Question

Find the values of the indicated roots.

$\sqrt[\mathbf{3}]{\mathbf{2}\mathbf{i}\mathbf{-}\mathbf{2}}$

Short answer

The values of $\sqrt[3]{2\mathrm{i}-2}$are as follows:

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

Step-by-step solution

Given Information

The given expression is$\sqrt[3]{2\mathrm{i}-2}$

Definition of the complex number

Complex numbers comprise real numbers and imaginary numbers; a complex can be written in the form of:

$\mathit{z}\mathbf{=}\mathit{a}\mathbf{+}\mathit{i}\mathit{b}$

Here a and b are real numbers, and i is the imaginary number which is known as iota, whose value is $\sqrt{\mathbf{-}\mathbf{1}}$.

Solving the Equation

Let$z=2i-2$.

The exponential form of z is given by$z=r\times exp\left({\theta }_i\right)$ …….(1)

Find the modulus of the complex number z.

$r=\sqrt{2^2+2^2}=2\sqrt{2}$

Find the angle of the complex number z.

$\theta =arc\tan \left(-1\right)=-\frac{\mathrm{\pi }}{4}$

Find the angle in the 2^nd quadratic.

$\theta =\mathrm{\pi }-\frac{\mathrm{\pi }}{4}=\frac{3\mathrm{\pi }}{4}$

Hence the root is given by:

$z_k=r^{\frac{1}{n}}exp\left({\theta }_{k}i\right)$ ……. (2)

Value of angle${\theta }_k$ is${\theta }_k=\frac{{\displaystyle \frac{3\mathrm{\pi }}{4}}+2\mathrm{\pi k}}{3}$ ……. (3)

Find roots for different values of z and $\theta$.

Solve z and$\theta$ for $k=0,1$.

$$\begin{gathered} {\theta }_0=\frac{\mathrm{\pi }}{4} \\ {z}_0=\sqrt{2}exp\left(\mathrm{\pi }/4\right) \\ {\theta }_1=\frac{11\mathrm{\pi }}{12} \\ {z}_1=\sqrt{2}exp\left(11\mathrm{\pi }/12\right) \end{gathered}$$

Solve z and $\theta$for $k=2$.

$$\begin{gathered} {\theta }_2=\frac{19\mathrm{\pi }}{12} \\ {z}_2=\sqrt{2}exp\left(19\mathrm{\pi }/12\right) \end{gathered}$$

Solving the Cartesian form of root

Solve for$Z_0$.

$$\begin{gathered} z_0=\sqrt{2}\left[\cos \left(\frac{\mathrm{\pi }}{4}\right)+i\sin \left(\frac{\mathrm{\pi }}{4}\right)\right] \\ =1+i \end{gathered}$$

Solve for $z_1$.

$$\begin{gathered} z_1=\sqrt{2}\left[\cos \left(\frac{11\mathrm{\pi }}{12}\right)+\sin \left(\frac{11\mathrm{\pi }}{12}\right)\right] \\ =\frac{\left(-1+\sqrt{3}\right)+i\left(-1+\sqrt{3}\right)}{2} \\ =-1.366+0.366i \end{gathered}$$

Solve for $z_2$.

$$\begin{gathered} z_2=\sqrt{2}\left[\cos \left(\frac{19\mathrm{\pi }}{12}\right)+i\sin \left(\frac{19\mathrm{\pi }}{12}\right)\right] \\ =\frac{\left(-1+\sqrt{3}\right)+i\left(-1-\sqrt{3}\right)}{2} \\ =0.366-1.366i \end{gathered}$$

Hence, the values of the complex number$\sqrt[3]{2i-2}$ are as follows:

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