Find the real and imaginary parts $\mathit{u}\left(x,y\right)$ and$\mathit{v}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$ of the following functions.

$\mathit{c}\mathit{o}\mathit{s}\overline{\mathbf{z}}$

Complex numbersare expressed in the form of $\mathit{z}\mathbf{}\mathbf{=}\mathbf{}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathit{i}\mathit{y}$, where x,y are real numbers, and i is an imaginary number.

Similarly, the function of z is represented as follows:

$\mathit{f}\left(z\right)\mathbf{=}\mathit{f}\left(x+iy\right)\mathbf{=}\mathit{u}\left(x,y\right)\mathbf{+}\mathit{i}\mathit{v}\left(x,y\right)$, where$\mathit{u}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$ is the real part and$\mathit{v}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$ is the imaginary part.

The conjugate of a complex number is described as the number with the same real part as the original number and an imaginary part opposite in sign but equal in magnitude. The conjugate of$\mathit{z}\mathbf{}\mathbf{=}\mathbf{}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathit{i}\mathit{y}$ is denoted as$\mathit{z}\mathbf{}\mathbf{=}\mathbf{}\mathit{x}\mathbf{}\mathbf{-}\mathbf{}\mathit{i}\mathit{y}$ .

Given the function is $\cos \overline{z}$.

The complex number z can be written as$z=x+iy$, where x is a real part and y is an imaginary part.

Substitute the complex number and simplify.

$$\begin{gathered} \cos \overline{z}=\cos \left(\overline{x+iy}\right) \\ =\cos \left(x-iy\right) \end{gathered}$$

Use Euler’s formula $\cos x=\frac{e^{ix}+e^{-ix}}{2}$, and simplify.

localid="1664353381399" $$\begin{gathered} \cos \overline{z}=\cos \left(x-iy\right) \\ =\frac{e^{i\left(x-iy\right)}+e^{-i\left(x-iy\right)}}{2} \\ =\frac{e^{\left(ix-y\right)}+e^{-\left(ix-y\right)}}{2} \\ =\frac{e^y·e^{ix}+e^{-y}·e^{-ix}}{2} \\ =\frac{e^y(\cos x+i\sin x)+e^{-y}(\cos x-i\sin x)}{2} \end{gathered}$$

Simplify the expression further as follows:

$$\begin{gathered} \cos \overline{z}=\frac{e^y(\cos x+i\sin x)+e^{-y}(\cos x-i\sin x)}{2} \\ =\frac{\cos x\left(e^y+e^{-y}\right)+i\sin x\left(e^y-e^{-y}\right)}{2} \\ =\frac{\cos x\left(e^y+e^{-y}\right)}{2}+\frac{i\sin x\left(e^y+e^{-y}\right)}{2} \end{gathered}$$

Hence, the real part is$\frac{\cos x\left(e^y+e^{-y}\right)}{2}$and imaginary part is$\frac{\sin x\left(e^y-e^{-y}\right)}{2}$.