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# 6.7: De Moivre’s and the nth Root Theorems

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Use De Moivre’s Theorem to find the powers of complex numbers in polar form.
• Find the nth\begin{align*}n^{th}\end{align*} roots of complex numbers in polar form.

## De Moivre’s Theorem

The basic operations of addition, subtraction, multiplication and division of complex numbers have all been explored in this chapter. The addition and subtraction of complex numbers lent themselves best to numbers expressed in standard form. However multiplication and division were easily performed when the complex numbers were in polar form. Another operation that is performed using the polar form of complex numbers is the process of raising a complex number to a power.

The polar form of a complex number is r(cosθ+isinθ)\begin{align*}r(\cos \theta + i \sin \theta)\end{align*}. If we allow z\begin{align*}z\end{align*} to equal the polar form of a complex number, it is very easy to see the development of a pattern when raising a complex number in polar form to a power. To discover this pattern, it is necessary to perform some basic multiplication of complex numbers in polar form. Recall #3 and #7 from the Review Questions in the previous section.

If z=r(cosθ+isinθ)\begin{align*}z = r(\cos \theta + i \sin \theta)\end{align*} and z2=zz\begin{align*}z^2 = z \cdot z\end{align*} then:

z2z2z2=r(cosθ+isinθ)r(cosθ+isinθ)=r2[cos(θ+θ)+isin(θ+θ)]=r2(cos2θ+isin2θ)\begin{align*}z^2 &= r(\cos \theta + i \sin \theta) \cdot r(\cos \theta + i \sin \theta)\\ z^2 &= r^2 [\cos (\theta + \theta) + i \sin (\theta + \theta)]\\ z^2 &= r^2 (\cos 2\theta + i \sin 2\theta)\end{align*}

Likewise, if z=r(cosθ+isinθ)\begin{align*}z = r(\cos \theta + i \sin \theta)\end{align*} and z3=z2z\begin{align*}z^3 = z^2 \cdot z\end{align*} then:

z3z3z3=r2(cos2θ+isin2θ)r(cosθ+isinθ)=r3[cos(2θ+θ)+isin(2θ+θ)]=r3(cos3θ+isin3θ)\begin{align*}z^3 &= r^2(\cos 2\theta + i \sin 2\theta) \cdot r(\cos \theta + i \sin \theta)\\ z^3 &= r^3 [\cos (2\theta + \theta) + i \sin (2\theta + \theta)]\\ z^3 &= r^3 (\cos 3\theta + i \sin 3\theta)\end{align*}

Again, if z=r(cosθ+isinθ)\begin{align*}z = r(\cos \theta + i \sin \theta)\end{align*} and z4=z3z\begin{align*}z^4 = z^3 \cdot z\end{align*} then

z4=r4(cos4θ+isin4θ)\begin{align*}z^4 = r^4(\cos 4\theta + i \sin 4\theta)\end{align*}

These examples suggest a general rule valid for all powers of z\begin{align*}z\end{align*}, or n\begin{align*}n\end{align*}. We offer this rule and assume its validity for all n\begin{align*}n\end{align*} without formal proof, leaving that for later studies. The general rule for raising a complex number in polar form to a power is called De Moivre’s Theorem, and has important applications in engineering, particularly circuit analysis. The rule is as follows:

zn=[r(cosθ+isinθ)]n=rn(cosnθ+isinnθ)\begin{align*}z^n = [r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta)\end{align*}

Where z=r(cosθ+isinθ)\begin{align*}z = r(\cos \theta + i \sin \theta)\end{align*} and let n\begin{align*}n\end{align*} be a positive integer.

Notice what this rule looks like geometrically. A complex number taken to the n\begin{align*}n\end{align*}th power has two motions: First, its distance from the origin is taken to the n\begin{align*}n\end{align*}th power; second, its angle is multiplied by n\begin{align*}n\end{align*}. Conversely, the roots of a number have angles that are evenly spaced about the origin.

