Solved Problem on Cauchy-Riemann Equations

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

Condition 1: The function w 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) and v(x, y)

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

Calculating the partial derivatives

\[ \begin{array}{l} \dfrac{\partial u}{\partial x}=-\operatorname{e}^{y}\operatorname{sen}x\\[5pt] \dfrac{\partial v}{\partial y}=\operatorname{e}^{y}\operatorname{sen}x\\[5pt] \dfrac{\partial u}{\partial y}=\operatorname{e}^{y}\cos x\\[5pt] \dfrac{\partial v}{\partial x}=\operatorname{e}^{y}\cos x \end{array} \]

Condition 2: The derivatives are continuous everywhere in the complex plane.

Applying the Cauchy-Riemann Equations

\[ \begin{gather} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\ -\operatorname{e}^{y}\operatorname{sen}x\neq\operatorname{e}^{y}\operatorname{sen}x \end{gather} \]

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

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

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

The Cauchy-Riemann Equations are not satisfiedm but in the first condition, if we do

\[ -\operatorname{e}^{y}\operatorname{sen}x=\operatorname{e}^{y}\operatorname{sen}x \]

will only be true, if

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

In the second condition, if we do

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

will only be true, if

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

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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