Question

Verify the results given for the roots in Example 4. You can find the exact values in terms of $\sqrt{\mathbf{3}}$ by using trigonometric addition formulas or more easily by using a computer to solve ${\mathit{z}}^{\mathbf{6}}\mathbf{=}\mathbf{-}\mathbf{8}\mathit{i}$. (You still may have to do a little work by hand to put the computer’s solution into the given form.)

Short answer

The value of the$z^6=-8i$are,

$$\begin{gathered} z_0=1+i \\ {z}_1=-0.366+1.366i \\ {z}_2=-1.366+0.366i \\ {z}_3=-1-i \\ {z}_4=0.366-1.366i \\ {z}_5=1.366-0.366i \end{gathered}$$

Step-by-step solution

Given Information.

The given equation is $z^6=-8i$.

Definition of Power series

A power series is an infinite series that looks like :

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}{\left(x-c\right)}^{\mathbf{n}}\mathbf{=}{\mathit{a}}_{\mathbf{0}}\mathbf{+}{\mathit{a}}_{\mathbf{1}}\left(x-c\right)\mathbf{+}{\mathit{a}}_{\mathbf{2}}{\left(x-c\right)}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}$
Where${\mathit{a}}_{\mathbf{n}}$ represents thecoefficient of the nthterm and cis a constant.

Write in exponential form.

Write in the exponential form of $z^6=-8i$.

$$\begin{gathered} z^6=-8i \\ {z}^6=-8e^{\left(3\pi i/2\right)} \\ {z}^6={\left(8e^{\left(3\pi i/2\right)}\right)}^{1/6} \\ {z}_k=R{e}^{\left({\theta }_{k}i\right)} \end{gathered}$$

All the roots have the same radius.

$$\begin{gathered} R={\left(8\right)}^{1/6} \\ R=\sqrt{2} \end{gathered}$$

Write the general form of the angle.

${\theta }_k=\frac{{\displaystyle \frac{3\pi }{2}}+2\pi k}{n}$

Substitute the value of .

Put , k = 0and we get,

$$\begin{gathered} {\theta }_o=\frac{\pi }{4} \\ {z}_o=\sqrt{2}{e}^{\left(\pi i/4\right)} \end{gathered}$$

Put , k = 1 and we get,

$$\begin{gathered} {\theta }_1=\frac{7\pi }{12} \\ {z}_1=\sqrt{2}{e}^{\left(7\pi i/12\right)} \end{gathered}$$

Put , k = 2and we get,

$$\begin{gathered} {\theta }_2=\frac{11\pi }{12} \\ {z}_2=\sqrt{2}{e}^{\left(11\pi i/12\right)} \end{gathered}$$

Put , k = 3 and we get,

$$\begin{gathered} {\theta }_3=\frac{5\pi }{4} \\ {z}_3=\sqrt{2}{e}^{\left(5\pi i/4\right)} \end{gathered}$$

Put k = 4 , and we get,

$$\begin{gathered} {\theta }_4=\frac{19\pi }{12} \\ {z}_4=\sqrt{2}{e}^{\left(19\pi i/12\right)} \end{gathered}$$

Put k = 5, and we get,

$$\begin{gathered} {\theta }_5=\frac{23\pi }{12} \\ {z}_5=\sqrt{2}{e}^{\left(23\pi i/12\right)} \end{gathered}$$

Write the rectangular form of the roots

Put the value in the formula to find the rectangular form of roots as:

$$\begin{gathered} z_0={\sqrt{2e}}^{\left(\pi i/4\right)} \\ {z}_0=\sqrt{2}\left(\cos \left(\pi /4\right)+i\sin \left(\pi /4\right)\right) \\ {z}_0=1+i \end{gathered}$$

Find another root as:

$$\begin{gathered} z_1={\sqrt{2e}}^{\left(7\pi i/12\right)} \\ {z}_1=\sqrt{2}\left(\cos \left(7\pi /12\right)+i\sin \left(7\pi /12\right)\right) \\ {z}_1=\frac{1-\sqrt{3}}{2}+i\frac{1+\sqrt{3}}{2} \\ {z}_1=-0.366+1.366i \end{gathered}$$

Find another root as:

$$\begin{gathered} z_2={\sqrt{2e}}^{\left(11\pi i/12\right)} \\ {z}_2=\sqrt{2}\left(\cos \left(11\pi /12\right)+i\sin \left(11\pi /12\right)\right) \\ {z}_2=\frac{1+\sqrt{3}}{2}+i\frac{-1+\sqrt{3}}{2} \\ {z}_2=-1.366+0.366i \end{gathered}$$

Write the rectangular form of the roots.

Find another root as:

$$\begin{gathered} z_3={\sqrt{2e}}^{\left(5\pi i/4\right)} \\ {z}_3=\sqrt{2}\left(\cos \left(5\pi /4\right)+i\sin \left(5\pi /4\right)\right) \\ {z}_3=-1-i \end{gathered}$$

Find another root as:

$$\begin{gathered} z_4={\sqrt{2e}}^{\left(19\pi i/12\right)} \\ {z}_4=\sqrt{2}\left(\cos \left(19\pi /12\right)+i\sin \left(19\pi /12\right)\right) \\ {z}_4=\frac{-1+\sqrt{3}}{2}-i\frac{1+\sqrt{3}}{2} \\ {z}_4=0.366-1.366i \end{gathered}$$

Find another root as:

$$\begin{gathered} z_5=\sqrt{2}ex{p}^{\left(23\pi i/12\right)} \\ {z}_5=\sqrt{2}\left(\cos \left(23\pi /12\right)+i\sin \left(23\pi /12\right)\right) \\ {z}_5=\frac{1+\sqrt{3}}{2}+i\frac{1-\sqrt{3}}{2} \\ {z}_5=1.366-0.366i \end{gathered}$$

Therefore, the roots of z are,

$$\begin{gathered} z_0=1+i \\ {z}_1=-0.366+1.366i \\ {z}_2=-1.366+0.366i \\ {z}_3=-1-i \\ {z}_4=0.366-1.366i \\ {z}_5=1.366-0.366i \end{gathered}$$