21.2: Multiplication and division of complex numbers
- Page ID
- 54469
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)One interesting feature of the polar form of a complex number is that the multiplication and division are very easy to perform.
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
\[\begin{aligned}
\dfrac{r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right)}{r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right)}&=r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right) \cdot \dfrac{1}{r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right)} \\
&=r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right) \cdot \dfrac{1}{r_{2}}\left(\cos \left(-\theta_{2}\right)+i \sin \left(-\theta_{2}\right)\right)\\
&=\dfrac{r_{1}}{r_{2}} \cdot\left(\cos \left(\theta_{1}-\theta_{2}\right)+i \sin \left(\theta_{1}-\theta_{2}\right)\right)
\end{aligned} \nonumber \]
Multiply or divide the complex numbers, and write your answer in polar and standard form.
- \(5\left(\cos \left(11^{\circ}\right)+i \sin \left(11^{\circ}\right)\right) \cdot 8\left(\cos \left(34^{\circ}\right)+i \sin \left(34^{\circ}\right)\right)\)
- \(3\left(\cos \left(\dfrac{5 \pi}{8}\right)+i \sin \left(\dfrac{5 \pi}{8}\right)\right) \cdot 12\left(\cos \left(\dfrac{7 \pi}{8}\right)+i \sin \left(\dfrac{7 \pi}{8}\right)\right)\)
- \(\dfrac{32\left(\cos \left(\dfrac{\pi}{4}\right)+i \sin \left(\dfrac{\pi}{4}\right)\right)}{8\left(\cos \left(\dfrac{7 \pi}{12}\right)+i \sin \left(\dfrac{7 \pi}{12}\right)\right)}\)
- \(\dfrac{4\left(\cos \left(203^{\circ}\right)+i \sin \left(203^{\circ}\right)\right)}{6\left(\cos \left(74^{\circ}\right)+i \sin \left(74^{\circ}\right)\right)}\)
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 \(a + bi\).
- \[\begin{aligned}
5\left(\cos \left(11^{\circ}\right)+i \sin \left(11^{\circ}\right)\right) \cdot 8\left(\cos \left(34^{\circ}\right)+i \sin \left(34^{\circ}\right)\right) &=5 \cdot 8 \cdot\left(\cos \left(11^{\circ}+34^{\circ}\right)+i \sin \left(11^{\circ}+34^{\circ}\right)\right)=40\left(\cos \left(45^{\circ}\right)+i \sin \left(45^{\circ}\right)\right) \\
&=40\left(\dfrac{\sqrt{2}}{2}+i \dfrac{\sqrt{2}}{2}\right)\\
&=40 \dfrac{\sqrt{2}}{2}+i \cdot 40 \dfrac{\sqrt{2}}{2}\\
&=20 \sqrt{2}+20 \sqrt{2} i
\end{aligned} \nonumber \] - 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
\[36\left(\cos\left(\dfrac{3\pi}{2}\right)+i\sin\left(\dfrac{3\pi}{2}\right)\right) = 36 (0+i\cdot (-1))=-36i \nonumber \]
- 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
\[\begin{aligned} \dfrac{32\left(\cos\left(\dfrac{\pi}{4}\right)+i\sin\left(\dfrac{\pi}{4}\right)\right)}{8\left(\cos\left(\dfrac{7\pi}{12}\right)+i\sin\left(\dfrac{7\pi}{12}\right)\right)}&= 4\left(\cos\left(\dfrac{-\pi}{3}\right)+i\sin\left(\dfrac{-\pi}{3}\right)\right)\\ &=4\cdot \left(\dfrac 1 2-i \dfrac{\sqrt{3}}{2}\right)\\&=2-2\sqrt{3} \cdot i \end{aligned} \nonumber \]
- 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.