Example 1: Find. [2(cos120+isin120)]5\begin{align*}[2(\cos 120^\circ + i \sin 120^\circ)]^5\end{align*}

Solution: θ=120=2π3 rad\begin{align*}\theta=120^\circ=\frac{2\pi}{3} \ \text{rad}\end{align*}, using De Moivre’s Theorem:

\begin{align*}z^n=[r(\cos \theta+i \sin \theta)]^n &= r^n(\cos n\theta + i \sin n\theta)\\ [2(\cos 120^\circ + i \sin 120^\circ)]^5 &= 2^5 \left[\cos 5\frac{2\pi}{3}+i \sin 5\frac{2\pi}{3}\right]\\ &= 32\left (\cos \frac{10\pi}{3}+i \sin \frac{10\pi}{3} \right )\\ &= 32 \left(-\frac{1}{2}+-\frac{i\sqrt{3}}{2}\right)\\ &=-16-16i \sqrt{3}\end{align*}

Example 2: Find \begin{align*}\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{10}\end{align*}

Solution: Change into polar form.

\begin{align*}r &= \sqrt{x^2+y^2} && \theta = \tan^{-1} \left(\frac{\sqrt{3}}{2} \cdot -\frac{2}{1} \right)=-\frac{\pi}{3} + \pi=\frac{2\pi}{3}\\ r &= \sqrt{\left(\frac{-1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\\ r &= \sqrt{\frac{1}{4}+\frac{3}{4}}\\ r &= \sqrt{1}=1\end{align*}

The polar form of \begin{align*}\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)\end{align*} is \begin{align*}1 \left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right)\end{align*}

Now use De Moivre’s Theorem:

\begin{align*}z^n = [r(\cos \theta + i \sin \theta)]^n &= r^n(\cos n\theta+i \sin n\theta)\\ \left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{10} &= 1^{10}\left[\cos 10\left(\frac{2\pi}{3}\right)+i \sin 10 \left(\frac{2\pi}{3}\right)\right]\\ \left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{10} &= 1 \left(\cos \frac{20\pi}{3}+i \sin \frac{20\pi}{3}\right)\\ \left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{10} &= -\frac{1}{2}+i \frac{\sqrt{3}}{2} \rightarrow \ \text{Standard Form}\end{align*}

## nth Roots

We have explored all of the basic operations of arithmetic as they apply to complex numbers in standard form and in polar form. The last discovery is that of taking roots of complex numbers in polar form. Using De Moivre’s Theorem we can develop another general rule –one for finding the \begin{align*}n^{th}\end{align*} root of a complex number written in polar form.

As before, let \begin{align*}z = r(\cos \theta + i \sin \theta)\end{align*} and let the \begin{align*}n^{th}\end{align*} root of \begin{align*}z\end{align*} be \begin{align*}v = s (\cos \alpha + i \sin \alpha)\end{align*}. So, in general, \begin{align*}\sqrt[n]{z}=v\end{align*} and \begin{align*}v^n=z\end{align*}.

\begin{align*}\sqrt[n]{z} &= v\\ \sqrt[n]{r(\cos \theta+i \sin \theta)} &=s(\cos \alpha + i \sin \alpha)\\ [r(\cos \theta+i \sin \theta)]^{\frac{1}{n}} &= s(\cos \alpha +i \sin \alpha)\\ r^{\frac{1}{n}}\left(\cos \frac{1}{n} \theta+i \sin \frac{1}{n}\theta \right) &= s(\cos \alpha+i \sin \alpha)\\ r^{\frac{1}{n}}\left(\cos \frac{\theta}{n}+i \sin \frac{\theta}{n} \right) &= s(\cos \alpha+i \sin \alpha)\end{align*}

From this derivation, we can conclude that \begin{align*}r^{\frac{1}{n}}=s\end{align*} or \begin{align*}s^n=r\end{align*} and \begin{align*}\alpha=\frac{\theta}{n}\end{align*}. Therefore, for any integer \begin{align*}k (0, 1, 2, \ldots n -1)\end{align*}, \begin{align*}v\end{align*} is an \begin{align*}n^{th}\end{align*} root of \begin{align*}z\end{align*} if \begin{align*}s=\sqrt[n]{r}\end{align*} and \begin{align*}\alpha=\frac{\theta+2\pi k}{n}\end{align*}. Therefore, the general rule for finding the \begin{align*}n^{th}\end{align*} roots of a complex number if \begin{align*}z = r(\cos \theta + i \sin \theta)\end{align*} is: \begin{align*}\sqrt[n]{r} \left(\cos \frac{\theta+2\pi k}{n}+i \sin \frac{\theta+2\pi k}{n}\right)\end{align*}. Let’s begin with a simple example and we will leave \begin{align*}\theta\end{align*} in degrees.

