Question

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

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

Short answer

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

Step-by-step solution

Given Information.

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

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.

Find the value of sin(i ln1-i1+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 $z_1=1-i$.

localid="1658902575018" $$\begin{gathered} r_1=\left|1-i\right| \\ =\sqrt{2} \\ {\theta }_1=arc\tan \left(-1\right) \\ =-\frac{\pi }{4} \end{gathered}$$

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 .