21.2: Multiplication and division of complex numbers

One interesting feature of the polar form of a complex number is that the multiplication and division are very easy to perform.

Proposition: Product and Quotient of Complex numbers

Let \(r_1(\cos(\theta_1)+i\sin(\theta_1))\) and \(r_2(\cos(\theta_2)+i\sin(\theta_2))\) be two complex numbers in polar form. Then, the product and quotient of these are given by

\[\label{EQU:Product} r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right) \cdot r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right) =r_{1} r_{2} \cdot\left(\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right) \]

\[\label{EQU:Quotient}\dfrac{r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right)}{r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right)}=\dfrac{r_{1}}{r_{2}} \cdot\left(\cos \left(\theta_{1}-\theta_{2}\right)+i \sin \left(\theta_{1}-\theta_{2}\right)\right) \]

Proof

The proof uses the addition formulas for trigonometric functions \(\sin(\alpha+\beta)\) and \(\cos(\alpha+\beta)\) from proposition [PROP:trig-add-subt-formulas].

\[\begin{aligned}
r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right) \cdot r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right) &=r_{1} r_{2} \cdot\left(\cos \left(\theta_{1}\right) \cos \left(\theta_{2}\right)+i \cos \left(\theta_{1}\right) \sin \left(\theta_{2}\right)+i \sin \left(\theta_{1}\right) \cos \left(\theta_{2}\right)+i^{2} \sin \left(\theta_{1}\right) \sin \left(\theta_{2}\right)\right) \\
&=r_{1} r_{2} \cdot\left(\left(\cos \left(\theta_{1}\right) \cos \left(\theta_{2}\right)-\sin \left(\theta_{1}\right) \sin \left(\theta_{2}\right)\right)+i\left(\cos \left(\theta_{1}\right) \sin \left(\theta_{2}\right)+\sin \left(\theta_{1}\right) \cos \left(\theta_{2}\right)\right)\right) \\
&=r_{1} r_{2} \cdot\left(\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right)
\end{aligned} \nonumber \]

For the division formula, note, that the multiplication formula \(\ref{EQU:Product}\) gives

\[\begin{aligned}
r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right) \cdot \dfrac{1}{r_{2}}\left(\cos \left(-\theta_{2}\right)+i \sin \left(-\theta_{2}\right)\right) &=r_{2} \dfrac{1}{r_{2}}\left(\cos \left(\theta_{2}-\theta_{2}\right)+i \sin \left(\theta_{2}-\theta_{2}\right)\right) \\
&=1 \cdot(\cos 0+i \sin 0)\\
&=1 \cdot(1+i \cdot 0)\\
&=1 \\
\Longrightarrow \dfrac{1}{r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right)}&=\dfrac{1}{r_{2}}\left(\cos \left(-\theta_{2}\right)+i \sin \left(-\theta_{2}\right)\right)
\end{aligned} \nonumber \]

so that

\[\begin{aligned}
\dfrac{r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right)}{r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right)}&=r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right) \cdot \dfrac{1}{r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right)} \\
&=r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right) \cdot \dfrac{1}{r_{2}}\left(\cos \left(-\theta_{2}\right)+i \sin \left(-\theta_{2}\right)\right)\\
&=\dfrac{r_{1}}{r_{2}} \cdot\left(\cos \left(\theta_{1}-\theta_{2}\right)+i \sin \left(\theta_{1}-\theta_{2}\right)\right)
\end{aligned} \nonumber \]

Example \(\PageIndex{1}\)

Multiply or divide the complex numbers, and write your answer in polar and standard form.

1. \(5\left(\cos \left(11^{\circ}\right)+i \sin \left(11^{\circ}\right)\right) \cdot 8\left(\cos \left(34^{\circ}\right)+i \sin \left(34^{\circ}\right)\right)\)
2. \(3\left(\cos \left(\dfrac{5 \pi}{8}\right)+i \sin \left(\dfrac{5 \pi}{8}\right)\right) \cdot 12\left(\cos \left(\dfrac{7 \pi}{8}\right)+i \sin \left(\dfrac{7 \pi}{8}\right)\right)\)
3. \(\dfrac{32\left(\cos \left(\dfrac{\pi}{4}\right)+i \sin \left(\dfrac{\pi}{4}\right)\right)}{8\left(\cos \left(\dfrac{7 \pi}{12}\right)+i \sin \left(\dfrac{7 \pi}{12}\right)\right)}\)
4. \(\dfrac{4\left(\cos \left(203^{\circ}\right)+i \sin \left(203^{\circ}\right)\right)}{6\left(\cos \left(74^{\circ}\right)+i \sin \left(74^{\circ}\right)\right)}\)

Solution

We will multiply and divide the complex numbers using equations \(\ref{EQU:Product}\) and \(\ref{EQU:Quotient}\), respectively, and then convert them to standard notation \(a + bi\).

