21.3: Exercises
- Page ID
- 54470
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Plot the complex numbers in the complex plane.
- \(4+2i\)
- \(-3-5i\)
- \(6-2i\)
- \(-5+i\)
- \(-2i\)
- \(\sqrt{2}-\sqrt{2}i\)
- \(7\)
- \(i\)
- \(0\)
- \(2i-\sqrt{3}\)
- Answer
-
Add, subtract, multiply, and divide, as indicated.
- \((5-2i)+(-2+6i)\)
- \((-9-i)-(5-3i)\)
- \((3+2i)\cdot (4+3i)\)
- \((-2-i)\cdot (-1+4i)\)
- \(\dfrac{2+3i}{2+i}\)
- \((5+5i)\div (2-4i)\)
- Answer
-
- \(3+4 i\)
- \(-14+2 i\)
- \(6+17 i\)
- \(6-7 i\)
- \(\dfrac{7}{5}+\dfrac{4}{5} i\)
- \(-\dfrac{1}{2}+\dfrac{3}{2} i\)
Find the absolute value \(|a+bi|\) of the given complex number, and simplify your answer as much as possible.
- \(|4+3i|\)
- \(|1-2i|\)
- \(|-3i|\)
- \(|-2-6i|\)
- \(|\sqrt{8}-i|\)
- \(|-2\sqrt{3}-2i|\)
- \(|-5|\)
- \(|-\sqrt{17}+4\sqrt{2}i|\)
- Answer
-
- \(5\)
- \(\sqrt{5}\)
- \(3\)
- \(2 \sqrt{10}\)
- \(3\)
- \(4\)
- \(5\)
- \(7\)
Convert the complex number into polar form \(r(\cos(\theta)+i\sin(\theta))\).
- \(2+2i\)
- \(4\sqrt{3}+4i\)
- \(3-2i\)
- \(-5+5i\)
- \(4-3i\)
- \(-4+3i\)
- \(-\sqrt{5}-\sqrt{15}i\)
- \(\sqrt{7}-\sqrt{21}i\)
- \(-5-12i\)
- \(6i\)
- \(-10\)
- \(-\sqrt{3}+3i\)
- Answer
-
- \(2 \sqrt{2}\left(\cos \left(\dfrac{\pi}{4}\right)+i \sin \left(\dfrac{\pi}{4}\right)\right)\)
- \(8\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)\)
- approximately \(\sqrt{13}(\cos (-.588)+i \sin (-.588))\) or \(\sqrt{13}(\cos (-.187 \pi)+i \sin (-.187 \pi))\)
- \(5 \sqrt{2}\left(\cos \left(\dfrac{3 \pi}{4}\right)+i \sin \left(\dfrac{3 \pi}{4}\right)\right)\)
- approximately \(5(\cos (-.644)+i \sin (-.644))\) or \(5(\cos (-.205 \pi)+i \sin (-.205 \pi))\)
- approximately \(5(\cos (2.498)+i \sin (2.498))\) or \(5(\cos (.795 \pi)+i \sin (.795 \pi))\)
- \(2 \sqrt{5}\left(\cos \left(\dfrac{4 \pi}{3}\right)+i \sin \left(\dfrac{4 \pi}{3}\right)\right)\)
- \(2 \sqrt{7}\left(\cos \left(-\dfrac{\pi}{3}\right)+i \sin \left(-\dfrac{\pi}{3}\right)\right)\)
- approximately \(13(\cos (4.318)+i \sin (4.318))\) or \(13(\cos (1.374 \pi)+i \sin (1.374 \pi))\)
- \(6\left(\cos \left(\dfrac{\pi}{2}\right)+i \sin \left(\dfrac{\pi}{2}\right)\right)\)
- \(10(\cos (\pi)+i \sin (\pi))\)
- \(2 \sqrt{3}\left(\cos \left(\dfrac{2 \pi}{3}\right)+ i \sin \left(\dfrac{2 \pi}{3}\right)\right)\)
Convert the complex number into the standard form \(a+bi\).
- \(6(\cos(134^\circ)+i\sin(134^\circ))\)
- \(\dfrac 1 2 \left(\cos\left(\dfrac \pi {17}\right)+i\sin\left(\dfrac \pi {17}\right)\right)\)
- \(2(\cos(270^\circ)+i\sin(270^\circ))\)
- \(\cos\left(\dfrac{\pi} 6\right)+i\sin\left(\dfrac{\pi}6\right)\)
- \(10\left(\cos\left(\dfrac{7\pi}{6}\right)+i\sin\left(\dfrac{7\pi}{6}\right)\right)\)
- \(6 \left(\cos\left(-\dfrac{5\pi}{12}\right)+i\sin\left(-\dfrac{5\pi}{12}\right)\right)\)
- Answer
-
- approximately \(-4.168+4.316 i\)
- approximately \(.491+0.0919 i\)
- \(-2 i\)
- \(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2} i\)
- \(-5 \sqrt{3}-5 i\)
- approximately \(1.553-5 / 796 i\)
Multiply the complex numbers and write the answer in standard form \(a+bi\).