Example 3: Find the two square roots of \begin{align*}2i\end{align*}.

Solution: Express \begin{align*}2i\end{align*} in polar form.

\begin{align*}& r=\sqrt{x^2+y^2} && \cos \theta=0\\ & r=\sqrt{(0)^2+(2)^2} && \qquad \theta=90^\circ\\ & r=\sqrt{4}=2\end{align*}

\begin{align*}(2i)^{\frac{1}{2}}=2^{\frac{1}{2}} \left(\cos \frac{90^\circ}{2}+i \sin \frac{90^\circ}{2}\right)=\sqrt{2}(\cos 45^\circ +i \sin 45^\circ)=1+i\end{align*}

To find the other root, add \begin{align*}360^\circ\end{align*} to \begin{align*}\theta\end{align*}.

\begin{align*}(2i)^{\frac{1}{2}}=2^{\frac{1}{2}} \left(\cos \frac{450^\circ}{2}+i \sin \frac{450^\circ}{2}\right)=\sqrt{2}(\cos 225^\circ +i \sin 225^\circ)=-1-i\end{align*}

Example 4: Find the three cube roots of \begin{align*}-2-2i \sqrt{3}\end{align*}

Solution: Express \begin{align*}-2-2i \sqrt{3}\end{align*} in polar form:

\begin{align*}r &=\sqrt{x^2+y^2}\\ r &= \sqrt{(-2)^2+(-2\sqrt{3})^2}\\ r &= \sqrt{16}=4 && \theta = \tan^{-1} \left(\frac{-2\sqrt{3}}{-2}\right)=\frac{4\pi}{3}\end{align*}

\begin{align*}& \sqrt[n]{r}\left( \cos \frac{\theta + 2\pi k}{n}+i \sin \frac{\theta + 2\pi k}{n}\right)\\ \sqrt[3]{-2-2i \sqrt{3}} &= \sqrt[3]{4} \left(\cos \frac{\frac{4 \pi}{3} + 2\pi k}{3}+i \sin \frac{\frac{4\pi}{3} +2\pi k}{3}\right) k=0, 1, 2\end{align*}

\begin{align*}z_1 &= \sqrt[3]{4}\left[ \cos \left(\frac{4\pi}{9}+\frac{0}{3}\right)+i \sin \left(\frac{4\pi}{9}+\frac{0}{3}\right)\right] && k=0\\ &= \sqrt[3]{4}\left[\cos \frac{4\pi}{9}+i \sin \frac{4\pi}{9}\right]\\ z_2 &= \sqrt[3]{4}\left[ \cos \left(\frac{4\pi}{9}+\frac{2\pi}{3}\right)+i \sin \left(\frac{4\pi}{9}+\frac{2\pi}{3}\right)\right] && k=1\\ &= \sqrt[3]{4}\left[\cos \frac{10\pi}{9}+i \sin \frac{10\pi}{9}\right]\\ z_3 &= \sqrt[3]{4}\left[ \cos \left(\frac{4\pi}{9}+\frac{4\pi}{3}\right)+i \sin \left(\frac{4\pi}{9}+\frac{4\pi}{3}\right)\right] && k=2\\ &= \sqrt[3]{4}\left[\cos \frac{16\pi}{9}+i \sin \frac{16\pi}{9}\right]\end{align*}

In standard form: \begin{align*}z_1=0.276+1.563i, z_2=-1.492-0.543i, z_3=1.216-1.02i\end{align*}.

## Solve Equations

The roots of a complex number are cyclic in nature. This means that when the roots are plotted on the complex plane, the \begin{align*}n^{th}\end{align*} roots are equally spaced on the circumference of a circle.

Since you began Algebra, solving equations has been an extensive topic. Now we will extend the rules to include complex numbers. The easiest way to explore the process is to actually solve an equation. The solution can be obtained by using De Moivre’s Theorem.

Example 5:

Consider the equation \begin{align*}x^5 - 32 = 0\end{align*}. The solution is the same as the solution of \begin{align*}x^5 = 32\end{align*}. In other words, we must determine the fifth roots of 32.

