Q4P - Mathematical Methods in Physical Sciences

Question

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

${\mathbf{sinh}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{i}\mathbf{/}\mathbf{2}\mathbf{)}$

Step-by-Step Solution

Answer

The values of $z_1$ and $z_2$ are $i\left(\frac{\pi }{6}+2n\pi \right)$ and $i\left(\frac{5\pi }{6}+2n\pi \right)$ respectively

Step 1: Given Information

Given expression is ${\sinh }^{-1}(i/2)$.

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,${\sinh }^{-1}(i/2)$ .

Write the exponential form of the sinh(x).

$$\begin{gathered} \sinh \left(\mathrm{z}\right)=\frac{\mathrm{i}}{2} \\ \frac{{\mathrm{e}}^{\left(\mathrm{z}\right)}-{\mathrm{e}}^{\left(-\mathrm{z}\right)}}{2}=\frac{\mathrm{i}}{2} \end{gathered}$$

 

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

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

The coefficient of equation is as follows.

a = 1

b = -i

c = -1  

Use quadratic formula to find roots.

$$\begin{gathered} u=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ {u}_1=\frac{i+\sqrt{3}}{2} \\ {u}_2=\frac{i-\sqrt{3}}{2} \end{gathered}$$

Step 4: Find value of

Find $z_1$ by putting value of $u_1$

$$\begin{gathered} {\mathrm{z}}_1=\mathrm{In}\left(u_1\right) \\ =\mathrm{In}\left(r\right)+i\left(\theta +2n\pi \right) n=0,1,2,3,.... \\ =\mathrm{In}\left|\frac{\mathrm{i}+\sqrt{3}}{2}\right|+i\left(\theta +2n\pi \right) \\ =\mathrm{In}\left(1\right)+i\left(\frac{\pi }{6}+2n\pi \right) \\ =i\left(\frac{\pi }{6}+2n\pi \right) \end{gathered}$$

 

Find $z_2$ by putting value of $u_2$

$$\begin{gathered} {\mathrm{z}}_2=\mathrm{In}\left(u_2\right) \\ =\mathrm{In}\left(r\right)+i\left(\theta +2n\pi \right) n=0,1,2,3,.... \\ =\mathrm{In}\left|\frac{\mathrm{i}+\sqrt{3}}{2}\right|+i\left(\theta +2n\pi \right) \\ =\mathrm{In}\left(1\right)+i\left(\frac{5\pi }{6}+2n\pi \right) \\ =i\left(\frac{5\pi }{6}+2n\pi \right) \end{gathered}$$ 

Therefore, values of $z_1$ and $z_2$ are $i\left(\frac{\pi }{6}+2n\pi \right)$ and $i\left(\frac{5\pi }{6}+2n\pi \right)$ respectively.