Q20P - Mathematical Methods in Physical Sciences

Question

Evaluate each of the following in x + iy form, and compare with a computer solution.

$\mathbf{sin}\left(i \ln \frac{1-i}{1+i}\right)$

Step-by-Step Solution

Answer

The value of $\sin \left(i \ln \frac{1-i}{1+i}\right)$is, 1. 

Step 1: Given Information.

The given number is, $\sin \left(i \ln \frac{1-i}{1+i}\right)$.

Step 2: Definition of Complex Number.

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

Step 3: Find the value of s i n ( i   l n 1 - i 1 + i ) .

Let the complex number as.$u=\sin \left(i \ln \frac{1-i}{1+i}\right)$

Use the identity $\sin \left(iz\right)=i \sin h\left(z\right)$  .

$\mathrm{u}=\mathrm{i} \sinh \left(\ln \frac{1-\mathrm{i}}{1+\mathrm{i}}\right)$

 

Let,$z_1=1-i$

Find the argument and the angle of  $z_1=1-i$.

$$\begin{gathered} r_1=\left|1-i\right| \\ =\sqrt{2} \\ {\theta }_1=arc\tan \left(-1\right) \\ =-\frac{\pi }{4} \end{gathered}$$

The angle is in $4^{th}$ quadrant; therefore it is valid for using in the solution.

 

Let,$z_2=1+i$

$$\begin{gathered} r_2=\left|1-i\right| \\ =\sqrt{2} \\ {\theta }_2=arc\tan \left(1\right) \\ =\frac{\pi }{4} \end{gathered}$$

The angle is in $1^{st}$ quadrant therefore it is valid for using in the solution.

$$\begin{gathered} \mathrm{u}=\mathrm{isinh}\left(\mathrm{In}\left(\frac{\sqrt{2}\exp \left(-\mathrm{\pi }/4\right)}{\sqrt{2}\exp \left(\mathrm{\pi }/4\right)}\right)\right) \\ =\mathrm{isinh}\left[\mathrm{In}\left(\exp \left(\frac{-\mathrm{\pi i}}{2}\right)\right)\right] \\ =\mathrm{isinh}\left[\mathrm{In}\left(1\right)-\frac{\mathrm{\pi i}}{2}\pm 2\mathrm{ni\pi }\right] \\ =-\sin \left[-\frac{\mathrm{\pi i}}{2}\pm 2\mathrm{ni\pi }\right] \\ =1 \\ \mathrm{n}=0,1,2,3,... \end{gathered}$$

Hence, the value of $\sin \left(i \ln \frac{1-i}{1+i}\right)$ is 1 .