Solution:

\begin{align*}x^5-32 &= 0 \ \text{and} \ x^5=32.\\ r &= \sqrt{x^2+y^2}\\ r &= \sqrt{(32)^2+(0)^2}\\ r &= 32\\ \theta &= \tan^{-1} \left(\frac{0}{32}\right)=0\end{align*}

Write an expression for determining the fifth roots of \begin{align*}32 = 32 + 0i\end{align*}

\begin{align*}32^{\frac{1}{5}} &= [32( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{5}}\\ &= 2\left(\cos \frac{2\pi k}{5}+i \sin \frac{2\pi k}{5}\right)k=0, 1, 2, 3, 4\\ x_1 &= 2\left(\cos \frac{0}{5}+i \sin \frac{0}{5}\right) \rightarrow 2(\cos 0+i \sin 0)=2 && for \ k=0\\ x_2 &= 2\left(\cos \frac{2\pi}{5}+i \sin \frac{2\pi}{5}\right) \approx 0.62 + 1.9i && for \ k=1\\ x_3 &= 2\left(\cos \frac{4\pi}{5}+i \sin \frac{4\pi}{5}\right) \approx -1.62 + 1.18i && for \ k=2\\ x_4 &= 2\left(\cos \frac{6\pi}{5}+i \sin \frac{6\pi}{5}\right) \approx -1.62-1.18i && for \ k=3\\ x_5 &= 2\left(\cos \frac{8\pi}{5}+i \sin \frac{8\pi}{5}\right) \approx 0.62-1.9i && for \ k=4\end{align*}

## The Geometry of Complex Roots

In the previous example, we have one real and four complex roots. Plot these in the complex plane.

The \begin{align*}n^{th}\end{align*} roots of a complex number, when graphed on the complex plane, are equally spaced around a circle. So, instead of having all the roots, all that is necessary to graph the roots is one of them and the radius of the circle. For this particular example, the roots are \begin{align*}\frac{2\pi}{5}\end{align*} or \begin{align*}72^\circ\end{align*} apart (look in the root equation in the example, \begin{align*}\theta\end{align*} increased by \begin{align*}\frac{2\pi}{5}\end{align*}). This goes along with what we know about regular pentagons. The roots are \begin{align*}\frac{2\pi}{n}\end{align*} degrees apart.

Example 6: Calculate the three cube roots of 1 and represent them graphically.

Solution: In standard form, \begin{align*}1 = 1 + 0i \ r = 1\end{align*} and \begin{align*}\theta = 0\end{align*}. The polar form is \begin{align*}1 + 0i = 1 [\cos (0 + 2\pi k) + i \sin (0 + 2\pi k)]\end{align*}. The expression for determining the cube roots of \begin{align*}1 + 0i\end{align*} is:

\begin{align*}(1+0i)^{\frac{1}{3}}=1^{\frac{1}{3}} \left(\cos \frac{0+2 \pi k}{3}+i \sin \frac{0+2 \pi k}{3} \right)\end{align*}

When \begin{align*}k = 0, k = 1\end{align*} and \begin{align*}k = 2\end{align*} the three cube roots of 1 are \begin{align*}1, -\frac{1}{2}+i \frac{\sqrt{3}}{2}, -\frac{1}{2} -i \frac{\sqrt{3}}{2}\end{align*}. When these three roots are represented graphically, the three points, on the circle with a radius of 1 (the cubed root of 1 is 1), form a triangle.

## Points to Consider

• If the roots can be determined, will some form of De Moivre’s Theorem be used?
• If the root of a complex number in polar form can be determined, can the solution to an exponential equation be found in the same way?
• Does the number of roots have anything to do with the shape of the graph?

