2+d^2}\in\mathbb{R}

The trick works because multiplying the denominator c+idc+id by its own conjugate c−idc-id gives the real number c2+d2c^2+d^2 (by the identity zzˉ=a2+b2z\bar z=a^2+b^2 from Section 1.2.6), clearing all trace of ii from the bottom of the fraction.

Worked illustration. If z1=3+2i, z2=1+iz_1=3+2i,\ z_2=1+i, then multiplying numerator and denominator by z2‾=1−i\overline{z_2}=1-i:

z1z2=3+2i1+i×1−i1−i=3−3i+2i−2i21+1=5−i2\frac{z_1}{z_2}=\frac{3+2i}{1+i}\times\frac{1-i}{1-i}=\frac{3-3i+2i-2i^2}{1+1}=\frac{5-i}{2}

so z1z2=52−12i\dfrac{z_1}{z_2}=\dfrac52-\dfrac12i.

Properties of division.

  1. 1i=1i×ii=i−1=−i\dfrac1i=\dfrac1i\times\dfrac{i}{i}=\dfrac{i}{-1}=-i — the reciprocal of ii is −i-i, a fact used constantly when simplifying expressions with ii in a denominator. …
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