# 21.2: Multiplication and division of complex numbers

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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

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

1. 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

1. 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.

This page titled 21.2: Multiplication and division of complex numbers is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.