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25.1 Multiplication/Division of Complex Numbers

Let \(\, z_1 = r_1\left (\cos \theta _1 + i \sin \theta _1\right )\,\) and \(\, z_2 = r_2\left (\cos \theta _2 + i \sin \theta _2\right )\).

\begin {align*} z_1z_2 & = \left (r_1\left (\cos \theta _1 + i \sin \theta _1\right )\right )\left (r_2\left (\cos \theta _2 + i \sin \theta _2\right )\right )\\ & = r_1r_2\left [\cos \theta _1\cos \theta _2 + i\cos \theta _1\sin \theta _2 + i\sin \theta _1\cos \theta _2 + i^2 \sin \theta _1\sin \theta _2\right ]\\ & = r_1r_2\left [\cos \theta _1\cos \theta _2 + i\left (\cos \theta _1\sin \theta _2 + \sin \theta _1\cos \theta _2\right ) - \sin \theta _1\sin \theta _2\right ]\\ & = r_1r_2\left [\left (\cos \theta _1\cos \theta _2 - \sin \theta _1\sin \theta _2\right ) + i\left (\cos \theta _1\sin \theta _2 + \sin \theta _1\cos \theta _2\right )\right ]\\ & = r_1r_2\left [\cos \left (\theta _1 + \theta _2\right ) + i\sin \left (\theta _1 + \theta _2\right )\right ]\\ \end {align*}

\begin {align*} \frac {z_1}{z_2} & = \frac {r_1\left (\cos \theta _1 + i\sin \theta _1\right )}{r_2\left ( \cos \theta _2 + i\sin \theta _2\right )}\\\\ & = \frac {r_1}{r_2}\cdot \frac {\left (\cos \theta _1 + i\sin \theta _1\right )}{\left ( \cos \theta _2 + i\sin \theta _2\right )}\cdot \frac {\left ( \cos \theta _2 - i\sin \theta _2\right )}{\left ( \cos \theta _2 - i\sin \theta _2\right )}\\\\ & = \frac {r_1}{r_2}\cdot \frac {\cos \theta _1\cos \theta _2 - i\cos \theta _1\sin \theta _2 + i\sin \theta _1\cos \theta _2 + \sin \theta _1\sin \theta _2}{\cos ^2\theta _2 + \sin ^2\theta }\\\\ & = \frac {r_1}{r_2}\cdot \left [\left (\cos \theta _1\cos \theta _2 + \sin \theta _1\sin \theta _2\right ) + i\left (\sin \theta _1\cos \theta _2 - \cos \theta _1\sin \theta _2\right )\right ]\\\\ & = \frac {r_1}{r_2}\left [\cos \left (\theta _1-\theta _2\right ) + i\sin \left (\theta _1-\theta _2\right )\right ]\\\\ \end {align*}

\(*\)
\(\quad \displaystyle {z_1z_2 = r_1r_2 e^{i(\theta _1 + \theta _2)}}\)
\(*\)
\(\quad \displaystyle {\frac {z_1}{z_2} = \frac {r_1}{r_2}e^{i(\theta _1 - \theta _2)}}\)
\(*\)
\(\quad \left |z_1z_2\right | = \left |z_1\right |\left |z_2\right |\,\), \(\, \) arg\((z_1z_2) = \) arg\((z_1) + \) arg\((z_2)\)
\(*\)
\(\quad \left |\frac {z_1}{z_2}\right | = \frac {\left |z_1\right |}{\left |z_2\right |}\,\), \(\,\) arg\(\left (\frac {z_1}{z_2}\right ) = \) arg\((z_1) - \) arg\((z_2)\)

Example 25.3. Given that \(\, z_1 = 9\left ( \cos \frac {7\pi }{12} + i \sin \frac {7\pi }{12}\right )\,\) and \(\, z_2 = 5\left ( \cos \frac {2\pi }{3} + i \sin \frac {2\pi }{3}\right )\).

