91Ó°ÊÓ

The given expression is$\cos \left[2iIn\frac{1-i}{1+i}\right]$ .

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

Consider $w=\cos \left[2iIn\frac{1-i}{1+i}\right]$. …. (1)

Use cos$\left(zi\right)$=cosh(z) to rewrite the equation (1).

$w=\cos h\left(2In\frac{1-i}{1+i}\right)$

Let be, $1-i$be, $z_1$.

$z_1=1-i$

The value of modulus of $z_1$is,$\sqrt{1^2+1^2}=\sqrt{2}$.

The argument of is $\left(\frac{-1}{1}\right)=-\frac{\mathrm{Ï€}}{4}$.

Let $1-i$be $z_1$.

$z_1=1-i$

The value of modulus of$z_1$is ,$\sqrt{1^2+1^2}=\sqrt{2}$.

The argument of $z_1$is arctan$\left(\frac{-1}{1}\right)=-\frac{\mathrm{Ï€}}{4}$.

Let 1 + i be$z_2$

$z_2=1+i$

The value of modulus of$z_2$ is$\sqrt{1^2+1^2}=\sqrt{2}$ .

The argument of$z_2$ is arctan$\left(\frac{1}{1}\right)=\frac{\mathrm{Ï€}}{4}$ .

Put the values of $z_1$and$z_2$in equation (1).

$$\begin{gathered} w=\cos h\left(2In\frac{\sqrt{2}{e}^{-\mathrm{Ï€¾\pm }/4}}{\sqrt{2}{e}^{\mathrm{Ï€¾\pm }/4}}\right) \\ =\cos h\left(2In\left(e^{-\mathrm{Ï€¾\pm }/2}\right)\right) \\ =\cos h\left[2i\left(-\frac{\mathrm{Ï€}}{2}Â\pm 2n\mathrm{Ï€}\right)\right] \\ =\cos \left(-\mathrm{Ï€}Â\pm 4\mathrm{²ÔÏ€}\right) \\ W=\cos \left(-\mathrm{Ï€}\right) \\ W=-1 \end{gathered}$$

Hence the value of $\cos \left[2iIn\frac{1-i}{1+i}\right]is-1$