91影视

The given function is $e^z$.

For the complex function $\mathit{f}\mathbf{\left(}\mathit{z}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{(}\mathit{x}\mathbf{+}\mathit{i}\mathit{y}\mathbf{)}\mathbf{=}\mathit{u}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{+}\mathit{i}\mathit{v}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$,where $\mathit{u}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$is the real part and$\mathit{v}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$is the imaginary part, the conditions are

$\frac{\mathbf{鈭�}\mathbf{u}}{\mathbf{鈭�}\mathbf{x}}\mathbf{=}\frac{\mathbf{鈭�}\mathbf{v}}{\mathbf{鈭�}\mathbf{y}}$and role="math" localid="1653395213752" $\frac{\mathbf{鈭�}\mathbf{v}}{\mathbf{鈭�}\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{鈭�}\mathbf{u}}{\mathbf{鈭�}\mathbf{y}}$to be analytic.

Substitute $z=x+iy$in $e^z$and simplify.

$$\begin{gathered} e^z=e^{x+iy} \\ =e^x{e}^{iy} \end{gathered}$$

Use Euler鈥檚 formula $e^{ix}=\cos x+i\sin x$and simplify further.

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

The above equation is in the form of $f\left(z\right)=f(x+iy)=u(x,y)+iv(x,y)$such that $u(x,y)=e^x\cos y$and $v(x,y)=e^x\sin y$.

Hence, the real part of the given function is $u(x,y)=e^x\cos y$, and the imaginary part is $v(x,y)=e^x\sin y$.

Substitute the values of $u$and $v$in $\frac{鈭�u}{鈭�x}=\frac{鈭�v}{鈭�y}$and $\frac{鈭�v}{鈭�x}=-\frac{鈭�u}{鈭�y}$and simplify.

role="math" localid="1653396078607" $$\begin{gathered} \frac{鈭�u}{鈭�x}=e^x\cos y \\ \frac{鈭�u}{鈭�y}=-e^x\sin y \\ \frac{鈭�v}{鈭�x}=e^x\sin y \\ \frac{鈭�v}{鈭�y}=e^x\cos y \end{gathered}$$

Here, $\frac{鈭�u}{鈭�x}=\frac{鈭�v}{鈭�y}$and $\frac{鈭�v}{鈭�x}=-\frac{鈭�u}{鈭�y}$, that is, this function satisfies the Cauchy-Riemann condition

Therefore, the given function is analytic.