Q23P - Mathematical Methods in Physical Sciences

Question

Find the values of the indicated roots:

$\sqrt[\mathbf{4}]{\mathbf{8}\mathbf{i}\sqrt{\mathbf{3}}\mathbf{-}\mathbf{8}}$

Step-by-Step Solution

Answer

The values of the complex number $\sqrt[4]{8\mathrm{i}\sqrt{3}-8}$are as follows:

$$\begin{gathered} z_0=\sqrt{3}+i \\ {z}_1=-1+\sqrt{3i} \\ {z}_2=-\sqrt{3}-i \\ {z}_3=1-\sqrt{3i} \end{gathered}$$ 

Step 1: Given Information

The given expression is $\sqrt[4]{8\mathrm{i}\sqrt{3}-8}$.

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 $z=8i\sqrt{3}-8$.

The exponential form of z is given by $z=r\times e^{\theta i}$    

 

Find the modulus of the complex number z.

 

$$\begin{gathered} r=\sqrt{{\left(8\sqrt{3}\right)}^2+8^2} \\ =16 \end{gathered}$$ 

 

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 2^nd quadratic.

 

$$\begin{gathered} \theta =\mathrm{\pi }-\frac{\mathrm{\pi }}{3} \\ =\frac{2\mathrm{\pi }}{3} \end{gathered}$$ 

 

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

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

 

Step 4: Find the Roots

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

 

Solve z and $\theta$ for $k=0,1$.

$$\begin{gathered} {\theta }_0=\frac{\mathrm{\pi }}{6} \\ {z}_0=2e^{\mathrm{\pi }/6} \\ {\theta }_1=\frac{2\mathrm{\pi }}{3} \\ {z}_1=2e^{\left(2\mathrm{\pi }/3\right)} \end{gathered}$$ 

 

Solve z and $\theta$ for $k=2,3$.

$$\begin{gathered} {\theta }_2=\frac{7\mathrm{\pi }}{6} \\ {z}_0=2e^{\left(7\mathrm{\pi }/6\right)} \\ {\theta }_3=\frac{5\mathrm{\pi }}{3} \\ {z}_3=2e^{5\mathrm{\pi }/3} \end{gathered}$$ 

Step 5: Solving the Cartesian form of root

Solve for $z_0$.

$$\begin{gathered} z_0=2\left[\cos \left(\frac{\mathrm{\pi }}{6}\right)+i \sin \left(\frac{\mathrm{\pi }}{6}\right)\right] \\ =\sqrt{3}+i \end{gathered}$$ 

 

Solve for $z_1$ .

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

Solve for $z_2$ .

$$\begin{gathered} z_2=2\left[\cos \left(\frac{7\mathrm{\pi }}{6}\right)+i \sin \left(\frac{7\mathrm{\pi }}{6}\right)\right] \\ =\sqrt{3} -i \end{gathered}$$ 

 

Solve for $z_3$.

$$\begin{gathered} z_3=2\left[\cos \left(\frac{5\mathrm{\pi }}{3}\right)+i \sin \left(\frac{5\mathrm{\pi }}{3}\right)\right] \\ =1-\sqrt{3}i \end{gathered}$$ 

 

Hence, the values of the complex number $\sqrt[4]{8i\sqrt{3}-8}$ are as follows:

$$\begin{gathered} z_0=\sqrt{3}+i \\ {z}_1=-1+\sqrt{3i} \\ {z}_2=-\sqrt{3}-i \\ {z}_3=1-\sqrt{3i} \end{gathered}$$