Q22P - Mathematical Methods in Physical Sciences

Question

Find the values of the indicated roots.

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

Step-by-Step Solution

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 1: Given Information

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

Step 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 $\sqrt{\mathbf{-}\mathbf{1}}$.

Step 3: 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 $k=2$.

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

Step 4: 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:

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