3.6 Division

Definition 3.11 (Quotient). Let \(z_1=x_1+iy_1\) and \(z_2=x_2+iy_2\) with \(z_2\neq 0\). Then \(\dfrac {z_1}{z_2}=\dfrac {z_1\overline {z_2}}{z_2\overline {z_2}}\), that is \begin {align*} \frac {x_1+iy_1}{x_2+iy_2} & =\frac {(x_1+iy_1)(x_2-iy_2)}{(x_2+iy_2)(x_2-iy_2)}\\\\ & = \frac {x_1x_2+y_1y_2+i(x_2y_1-x_1y_2)}{x_2^2+y_2^2}\\\\ & = \frac {x_1x_2+y_1y_2}{x_2^2+y^2_2}+i\,\frac {x_2y_1-x_1y_2}{x^2_2+y^2_2}\\ \end {align*}

Note 3.12. This manoeuvre is worth naming, because it recurs constantly. Multiplying above and below by the conjugate of the denominator turns the denominator into \(z_2\overline {z_2}=\left |z_2\right |^{2}\), which is real — and once the denominator is real the quotient splits cleanly into real and imaginary parts. It is the same idea as rationalising a surd such as \(\frac {1}{3-\sqrt {2}}\), and works for the same reason.

Example 3.13. Express \(\dfrac {2+i}{3-2i}\) in the form \(a+ib\).

Solution. Multiply above and below by the conjugate of the denominator: \begin {align*} \frac {2+i}{3-2i} & = \frac {(2+i)(3+2i)}{(3-2i)(3+2i)} = \frac {(6-2)+i(4+3)}{3^{2}+2^{2}}\\ & = \frac {4+7i}{13} = \frac {4}{13}+\frac {7}{13}i . \end {align*}

Example 3.14.

Express each of the following complex numbers in the form \(a+ib\) where \(a\) and \(b\) are real numbers. Hence find \(|z|\) in each case.

(a).: \((4+i)^{2}\)
(b).: \(\dfrac {2-5i}{3+2i}\)
(c).: \(\dfrac {3+5i}{4-3i}\)

Solution.

(a). Expand and use \(i^{2}=-1\): \begin {align*} z=(4+i)^{2} & = 16+4i+4i+i^{2}\\ & = 16+8i-1\\ & = 15+8i, \end {align*}

so \(a=15\) and \(b=8\), and \[|z|=\sqrt {15^{2}+8^{2}}=\sqrt {225+64}=\sqrt {289}=17 .\]

(b). Multiply above and below by the conjugate of the denominator: \begin {align*} z=\frac {2-5i}{3+2i} & = \frac {(2-5i)(3-2i)}{(3+2i)(3-2i)}\\ & = \frac {(6-10)-i(15+4)}{3^{2}+2^{2}}\\ & = \frac {-4-19i}{13} = -\frac {4}{13}-\frac {19}{13}i, \end {align*}

so \(a=-\frac {4}{13}\) and \(b=-\frac {19}{13}\), and \[|z|=\sqrt {\left (-\frac {4}{13}\right )^{2}+\left (-\frac {19}{13}\right )^{2}} =\frac {1}{13}\sqrt {16+361}=\frac {\sqrt {377}}{13} .\]

(c). The same step again: \begin {align*} z=\frac {3+5i}{4-3i} & = \frac {(3+5i)(4+3i)}{(4-3i)(4+3i)}\\ & = \frac {(12-15)+i(9+20)}{4^{2}+3^{2}}\\ & = \frac {-3+29i}{25} = -\frac {3}{25}+\frac {29}{25}i, \end {align*}

so \(a=-\frac {3}{25}\) and \(b=\frac {29}{25}\), and \[|z|=\sqrt {\left (-\frac {3}{25}\right )^{2}+\left (\frac {29}{25}\right )^{2}} =\frac {1}{25}\sqrt {9+841}=\frac {\sqrt {850}}{25}=\frac {\sqrt {34}}{5} .\]

Note 3.15. The moduli in (b) and (c) can be had without computing \(z\) at all, because the modulus is multiplicative: \[\left |z_1z_2\right |=\left |z_1\right |\left |z_2\right |, \qquad \left |\frac {z_1}{z_2}\right |=\frac {\left |z_1\right |}{\left |z_2\right |} .\] So for (b), \[\left |\frac {2-5i}{3+2i}\right |=\frac {\sqrt {4+25}}{\sqrt {9+4}} =\frac {\sqrt {29}}{\sqrt {13}}=\frac {\sqrt {377}}{13},\] and for (c), since \(\left |4-3i\right |=5\), \[\left |\frac {3+5i}{4-3i}\right |=\frac {\sqrt {9+25}}{5}=\frac {\sqrt {34}}{5} .\] Both in one line, and with no chance of an arithmetic slip in the real and imaginary parts. Simplify surds at the end: \(\sqrt {850}=\sqrt {25\times 34} =5\sqrt {34}\), so \(\frac {\sqrt {850}}{25}=\frac {\sqrt {34}}{5}\).

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