1. \[\begin{aligned}
   5\left(\cos \left(11^{\circ}\right)+i \sin \left(11^{\circ}\right)\right) \cdot 8\left(\cos \left(34^{\circ}\right)+i \sin \left(34^{\circ}\right)\right) &=5 \cdot 8 \cdot\left(\cos \left(11^{\circ}+34^{\circ}\right)+i \sin \left(11^{\circ}+34^{\circ}\right)\right)=40\left(\cos \left(45^{\circ}\right)+i \sin \left(45^{\circ}\right)\right) \\
   &=40\left(\dfrac{\sqrt{2}}{2}+i \dfrac{\sqrt{2}}{2}\right)\\
   &=40 \dfrac{\sqrt{2}}{2}+i \cdot 40 \dfrac{\sqrt{2}}{2}\\
   &=20 \sqrt{2}+20 \sqrt{2} i
   \end{aligned} \nonumber \]
2. Similarly, we obtain the next product.

\[3\left(\cos \left(\dfrac{5 \pi}{8}\right)+i \sin \left(\dfrac{5 \pi}{8}\right)\right) \cdot 12\left(\cos \left(\dfrac{7 \pi}{8}\right)+i \sin \left(\dfrac{7 \pi}{8}\right)\right) = 36\left(\cos \left(\dfrac{5 \pi}{8}+\dfrac{7 \pi}{8}\right)+i \sin \left(\dfrac{5 \pi}{8}+\dfrac{7 \pi}{8}\right)\right) \nonumber \]

Now, \(\dfrac{5\pi}{8}+\dfrac{7\pi}{8}=\dfrac{5\pi+7\pi}{8}=\dfrac{12\pi}{8}=\dfrac{3\pi}2\), and \(\cos\left(\dfrac{3\pi}2\right)=0\) and \(\sin\left(\dfrac{3\pi}2\right)=-1\). Therefore, we obtain that the product is

\[36\left(\cos\left(\dfrac{3\pi}{2}\right)+i\sin\left(\dfrac{3\pi}{2}\right)\right) = 36 (0+i\cdot (-1))=-36i \nonumber \]

3. For the quotient, we use the subtraction formula \(\ref{EQU:Quotient}\).

\[\dfrac{32\left(\cos\left(\dfrac{\pi}{4}\right)+i\sin\left(\dfrac{\pi}{4}\right)\right)}{8\left(\cos\left(\dfrac{7\pi}{12}\right)+i\sin\left(\dfrac{7\pi}{12}\right)\right)}=\dfrac{32}{8}\left(\cos\left(\dfrac{\pi}{4}-\dfrac{7\pi}{12}\right)+i\sin\left(\dfrac{\pi}{4}-\dfrac{7\pi}{12}\right)\right) \nonumber \]

Now, the difference in the argument of \(\cos\) and \(\sin\) is given by

\[\dfrac{\pi}{4}-\dfrac{7\pi}{12}=\dfrac{3\pi-7\pi}{12}=\dfrac{-4\pi}{12}=\dfrac{-\pi}{3} \nonumber \]

and \(\cos\left(-\dfrac{\pi}3\right)=\cos\left(\dfrac{\pi}3\right)=\dfrac 1 2\) and \(\sin\left(-\dfrac{\pi}3\right)=-\sin\left(\dfrac{\pi}3\right)=-\dfrac{\sqrt{3}}{2}\). With this, we obtain

\[\begin{aligned} \dfrac{32\left(\cos\left(\dfrac{\pi}{4}\right)+i\sin\left(\dfrac{\pi}{4}\right)\right)}{8\left(\cos\left(\dfrac{7\pi}{12}\right)+i\sin\left(\dfrac{7\pi}{12}\right)\right)}&= 4\left(\cos\left(\dfrac{-\pi}{3}\right)+i\sin\left(\dfrac{-\pi}{3}\right)\right)\\ &=4\cdot \left(\dfrac 1 2-i \dfrac{\sqrt{3}}{2}\right)\\&=2-2\sqrt{3} \cdot i \end{aligned} \nonumber \]

4. Finally, we calculate

\[\dfrac{4\left(\cos(203^\circ)+i\sin(203^\circ)\right)}{6(\cos(74^\circ)+i\sin(74^\circ))} = \dfrac{2}{3}\cdot \Big(\cos(129^\circ)+i\sin(129^\circ)\Big) \nonumber \]

Since we do not have exact values of \(\cos\) and \(\sin\) for the angle \(129^\circ\), we approximate the complex number in standard form with the calculator.

\[\begin{aligned} \dfrac{2}{3}\cdot \Big(\cos(129^\circ)+i\sin(129^\circ)\Big) &= \dfrac{2}{3}\cdot\cos(129^\circ)+i\cdot \dfrac{2}{3}\cdot\sin(129^\circ) \\ &\approx -0.420+ 0.518\cdot i\end{aligned} \nonumber \]

Note that here we approximated the solution to the nearest thousandth.