## Review Questions

1. Show that \begin{align*}z^3 = 1\end{align*}, if \begin{align*}z=-\frac{1}{2}+i \frac{\sqrt{3}}{2}\end{align*}
2. Evaluate:
1. \begin{align*}\left[\frac{\sqrt{2}}{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)\right]^8\end{align*}
2. \begin{align*}\left [3 \left (\sqrt{3}-i\sqrt{3} \right ) \right ]^4\end{align*}
3. \begin{align*}(\sqrt{5}-i)^7\end{align*}
4. \begin{align*}\left[3 \left(\cos \frac{\pi}{6}+i\sin \frac{\pi}{6}\right)\right]^{12}\end{align*}
3. Rewrite the following in rectangular form: \begin{align*}[2(\cos 315^\circ + i \sin 315^\circ)]^3\end{align*}
4. Find \begin{align*}\sqrt[3]{27i}\end{align*}.
5. Find the principal root of \begin{align*}(1 + i)^{\frac{1}{5}}\end{align*}. Remember the principal root is the positive root i.e. \begin{align*}\sqrt{9}=\pm 3\end{align*} so the principal root is +3.
6. Find the fourth roots of \begin{align*}81i\end{align*}.
7. Solve the equation \begin{align*}x^4 + 1 = 0\end{align*}. What shape do the roots make?
8. Solve the equation \begin{align*}x^3-64=0\end{align*}. What shape do the roots make?

