Q24P - Mathematical Methods in Physical Sciences

Question

Find the values of the indicated roots.

$\sqrt[\mathbf{8}]{\frac{\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{I}\sqrt{\mathbf{3}}}{\mathbf{2}}}$

Step-by-Step Solution

Answer

The values of the complex number $\sqrt[8]{\frac{-1-\mathrm{I}\sqrt{3}}{2}}$are:

$$\begin{gathered} z_0=0.866+0.5i \\ {z}_1=0.259-0.966i \\ {z}_2=-0.5-0.866i \\ {z}_3=-0.966+0.259i \end{gathered}$$ 

$$\begin{gathered} z_4=-0.259-0.5i, \\ {z}_5=-0.259-0.966i, \\ {z}_6=0.5-0.866i, \\ {z}_7=0.966-0.259i \end{gathered}$$ 

Step 1: Given Information

The given expression is$\sqrt[8]{\frac{-1-\mathrm{I}\sqrt{3}}{2}}$

Step 2: Definition of Complex Number

Complex numbers are represented in terms of 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 $z=\frac{-1-i\sqrt{3}}{2}$.

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

                                                                       

 

Find the modulus of the complex number z.

$$\begin{gathered} r=\sqrt{{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2+{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2} \\ =1 \end{gathered}$$

Find the angle of the complex number z.

 

$$\begin{gathered} \theta =arc\tan \left(\sqrt{3}\right) \\ =\frac{\mathrm{\pi }}{3} \end{gathered}$$ 

 

Find the angle in the 3^rd quadratic as:

$$\begin{gathered} \theta =\mathrm{\pi }+\frac{\mathrm{\pi }}{4} \\ =\frac{4\mathrm{\pi }}{3} \end{gathered}$$ 

 

Hence the root is $z_k=r^{\frac{1}{n}}exp\left({\theta }_{k}i\right)$ .

Angle ${\theta }_k$ is written as ${\theta }_k=\frac{{\displaystyle \frac{4\mathrm{\pi }}{3}}+2\mathrm{\pi k}}{8}$.

Step 4 : Roots in Exponential Form

Find the roots of the complex number z for different values of $\theta$  

 

Solve z and $\theta$ for $k=0,1$ .

 

 $$\begin{gathered} {\theta }_0=\frac{\mathrm{\pi }}{6} \\ {z}_0=e^{\left(\mathrm{\pi }/6\right)} \\ {\theta }_1=\frac{5\mathrm{\pi }}{12} \\ {z}_1=e^{\left(5\mathrm{\pi }/12\right)} \end{gathered}$$

 

 Solve z and $\theta$ for $k=2,3$.

$$\begin{gathered} {\theta }_2=\frac{2\mathrm{\pi }}{3} \\ {z}_2=e^{\left(2\mathrm{\pi }/3\right)} \\ {\theta }_3=\frac{11\mathrm{\pi }}{12} \\ z3=e^{11\mathrm{\pi }/12} \end{gathered}$$ 

 

Solve z and $\theta$ for $k=4,5$.

$$\begin{gathered} {\theta }_4=\frac{7\mathrm{\pi }}{6} \\ {z}_4=e^{7\mathrm{\pi }/6} \\ {\theta }_5=\frac{17\mathrm{\pi }}{12} \\ {z}_5=e^{17\mathrm{\pi }/12} \end{gathered}$$

Solve z and $\theta$ for $k=6,7$.

$$\begin{gathered} {\theta }_6=\frac{5\mathrm{\pi }}{3} \\ {z}_6=e^{5\mathrm{\pi }/3} \\ {\theta }_7=\frac{23\mathrm{\pi }}{12} \\ {z}_7=e^{23\mathrm{\pi }/12} \end{gathered}$$ 

Step 5: Solving the Cartesian form of root

Solve for $z_0$

$$\begin{gathered} z_0=\left[\cos \left(\frac{\mathrm{\pi }}{6}\right)+i \sin \left(\frac{\mathrm{\pi }}{6}\right)\right] \\ =0.866+0.5i \end{gathered}$$  

 

Solve for $z_1$ .

$$\begin{gathered} z_1=\left[\cos \left(\frac{5\mathrm{\pi }}{12}\right)+i \sin \left(\frac{5\mathrm{\pi }}{12}\right)\right] \\ =0.259+0.966i \end{gathered}$$ 

 

Solve for $z_2$ .

$$\begin{gathered} z_2=\left[\cos \left(\frac{2\mathrm{\pi }}{3}\right)+i \sin \left(\frac{2\mathrm{\pi }}{3}\right)\right] \\ =0.5+0.866i \end{gathered}$$ 

 

Solve for $z_3$ .

$$\begin{gathered} z_3=\left[\cos \left(\frac{11\mathrm{\pi }}{12}\right)+i \sin \left(\frac{11\mathrm{\pi }}{12}\right)\right] \\ =0.966+0.259i \end{gathered}$$

Solve for $z_4$ .

$$\begin{gathered} z_4=\left[\cos \left(\frac{7\mathrm{\pi }}{12}\right)+i \sin \left(\frac{7\mathrm{\pi }}{12}\right)\right] \\ =0.259-0.5i \end{gathered}$$ 

 

Solve for $z_5$ .

$$\begin{gathered} z_5=\left[\cos \left(\frac{17\mathrm{\pi }}{12}\right)+i \sin \left(\frac{17\mathrm{\pi }}{12}\right)\right] \\ =0.259+0.966i \end{gathered}$$ 

 

Solve for $z_6$ .

$$\begin{gathered} z_6=\left[\cos \left(\frac{5\mathrm{\pi }}{3}\right)+i \sin \left(\frac{5\mathrm{\pi }}{3}\right)\right] \\ =0.5-0.866i \end{gathered}$$ 

 

Solve for $z_7$ .

$$\begin{gathered} z_7=\left[\cos \left(\frac{23\mathrm{\pi }}{12}\right)+i \sin \left(\frac{23\mathrm{\pi }}{12}\right)\right] \\ =0.966-0.259i \end{gathered}$$

Hence, the values of the complex number $\sqrt[8]{\frac{-1-\mathrm{I}\sqrt{3}}{2}}$ are:

$$\begin{gathered} z_0=0.866+0.5i \\ {z}_1=0.259-0.966i \\ {z}_2=-0.5-0.866i \\ {z}_3=-0.966+0.259i \end{gathered}$$

$$\begin{gathered} z_4=-0.259-0.5i, \\ {z}_5=-0.259-0.966i, \\ {z}_6=0.5-0.866i, \\ {z}_7=0.966-0.259i \end{gathered}$$