Find \(\, z_1z_2\,\) in the form \(\, a + ib,\, a, b \in \mathbb {R}\).

Solution. \begin {align*} z_1z_2 & = r_1r_2\left [\cos \left (\theta _1 + \theta _2\right ) + i\sin \left (\theta _1 + \theta _2\right )\right ]\\\\ z_1z_2 & = 9(5)\left [\cos \big (\frac {7\pi }{12} + \frac {2\pi }{3} \Big ) + i\sin \Big (\frac {7\pi }{12}+ \frac {2\pi }{3} \Big )\right ]\\\\ & = 45 \left ( \cos \left (\frac {15\pi }{12}\right ) + i\sin \left (\frac {15\pi }{12}\right )\right ) = 45 \left ( \cos \left (\frac {5\pi }{4}\right ) + i\sin \left (\frac {5\pi }{4}\right )\right )\\\\ & = 45\left (-\frac {1}{\sqrt {2}} - i\frac {1}{\sqrt {2}}\right ) = \frac {45}{\sqrt {2}}\left (-1 - j\right )\\\\ & = \frac {-45}{\sqrt {2}}\left ( 1 + i\right )\\ & = \frac {-45\sqrt {2}}{2}\left (1 + i\right )\\ \end {align*}

Example 25.4. Find \(\, \frac {z_1}{z_2}\)

Solution. \begin {align*} \frac {z_1}{z_2} & = \frac {r_1\left (\cos \theta _1 + i\sin \theta _1\right )}{r_2\left ( \cos \theta _2 + i\sin \theta _2\right )}\\\\ & = \frac {r_1}{r_2}\left [\cos \left (\theta _1-\theta _2\right ) + i\sin \left (\theta _1-\theta _2\right )\right ]\\\\ & = \frac {9}{5}\left ( \cos \left (\frac {7\pi }{12} - \frac {2\pi }{3}\right ) + i \sin \left (\frac {7\pi }{12} - \frac {2\pi }{3}\right )\right )\\\\ & = \frac {9}{5}\left ( \cos \left ( - \frac {\pi }{12}\right ) + i \sin \left ( - \frac {\pi }{12}\right )\right ) = \frac {9}{5}\left ( \cos \left (\frac {\pi }{12}\right ) - i \sin \left (\frac {\pi }{12}\right )\right )\\\\ & = \frac {9}{5}\left ( \frac {1}{4} - i\frac {1}{2\sqrt {2}}\right )\\\\ & = \frac {9}{20}- i\frac {9}{10\sqrt {2}}\\\\\\\\\ \end {align*}

Example 25.5.

Express \(\,\displaystyle {e^{i\pi / 4}\,}\) in the form

1.
\(\quad \cos \theta + i\sin \theta \,, \, -\pi < \theta < \pi \)
2.
\(\quad a + ib\,, \, a,b\in \mathbb {R}\)

Solution.

1.
\(\quad \displaystyle {e^{i\pi /4} = \cos \left (-\frac {\pi }{4}\right ) + i\sin \left (-\frac {\pi }{4}\right ) = \cos \frac {\pi }{4} - i\sin \frac {\pi }{4}}\)
2.
\(\quad \frac {1}{\sqrt {2}} - \frac {i}{\sqrt {2}}\)

Now, \begin {align*} \left (\cos \theta + i \sin \theta \right )^2 & = \cos 2 \theta + i \sin 2\theta \\ \left (\cos \theta + i \sin \theta \right )^3 & = \left (\cos \theta + i \sin \theta \right )^2\left (\cos \theta + i \sin \theta \right )= \left (\cos 3 \theta + i \sin 3\theta \right )\\ \vdots & \\ \vdots & \\ \left (\cos \theta + i \sin \theta \right )^5 & = \left (\cos 5\theta + i \sin 5\theta \right )\\ \vdots \\ \vdots \\ \left (\cos \theta + i \sin \theta \right )^n & = \left (\cos n\theta + i \sin n\theta \right )\\\\ \end {align*}

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