Solved Problem on Cauchy-Riemann Equations

e) \( \displaystyle w=\operatorname{e}^{y}(\cos x+i\sin x) \)

Condition 1: The function w, given in the problem, is continuous everywhere in the complex plane.

The Cauchy-Riemann Equations are given by

\[ \bbox[#99CCFF,10px] {\begin{gather} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\[5pt] \frac{\partial u}{\partial y}=-{\frac{\partial v}{\partial x}} \end{gather}} \]

Identifying the functions u(x, y), real part, and v(x, y), imaginary part

\[ \begin{array}{l} w=\operatorname e^y(\cos x+i\sin x)\\[5pt] w=\operatorname e^y\cos x+i\operatorname e^y\sin x\\[5pt] u(x,y)=\operatorname e^y\cos x \\[5pt] v(x,y)=\operatorname e^y\sin x \end{array} \]

Calculating the partial derivatives

\[ \begin{align} & \dfrac{\partial u}{\partial x}=-\operatorname e^y\sin x \tag{I} \\[5pt] & \dfrac{\partial v}{\partial y}=\operatorname e^y\sin x \tag{II} \\[5pt] & \dfrac{\partial u}{\partial y}=\operatorname e^y\cos x \tag{III} \\[5pt] & \dfrac{\partial v}{\partial x}=\operatorname e^y\cos x \tag{IV} \end{align} \]

Condition 2: The derivatives (I), (II), (III) and (IV) are continuous everywhere in the complex plane.

Applying the Cauchy-Riemann Equations

\[ \begin{gather} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\[5pt] -\operatorname e^y\sin x\neq\operatorname e^y\sin x \tag{V} \end{gather} \]

\[ \begin{gather} \frac{\partial u}{\partial y}=-{\frac{\partial v}{\partial x}}\\[5pt] \operatorname e^y\cos x\neq-\operatorname e^y\cos x \tag{VI} \end{gather} \]

Condition 3: The function w does not satisfy the Cauchy-Riemann Equations.

The function w is continuous, the derivatives are continuous, but the function does not satisfy Cauchy-Riemann Equations.
The function w is not analytic in the complex plane .

The Cauchy-Riemann Equations are not satisfied, but if in the equation (V), we do

\[ \begin{gather} -\operatorname e^y\sin x=\operatorname e^y\sin x \end{gather} \]

this equality only holds if the expression is zero (in any other case we have a positive number equal to its negative value). The exponential (e^y) will never be equal to zero, the equality will only be true if the sine function is equal to zero

\[ \begin{gather} \sin x=0\\[5pt] x=\operatorname{arcsen}0\\[5pt] \qquad\qquad\qquad\qquad x=n\pi \quad \text{,} \quad n=0, 1, 2, ... \end{gather} \]

and in the equation (VI)

\[ \begin{gather} \operatorname e^y\cos x=-\operatorname e^y\cos x \end{gather} \]

will only be true, if

\[ \begin{gather} \cos x=0\\[5pt] x=\arccos 0\\[5pt] \qquad\qquad\qquad\qquad x=n\frac{\pi}{2} \quad \text{,} \quad n=0, 1, 2, ... \end{gather} \]

Since there is no value of x that satisfies both conditions simultaneously the function is not differentiable.
The function w is not differentiable in the complex plane .

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