To find the \(n^{th}\) root of a complex number in polar form, we use the \(n^{th}\) Root Theorem or De Moivre’s Theorem and raise the complex number to a power with a rational exponent. Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent. Viewed 1k times 0 $\begingroup$ How would one convert $(1+i)^n$ to polar form… QUOTIENTS OF COMPLEX NUMBERS IN POLAR FORM, If \(z_1=r_1(\cos \theta_1+i \sin \theta_1)\) and \(z_2=r_2(\cos \theta_2+i \sin \theta_2)\), then the quotient of these numbers is, \[\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}[\cos(\theta_1−\theta_2)+i \sin(\theta_1−\theta_2) ],\space z_2≠0\], \[\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\space cis(\theta_1−\theta_2),\space z_2≠0\]. Calculate the new trigonometric expressions and multiply through by \(r\). The first step toward working with a complex number in polar form is to find the absolute value. [latex]z_{1}=3\text{cis}\left(120^{\circ}\right)\text{; }z_{2}=\frac{1}{4}\text{cis}\left(60^{\circ}\right)[/latex], 26. Plotting a complex number \(a+bi\) is similar to plotting a real number, except that the horizontal axis represents the real part of the number, \(a\), and the vertical axis represents the imaginary part of the number, \(bi\). A complex number is [latex]a+bi[/latex]. Let us find \(r\). Finding Powers of Complex Numbers in Polar Form. If [latex]\tan \theta =\frac{5}{12}[/latex], and [latex]\tan \theta =\frac{y}{x}[/latex], we first determine [latex]r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{{12}^{2}+{5}^{2}}=13\text{. If [latex]{z}_{1}={r}_{1}\left(\cos {\theta }_{1}+i\sin {\theta }_{1}\right)[/latex] and [latex]{z}_{2}={r}_{2}\left(\cos {\theta }_{2}+i\sin {\theta }_{2}\right)[/latex], then the product of these numbers is given as: Notice that the product calls for multiplying the moduli and adding the angles. Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. [latex]|z|=\sqrt{{x}^{2}+{y}^{2}}[/latex], [latex]\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ |z|=\sqrt{{\left(\sqrt{5}\right)}^{2}+{\left(-1\right)}^{2}}\hfill \\ |z|=\sqrt{5+1}\hfill \\ |z|=\sqrt{6}\hfill \end{array}[/latex], [latex]\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ |z|=\sqrt{{\left(3\right)}^{2}+{\left(-4\right)}^{2}}\hfill \\ |z|=\sqrt{9+16}\hfill \\ \begin{array}{l}|z|=\sqrt{25}\\ |z|=5\end{array}\hfill \end{array}[/latex], [latex]\begin{array}{l}x=r\cos \theta \hfill \\ y=r\sin \theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}[/latex], [latex]\begin{array}{l}z=x+yi\hfill \\ z=r\cos \theta +\left(r\sin \theta \right)i\hfill \\ z=r\left(\cos \theta +i\sin \theta \right)\hfill \end{array}[/latex], [latex]\begin{array}{l}\hfill \\ x=r\cos \theta \hfill \\ y=r\sin \theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}[/latex], [latex]\begin{array}{l}z=x+yi\hfill \\ z=\left(r\cos \theta \right)+i\left(r\sin \theta \right)\hfill \\ z=r\left(\cos \theta +i\sin \theta \right)\hfill \end{array}[/latex], [latex]\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{0}^{2}+{4}^{2}}\hfill \\ r=\sqrt{16}\hfill \\ r=4\hfill \end{array}[/latex], [latex]\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{\left(-4\right)}^{2}+\left({4}^{2}\right)}\hfill \\ r=\sqrt{32}\hfill \\ r=4\sqrt{2}\hfill \end{array}[/latex], [latex]\begin{array}{l}\cos \theta =\frac{x}{r}\hfill \\ \cos \theta =\frac{-4}{4\sqrt{2}}\hfill \\ \cos \theta =-\frac{1}{\sqrt{2}}\hfill \\ \theta ={\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi }{4}\hfill \end{array}[/latex], [latex]z=12\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)[/latex], [latex]\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\\\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}[/latex], [latex]z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)[/latex], [latex]\begin{array}{l}z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)\hfill \\ \text{ }=\left(12\right)\frac{\sqrt{3}}{2}+\left(12\right)\frac{1}{2}i\hfill \\ \text{ }=6\sqrt{3}+6i\hfill \end{array}[/latex], [latex]\begin{array}{l}z=13\left(\cos \theta +i\sin \theta \right)\hfill \\ =13\left(\frac{12}{13}+\frac{5}{13}i\right)\hfill \\ =12+5i\hfill \end{array}[/latex], [latex]z=4\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)[/latex], [latex]\begin{array}{l}\hfill \\ \begin{array}{l}{z}_{1}{z}_{2}={r}_{1}{r}_{2}\left[\cos \left({\theta }_{1}+{\theta }_{2}\right)+i\sin \left({\theta }_{1}+{\theta }_{2}\right)\right]\hfill \\ {z}_{1}{z}_{2}={r}_{1}{r}_{2}\text{cis}\left({\theta }_{1}+{\theta }_{2}\right)\hfill \end{array}\hfill \end{array}[/latex], [latex]\begin{array}{l}{z}_{1}{z}_{2}=4\cdot 2\left[\cos \left(80^\circ +145^\circ \right)+i\sin \left(80^\circ +145^\circ \right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\cos \left(225^\circ \right)+i\sin \left(225^\circ \right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right]\hfill \\ {z}_{1}{z}_{2}=-4\sqrt{2}-4i\sqrt{2}\hfill \end{array}[/latex], [latex]\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\left[\cos \left({\theta }_{1}-{\theta }_{2}\right)+i\sin \left({\theta }_{1}-{\theta }_{2}\right)\right],{z}_{2}\ne 0\\ \frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\text{cis}\left({\theta }_{1}-{\theta }_{2}\right),{z}_{2}\ne 0\end{array}[/latex], [latex]\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{2}{4}\left[\cos \left(213^\circ -33^\circ \right)+i\sin \left(213^\circ -33^\circ \right)\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[\cos \left(180^\circ \right)+i\sin \left(180^\circ \right)\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[-1+0i\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}+0i\hfill \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}\hfill \end{array}[/latex], [latex]\begin{array}{l}{z}^{n}={r}^{n}\left[\cos \left(n\theta \right)+i\sin \left(n\theta \right)\right]\\ {z}^{n}={r}^{n}\text{cis}\left(n\theta \right)\end{array}[/latex], [latex]\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}}\hfill \\ r=\sqrt{2}\hfill \end{array}[/latex], [latex]\begin{array}{l}\tan \theta =\frac{1}{1}\hfill \\ \tan \theta =1\hfill \\ \theta =\frac{\pi }{4}\hfill \end{array}[/latex], [latex]\begin{array}{l}{\left(a+bi\right)}^{n}={r}^{n}\left[\cos \left(n\theta \right)+i\sin \left(n\theta \right)\right]\hfill \\ {\left(1+i\right)}^{5}={\left(\sqrt{2}\right)}^{5}\left[\cos \left(5\cdot \frac{\pi }{4}\right)+i\sin \left(5\cdot \frac{\pi }{4}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=4\sqrt{2}\left[\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=4\sqrt{2}\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=-4 - 4i\hfill \end{array}[/latex], [latex]{z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left[\cos \left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)+i\sin \left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)\right][/latex], [latex]\begin{array}{l}{z}^{\frac{1}{3}}={8}^{\frac{1}{3}}\left[\cos \left(\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3}\right)+i\sin \left(\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3}\right)\right]\hfill \\ {z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{2k\pi }{3}\right)+i\sin \left(\frac{2\pi }{9}+\frac{2k\pi }{3}\right)\right]\hfill \end{array}[/latex], [latex]{z}^{\frac{1}{3}}=2\left(\cos \left(\frac{2\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}\right)\right)[/latex], [latex]\begin{array}{l}{z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{6\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}+\frac{6\pi }{9}\right)\right]\begin{array}{cccc}& & & \end{array}\text{ Add }\frac{2\left(1\right)\pi }{3}\text{ to each angle. For the following exercises, convert the complex number from polar to rectangular form. Evaluate the cube root of z when [latex]z=8\text{cis}\left(\frac{7\pi}{4}\right)[/latex]. It states that, for a positive integer [latex]n,{z}^{n}[/latex] is found by raising the modulus to the [latex]n\text{th}[/latex] power and … For \(k=1\), the angle simplification is, \[\begin{align*} \dfrac{\dfrac{2\pi}{3}}{3}+\dfrac{2(1)\pi}{3} &= \dfrac{2\pi}{3}(\dfrac{1}{3})+\dfrac{2(1)\pi}{3}\left(\dfrac{3}{3}\right) \\ &=\dfrac{2\pi}{9}+\dfrac{6\pi}{9} \\ &=\dfrac{8\pi}{9} \end{align*}\]. 4.0 License form is to find the absolute value of the complex plane Theorem let =. That for centuries had puzzled the greatest minds in science of it [! Using a rational exponent …, n - 1 [ /latex ] is! 5: May 15, 2017, 11:35 AM: Shawn Plassmann ċ. K=0,1,2,3, …, n - 1 [ /latex ] x\ ) write [ latex ] r [ ]. The its power off the complex plane much simpler than they appear ’. 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