- \(4(\cos(27^\circ)+i\sin(27^\circ)) \cdot 10(\cos(33^\circ)+i\sin(33^\circ))\)
- \(7\left(\cos\left(\dfrac{2\pi}{9}\right)+i\sin\left(\dfrac{2\pi}{9}\right)\right) \cdot 6\left(\cos\left(\dfrac{\pi}{9}\right)+i\sin\left(\dfrac{\pi}{9}\right)\right)\)
- \(\left(\cos\left(\dfrac{13\pi}{12}\right)+i\sin\left(\dfrac{13\pi}{12}\right)\right) \cdot \left(\cos\left(\dfrac{-11\pi}{12}\right)+i\sin\left(\dfrac{-11\pi}{12}\right)\right)\)
- \(8\left(\cos\left(\dfrac{3\pi}{7}\right)+i\sin\left(\dfrac{3\pi}{7}\right)\right) \cdot 1.5\left(\cos\left(\dfrac{4\pi}{7}\right)+i\sin\left(\dfrac{4\pi}{7}\right)\right)\)
- \(0.2(\cos(196^\circ)+i\sin(196^\circ)) \cdot 0.5(\cos(88^\circ)+i\sin(88^\circ))\)
- \(4\left(\cos\left(\dfrac{7\pi}{8}\right)+i\sin\left(\dfrac{7\pi}{8}\right)\right) \cdot 0.25\left(\cos\left(\dfrac{-5\pi}{24}\right)+i\sin\left(\dfrac{-5\pi}{24}\right)\right)\)
- Answer
-
- \(40\left(\cos \left(60^{\circ}\right)+i \sin \left(60^{\circ}\right)\right)=20+20 \sqrt{3} i\)
- \(42\left(\cos \left(\dfrac{\pi}{3}\right)+i \sin \left(\dfrac{\pi}{3}\right) \right)=21+21 \sqrt{3} i\)
- \(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}+\dfrac{1}{2} i\)
- \(12(\cos (\pi)+i \sin (\pi))=-12\)
- \(.1\left(\cos \left(284^{\circ}\right)+i \sin \left(284^{\circ}\right)\right) \approx .0242-.0970 i\)
- \(\cos \left(\dfrac{2 \pi}{3}\right)+i \sin \left(\dfrac{2 \pi}{3}\right)=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2} i\)
Divide the complex numbers and write the answer in standard form \(a+bi\).
- \(\displaystyle\frac{18(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}))}{3(\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}))}\)
- \(\displaystyle\frac{10(\cos(254^\circ)+i\sin(254^\circ))}{15(\cos(164^\circ)+i\sin(164^\circ))}\)
- \(\displaystyle\frac{\sqrt{24}(\cos(\frac{11\pi}{14})+i\sin(\frac{11\pi}{14}))}{\sqrt{6}(\cos(\frac{2\pi}{7})+i\sin(\frac{2\pi}{7}))}\)
- \(\displaystyle\frac{\cos(\frac{8\pi}{5})+i\sin(\frac{8\pi}{5})}{2(\cos(\frac{\pi}{10})+i\sin(\frac{\pi}{10}))}\)
- \(\displaystyle\frac{42(\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4}))}{7(\cos(\frac{5\pi}{12})+i\sin(\frac{5\pi}{12}))}\)
- \(\displaystyle\frac{30(\cos(-175^\circ)+i\sin(-175^\circ))}{18(\cos(144^\circ)+i\sin(144^\circ))}\)
- Answer
-
- \(6(\cos (\pi / 3)+i \sin (\pi / 3))=3+3 \sqrt{3} i\)
- \(\dfrac{2}{3}\left(\cos \left(90^{\circ}\right)+i \sin \left(90^{\circ}\right)\right)=\dfrac{2}{3} i\)
- \(2(\cos (\pi / 2)+i \sin (\pi / 2))=2 i\)
- \(\dfrac{1}{2}\left(\cos \left(\dfrac{3 \pi}{2}\right)+i \sin \left(\dfrac{3 \pi}{2}\right)\right)=-\dfrac{1}{2} i\)
- \(6\left(\cos \left(\dfrac{4 \pi}{3}\right)+i \sin \left(\dfrac{4 \pi}{3}\right)\right)=-3-3 \sqrt{3} i\)
- \(\dfrac{5}{3}\left(\cos \left(-319^{\circ}\right)+i \sin \left(-319^{\circ}\right)\right) \approx 1.258+1.093 i\)