1. Express \begin{align*}z\end{align*} in polar form: \begin{align*}r &= \sqrt{x^2+y^2}\\ r &= \sqrt{\left(-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\\ r &= \sqrt{\frac{1}{4}+\frac{3}{4}}=1\\ \theta &= \tan^{-1} \left(-\frac{\sqrt{3}}{1}\right)=120^\circ\end{align*} The polar form is \begin{align*}z = 1 (\cos 120^\circ + i \sin 120^\circ)\end{align*} \begin{align*}z^n &= [r(\cos \theta + i \sin \theta)^n = r(\cos n\theta + i \sin n\theta)\\ z^3 &= 1^3 [\cos3 (120^\circ) + i \sin (120^\circ)]\\ z^3 &= 1(\cos 360^\circ + i \sin 360^\circ)\\ z^3 &= 1(1 + 0i)\\ z^3 &= 1\end{align*}
2. a. \begin{align*}\left[\frac{\sqrt{2}}{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)\right]^8=\left(\frac{\sqrt{2}}{2}\right)^8 \left(\cos \frac{8\pi}{4}+i \sin \frac{8\pi}{4}\right)=\frac{1}{16}\cos 2\pi+ \frac{i}{16} \sin 2\pi = \frac{1}{16}\end{align*} b. \begin{align*}\left[3(\sqrt{3}-i\sqrt{3})\right]^4 &= (3\sqrt{3}-3i\sqrt{3})^4\\ r &= \sqrt{(3\sqrt{3})^2+(3\sqrt{3})^2}=3\sqrt{6}, \tan \theta=\frac{3\sqrt{3}}{3\sqrt{3}}=1 \rightarrow 45^\circ\\ &= \left(3\sqrt{6} \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\right)^4=(3\sqrt{6})^4 \left(\cos \frac{4\pi}{4}+i \sin \frac{4\pi}{4}\right)\\ &= 81(36)[-1+i(0)]=-2936\end{align*} c. \begin{align*}(\sqrt{5}-i)^7 \rightarrow r &= \sqrt{(\sqrt{5})^2+(-1)^2}=\sqrt{6}, \tan \theta=-\frac{1}{\sqrt{5}} \rightarrow \theta=335.9^\circ\\ [\sqrt{6}(\cos 335.9^\circ+i \sin 335.9^\circ)]^7 &= (\sqrt{6})^7(\cos (7 \cdot 335.9^\circ)+i \sin (7 \cdot 335.9^\circ))\\ &= 216 \sqrt{6}(\cos 2351.3^\circ+i \sin 2351.3^\circ)\\ &= 216 \sqrt{6}(-0.981-0.196i)\\ &= -519.04-103.7i\end{align*} d. \begin{align*}\left[3\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\right]^{12}=3^{12}(\cos 2\pi+i \sin 2\pi)=531, 441\end{align*}
3. \begin{align*}r &= 2 \ \text{and} \ \theta=315^\circ \ \text{or} \ \frac{7\pi}{4}.\\ z^n &= [r(\cos \theta+i \sin \theta)]^n=r^n(\cos n\theta+i \sin n\theta)\\ z^3 &= 2^3 \left[(\cos 3 \left(\frac{7\pi}{4}\right)+i \sin 3 \left(\frac{7\pi}{4}\right)\right]\\ z^3 &= 8 \left(\cos \frac{21\pi}{4}+i \sin \frac{21\pi}{4}\right)\\ z^3 &= 8 \left(-\frac{\sqrt{2}}{2}-i \frac{\sqrt{2}}{2}\right)\\ z^3 &= -4\sqrt{2}-4i\sqrt{2}\end{align*} \begin{align*}\frac{21\pi}{4}\end{align*} is in the third quadrant so both are negative.
4. \begin{align*}&&& a=0 \ and \ b=27\\ & \sqrt[3]{27i} =(0+27i)^{\frac{1}{3}} && x=0 \ and \ y=27\\ & \text{Polar Form} && r=\sqrt{x^2+y^2} \qquad \qquad \theta=\frac{\pi}{2}\\ &&& r=\sqrt{(0)^2+(27)^2}\\ &&& r=27\\ &&& \sqrt[3]{27i} = \left[27 \left(\cos (\frac{\pi}{2} + 2 \pi k) +i \sin (\frac{\pi}{2} + 2 \pi k) \right)\right]^{\frac{1}{3}} \text{for } k = 0, 1, 2\\ &&& \sqrt[3]{27i} = \sqrt[3]{27} \left[\cos \left(\frac{1}{3}\right) \left(\frac{\pi}{2} + 2 \pi k\right)+i \sin \left(\frac{1}{3}\right) \left(\frac{\pi}{2} + 2 \pi k\right)\right] \text{for } k = 0, 1, 2\\ &&& \sqrt[3]{27i} = 3 \left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right) \text{for } k = 0\\ &&& \sqrt[3]{27i} = 3 \left(\cos \frac{5\pi}{6}+i \sin \frac{5\pi}{6}\right) \text{for } k = 1\\ &&& \sqrt[3]{27i} = 3 \left(\cos \frac{9\pi}{6}+i \sin \frac{9\pi}{6}\right) \text{for } k = 2\\ &&& \sqrt[3]{27i} = 3\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right), 3\left(\frac{-\sqrt{3}}{2}+\frac{1}{2}i \right), -3i\end{align*}
5. \begin{align*}& r=\sqrt{x^2+y^2} && \theta=\tan^{-1} \left(\frac{1}{1}\right)=\frac{\sqrt{2}}{2} && \text{Polar Form} = \sqrt{2} cis \frac{\pi}{4}\\ & r=\sqrt{(1)^2+(1)^2}\\ & r=\sqrt{2}\end{align*} \begin{align*}(1+i)^{\frac{1}{5}} &= \left[\sqrt{2} \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\right]^{\frac{1}{5}}\\ (1+i)^{\frac{1}{5}} &= \sqrt{2}^{\frac{1}{5}}\left[\cos \left(\frac{1}{5}\right) \left(\frac{\pi}{4}\right)+i \sin \left(\frac{1}{5}\right) \left(\frac{\pi}{4}\right)\right]\\ (1+i)^{\frac{1}{5}} &= \sqrt[10]{2} \left(\cos \frac{\pi}{20}+i \sin \frac{\pi}{20} \right)\end{align*} In standard form \begin{align*}(1+i)^{\frac{1}{5}}=(1.06+1.06i)\end{align*} and this is the principal root of \begin{align*}(1+i)^{\frac{1}{5}}\end{align*}.
