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.

$\mathit{s}\mathit{i}\mathit{n}\mathit{z}$

Short answer

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

Step-by-step solution

Definition of complex number

Complex numberis represented by $a+ib$,where a is the real number and b is the imaginary number

Solve complex number

Given the function is $\sin z$.

From the Euler’s formula,$e^{ix}=\cos x+i\sin x$ ,

$\cos x=\frac{e^{ix}+e^{-ix}}{2}$,$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$

Therefore$\sin z$ can be written as:

$\sin z=\frac{e^{iz}+e^{-iz}}{2}$

The complex number can be written as:

$z=x+iy$, where x is a real part and y is an imaginary part.

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

Find real and imaginary parts

Simplify the expression future.

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

Use Euler’s formula into the obtained expression

$$\begin{gathered} \frac{e^{ix}·e^{-y}+e^{-ix}·e^y}{2}=\frac{e^{-y}\left(\cos x+i\sin x\right)+e^y\left(\cos x-i\sin x\right)}{2} \\ =\frac{\cos x\left(e^{-y}+e^y\right)+i\sin x\left(e^{-y}-e^y\right)}{2} \\ =\frac{1}{2}\cos x\left(e^{-y}+e^y\right)+i\frac{1}{2}\sin x\left(e^{-y}-e^y\right) \end{gathered}$$

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