Question

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.

$\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{z}$

Short answer

The real part$u\left(x,y\right)$of the function is $\frac{1}{2}\cos y\left(e^x+e^{-x}\right)$, and the imaginary part $v\left(x,y\right)$of the function is,$\frac{1}{2}\sin y\left(e^x-e^{-x}\right)$ .

Step-by-step solution

Definition of complex number and formula of hyperbolic function

• Complex numberis represented by$a+ib$, where a is the real number and b is the imaginary number.
• Formula of hyperbolic cosine function:$\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{x}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}}{\mathbf{2}}$.
• Formula of hyperbolic sine function: $\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{x}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}}{\mathbf{2}}$.

Solve complex number

Given the function is $\cosh z$written as $\frac{e^z+e^{-z}}{2}$.

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

Therefore$\cosh z$ is simplified as $\frac{e^{x+iy}+e^{-\left(x+iy\right)}}{2}$.

Find real and imaginary parts.

Simplify the expression future.

$\frac{e^{x+iy}+e^{-\left(x+iy\right)}}{2}=\frac{e^x·e^{iy}+e^{-x}·e^{-iy}}{2}$

Use Euler’s formula into the obtained expression

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

Hence the real part is$\frac{1}{2}\cos y\left(e^x+e^{-x}\right)$ and imaginary part is $\frac{1}{2}\sin y\left(e^x-e^{-x}\right)$.