Q1P - Mathematical Methods in Physical Sciences

Question

Find each of the following in the x + iyform and compare a computer solution$\mathit{a}\mathit{r}\mathit{c}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$.

 

Step-by-Step Solution

Answer

The value of $\arcsin \left(2\right)$ is $\arcsin \left(2\right)\pm i\ln (2+\sqrt{3})+(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\pm 2n\pi )$Where.$n=0,1,2,3,\mathrm{...}$

Step 1: Given Information.

The given trigonometric number is.$\arcsin \left(2\right)$

 

Step 2: Definition of Complex Number.

The numbers that are presented in the form of$\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$ where,  x is the real numbers and y  is an imaginary number, those numbers are referred to as called Complex numbers.  

 

Step 3: Find the value of. a r c s i n ( 2 )

Let,$z=\arcsin \left(2\right)$

 $\begin{array}{c}\sin \left(z\right)=2\\ \frac{\exp \left(zi\right)-exp(-zi)}{2i}=2\end{array}$

 

Let the complex number as .

$\begin{array}{c}u-\frac{1}{u}=4i\\ u^2-4ui-1=0\mathrm{.......}\left(1\right)\end{array}$

Solve the quadratic equation (1).

$\begin{array}{c}a=1\\ b=-4i\\ c=-1\end{array}$

Use the quadratic formula to find roots of equation (1).

$\begin{array}{c}u=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-(-4i)\pm \sqrt{-16+4}}{2}\\ =\frac{4i\pm 2\sqrt{3}i}{2}\\ =2i\pm \sqrt{3}i\end{array}$

$\begin{array}{c}zi=\ln \left(u_1\right)\\ =ln\left(r\right)+i(\theta +2n\pi )\\ =\ln (2i+\sqrt{3}i)\\ =\ln (2+\sqrt{3})+i(\frac{\pi }{2}\pm 2n\pi )\end{array}$

 

Solve further.

$\begin{array}{c}n=0,1,2,3,\mathrm{...}\\ z_1=\frac{zi}{i}\\ =\frac{\ln (2+\sqrt{3})+i(\frac{\pi }{2}\pm 2n\pi )}{i}\\ =-i\ln (2+\sqrt{3})+(\frac{\pi }{2}\pm 2n\pi )\end{array}$

$\begin{array}{c}zi=\ln \left(u_2\right)\\ =ln\left(r\right)+i(\theta +2n\pi )\\ =\ln (2i-\sqrt{3}i)\\ =\ln (2-\sqrt{3})+i(\frac{\pi }{2}\pm 2n\pi )\end{array}$

Solve further.

$\begin{array}{c}n=0,1,2,3,\mathrm{...}\\ z_2=\frac{zi}{i}\\ =\frac{\ln (2-\sqrt{3})+i(\frac{\pi }{2}\pm 2n\pi )}{i}\\ =-i\ln (2-\sqrt{3})+(\frac{\pi }{2}\pm 2n\pi )\end{array}$

Put $z_1$and$z_2$ in one solution as the difference between them is of the sign of .

$\begin{array}{l}z=\pm i\ln (2+\sqrt{3})+(\frac{\pi }{2}\pm 2n\pi )\\ n=0,1,2,3,\mathrm{...}\end{array}$

 

Hence, the value of$\arcsin \left(2\right)$ is$\pm i\ln (2+\sqrt{3})+(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\pm 2n\pi )$ Where.$n=0,1,2,3,\mathrm{...}$