Q6P - Mathematical Methods in Physical Sciences

Question

Use the Cauchy-Riemann conditions to find out whether the functions in Problems 1.1 to 1.21 are analytic.

${\mathit{e}}^{\mathbf{z}}$

Step-by-Step Solution

Answer

The function $e^z$ is analytic.

Step 1: Given information

The given function is $e^z$.

Step 2: Concept of Cauchy-Riemann conditions

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

$\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{x}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{y}}$ and $\frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{y}}$ to be analytic.

Step 3: Substitute the value

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

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

Use Euler’s 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 $v(x,y)=e^x \sin y$.

Step 4: Apply Cauchy-Riemann conditions

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

$$\begin{gathered} \frac{\partial u}{\partial x}=e^x \cos y \\ \frac{\partial u}{\partial y}=-e^x \sin y \\ \frac{\partial v}{\partial x}=e^x \sin y \\ \frac{\partial v}{\partial y}=e^x \cos y \end{gathered}$$

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

 

Therefore, the given function is analytic.