Q5P - Mathematical Methods in Physical Sciences

Question

Question: Find each of the following in the x + iy form and compare a computer solution.

$\mathbf{arccos}\left(i\sqrt{8}\right)$

Step-by-Step Solution

Answer

The x + iy form is $z=\pm \left(\frac{\pi }{2}+2n\pi -In\left(2\sqrt{2}+3\right)\right)$.

Step 1: Given Information

Given expression is, $\arccos \left(\mathrm{i}\sqrt{8}\right)$.

Step 2: Definition of Trigonometric equation.

A trigonometric equation is one that has one or more trigonometric ratios with unknown angles.

Step 3: Use exponential form to expand the equation

Given the function is, $\arccos \left(\mathrm{i}\sqrt{8}\right)$.

 

Write the exponential form of the  cos(z).

$$\begin{gathered} z=\arccos \left(\mathrm{i}\sqrt{8}\right) \\ \cos \left(z\right)=i\sqrt{8} \\ \frac{e^{\left(zi\right)}+e^{\left(-zi\right)}}{2}=i\sqrt{8} \end{gathered}$$

 

Let $u=e^{\left(zi\right)}$ in equation (2).

$$\begin{gathered} u+\frac{1}{u}=4\sqrt{2}i \\ {u}^2-4\sqrt{2}iu+1=0 \end{gathered}$$

 

The coefficients of equation are as follows.

a = 1

b = -4$\sqrt{2}i$

c = 1 

Use quadratic formula to find roots.

$$\begin{gathered} u=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ =\frac{4\sqrt{2}i\pm \sqrt{-32-4}}{2} \\ =\frac{4\sqrt{2}i\pm 6i}{2} \\ =2\sqrt{2}i\pm 3i \end{gathered}$$

Step 4: Find value of z 1

Find $z_1$ by putting value of $u_1$

$$\begin{gathered} zi=In\left(u_1\right) \\ zi=In\left(r\right)+i\left(\theta +2n\pi \right) n=0,12,3,.... \\ =In\left|\frac{i+\sqrt{3}}{2}\right|+i\left(\theta +2n\pi \right) \\ =In\left(2\sqrt{2}+3i\right) \\ =In\left(2\sqrt{2}+3\right)+i\left(\frac{\pi }{2}+2n\pi \right) \end{gathered}$$

Use value of zi.

$$\begin{gathered} z_1=\frac{zi}{i} \\ =\frac{\mathrm{In}\left(2\sqrt{2}+3\right)+i\left(\frac{\pi }{2}\pm 2n\pi \right)}{i} \\ =-i\mathrm{In}\left(2\sqrt{2}+3\right)+i\left(\frac{\pi }{2}\pm 2n\pi \right) \\ =\left(\frac{\pi }{2}\pm 2n\pi \right)-1.76i \end{gathered}$$

 

Find $z_2$ by putting value of $u_2$

$$\begin{gathered} zi=In\left(u_1\right) \\ =In\left(r\right)+i\left(\theta +2n\pi \right) n=0,12,3,.... \\ =In\left|\frac{i+\sqrt{3}}{2}\right|+i\left(\theta +2n\pi \right) \\ =In\left(2\sqrt{2}-3i\right) \\ =In\left(-2\sqrt{2}+3\right)+i\left(-\frac{n}{2}+2n\pi \right) \end{gathered}$$

Use value of zi.

$$\begin{gathered} z_2=\frac{zi}{i} \\ =\frac{\mathrm{In}\left(-2\sqrt{2}+3\right)+i\left(-\frac{\pi }{2}\pm 2n\pi \right)}{i} \\ =-i\mathrm{In}\left(-2\sqrt{2}+3\right)+i\left(-\frac{\pi }{2}\pm 2n\pi \right) \\ =\left(-\frac{\pi }{2}\pm 2n\pi \right)+1.76i \end{gathered}$$

 

Combine $z_1$ and $z_2$ to find z.

$$\begin{gathered} z=\mp i\mathrm{In}\left(2\sqrt{2}+3\right)+\left(\pm \frac{\pi }{2}\pm 2n\pi \right) \\ z=\pm \left(\frac{\pi }{2}+2n\pi -i\mathrm{In}\left(2\sqrt{2}+3\right)\right) \end{gathered}$$

Therefore, the x + iy form is $\mathrm{z}=\pm \left(\frac{\mathrm{\pi }}{2}\pm 2\mathrm{n\pi }-\mathrm{In}\left(2\sqrt{2}+3\right)\right)$.