6. \begin{align*}81i\end{align*} in polar form is: \begin{align*}& r=\sqrt{0^2+81^2}=81, \tan \theta =\frac{81}{0}= und \rightarrow \theta=\frac{\pi}{2} \quad 81 \left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)\\ & \left[81 \left(\cos \left(\frac{\pi}{2}+2 \pi k \right)+i \sin \left(\frac{\pi}{2}+2\pi k\right)\right)\right]^{\frac{1}{4}}\\ & 3 \left(\cos \left(\frac{\frac{\pi}{2}+2\pi k}{4}\right)+i \sin \left(\frac{\frac{\pi}{2}+2 \pi k}{4}\right)\right)\\ & 3 \left(\cos \left(\frac{\pi}{8}+\frac{\pi k}{2}\right)+i \sin \left(\frac{\pi}{8}+\frac{\pi k}{2}\right)\right)\\ & z_1 =3 \left(\cos \left(\frac{\pi}{8}+\frac{0 \pi}{2}\right)+i \sin \left(\frac{\pi}{8}+\frac{0 \pi}{2}\right)\right)=3 \cos \frac{\pi}{8}+3i \sin \frac{\pi}{8}=2.77+1.15i\\ & z_2 =3 \left(\cos \left(\frac{\pi}{8}+\frac{\pi}{2}\right)+i \sin \left(\frac{\pi}{8}+\frac{\pi}{2}\right)\right)=3 \cos \frac{5 \pi}{8}+3i \sin \frac{5 \pi}{8}=-1.15+2.77i\\ & z_3 =3 \left(\cos \left(\frac{\pi}{8}+\frac{2\pi}{2}\right)+i \sin \left(\frac{\pi}{8}+\frac{2\pi}{2}\right)\right)=3 \cos \frac{9\pi}{8}+3i \sin \frac{9 \pi}{8}=-2.77-1.15i\\ & z_4 =3 \left(\cos \left(\frac{\pi}{8}+\frac{3\pi}{2}\right)+i \sin \left(\frac{\pi}{8}+\frac{3\pi}{2}\right)\right)=3 \cos \frac{13 \pi}{8}+3i \sin \frac{13 \pi}{8}=1.15-2.77i\end{align*}
7. \begin{align*}& x^4+1=0 && r=\sqrt{x^2+y^2}\\ & x^4=-1 && r=\sqrt{(-1)^2+(0)^2}\\ & x^4=-1+0i && r=1\\ & \theta = \tan^{-1} \left(\frac{0}{-1}\right)+\pi=\pi\end{align*} Write an expression for determining the fourth roots of \begin{align*}x^4 = -1 + 0i\end{align*} \begin{align*}(-1+0i)^{\frac{1}{4}} &= [1(\cos (\pi+2\pi k)+i \sin (\pi+2\pi k))]^{\frac{1}{4}}\\ (-1+0i)^{\frac{1}{4}} &= 1^{\frac{1}{4}} \left(\cos \frac{\pi+2\pi k}{4}+i \sin \frac{\pi+2\pi k}{4}\right)\\ x_1 &= 1 \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} && \text{for} \ k=0\\ x_2 &= 1 \left(\cos \frac{3\pi}{4}+i \sin \frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} && \text{for} \ k=1\\ x_3 &= 1 \left(\cos \frac{5\pi}{4}+i \sin \frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} && \text{for} \ k=2\\ x_4 &= 1 \left(\cos \frac{7\pi}{4}+i \sin \frac{7\pi}{4}\right)=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} && \text{for} \ k=3\end{align*} If a line segment is drawn from each root on the polar plane to its adjacent roots, the four roots will form the corners of a square.
8. \begin{align*}& x^3-64=0 \rightarrow x^3 = 64+0i\\ & 64 + 0i = 64(\cos (0 + 2 \pi k) + i \sin (0 + 2 \pi k))\\ & x = (x^3)^{\frac{1}{3}} = (64 + 0i)^{\frac{1}{3}} = \sqrt[3]{64} \left(\cos \left(\frac{0+2\pi k}{3}\right)+i \sin \left(\frac{0+2\pi k}{3}\right)\right)\\ & z_1 =4 \left(\cos \left(\frac{0+2\pi 0}{3}\right)+i \sin \left(\frac{0+2\pi 0}{3}\right)\right)=4 \cos 0 + 4i \sin 0 = 4 \text{ for } \ k=0\\ & z_2 =4 \left(\cos \left(\frac{0+2\pi}{3}\right)+i \sin \left(\frac{0+2\pi}{3}\right)\right)=4 \cos \frac{2\pi}{3}+4i \sin \frac{2\pi}{3}=-2 + 2i \sqrt{3} \text{ for } \ k=1\\ & z_3 =4 \left(\cos \left(\frac{0+4\pi}{3}\right)+i \sin \left(\frac{0+4\pi}{3}\right)\right)=4 \cos \frac{4\pi}{3}+4i \sin \frac{4\pi}{3}=-2-2i \sqrt{3} \text{ for } \ k=2\end{align*} If a line segment is drawn from each root on the polar plane to its adjacent roots, the three roots will form the vertices of an equilateral triangle.

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