Q21P - Mathematical Methods in Physical Sciences

Question

Follow steps (a), (b), (c) above to find all the values of the indicated roots.

$\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{i}\sqrt{\mathbf{3}}$  

Step-by-Step Solution

Answer

The value of $2+2\mathrm{i}\sqrt{3}$ is in the form of ${\mathrm{z}}_0 \mathrm{and} {\mathrm{z}}_1;{\mathrm{z}}_0=\sqrt{3}+\mathrm{i},{\mathrm{z}}_1=-\sqrt{3}-\mathrm{i}$.

Step 1: Given Information

The given expression is $2+2i\sqrt{3}.$

Step 2: Definition of 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: Find the value of r   and   θ

The Complex number is in the form $2+2\sqrt{3i}$.

 

$x=2,y=2\sqrt{3}$ 

 

The polar coordinates of the point are in the form of $z=r{e}^{i\theta }$.

 

$$\begin{gathered} r=\sqrt{x^2+y^2}=\sqrt{2^2+{\left(2\sqrt{3}\right)}^3}=4 \\ \theta =\frac{\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6} \end{gathered}$$ 

 

The equation $z=r{e}^{1\theta }$ can also be written in another form.

$$\begin{gathered} z^{\frac{1}{n}}={\left(r{e}^{1\theta }\right)}^{\frac{1}{n}} \\ {z}^{\frac{1}{n}}=r^{\frac{1}{n}}{e}^{\frac{1\theta }{n}} \\ {z}^{\frac{1}{n}=\sqrt[n]{r}}\left(\cos \frac{\theta }{n}+i s\mathrm{in}\frac{\theta }{n}\right) \end{gathered}$$                                                       ………..     (1)

When n =2, the equation becomes the $2^{nd}$root of the complex number.

$$\begin{gathered} z^{\frac{1}{2}}=r^{\frac{1}{2}}{e}^{\frac{1\theta }{2}} \\ r=4 \\ \theta =\frac{\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6} \end{gathered}$$ 

$$\begin{gathered} z_0=2e^{\frac{i\mathrm{\pi }}{6}} \\ =2\left(\cos \frac{\mathrm{\pi }}{6}+i \sin \frac{\mathrm{\pi }}{6}\right) \\ =\sqrt{3}+i \\ {z}_0=2e^{\frac{7i\mathrm{\pi }}{6}} \\ =2\left(\cos \frac{7\mathrm{\pi }}{6}+i \sin \frac{7\mathrm{\pi }}{6}\right) \\ =-\sqrt{3}-i \end{gathered}$$ 

 Hence, the value of $2+2i\sqrt{3}$ is in the form of $z_0$ and $z_1$:$z_0=\sqrt{3}+i,z_1=-\sqrt{3}-i$.