Solved Problem on Cauchy-Riemann Equations

c) \( \displaystyle w=\frac{x}{x^{2}+y^{2}}-i\frac{y}{x^{2}+y^{2}} \)

Condition 1: The function w is not continuous at the point z = 0, where (x, y) = (0, 0).

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)

\[ \begin{array}{l} u(x,y)=\dfrac{x}{x^{2}+y^{2}}\\ v(x,y)=-{\dfrac{y}{x^{2}+y^{2}}} \end{array} \]

Calculating the partial derivatives

\[ \begin{array}{l} \dfrac{\partial u}{\partial x}=\dfrac{1.\left(x^{2}+y^{2}\right)-{x.}(2x)}{\left(x^{2}+y^{2}\right)^{2}}=\dfrac{x^{2}+y^{2}-2x^{2}}{\left(x^{2}+y^{2}\right)^{2}}=\dfrac{-x^{2}+y^{2}}{\left(x^{2}+y^{2}\right)^{2}}\\[5pt] \dfrac{\partial v}{\partial y}=-{\dfrac{1.\left(x^{2}+y^{2}\right)-{y.}(2y)}{\left(x^{2}+y^{2}\right)^{2}}}=-{\dfrac{x^{2}+y^{2}-2y^{2}}{\left(x^{2}+y^{2}\right)^{2}}}=\dfrac{-x^{2}+y^{2}}{\left(x^{2}+y^{2}\right)^{2}}\\[5pt] \dfrac{\partial u}{\partial y}=-{\dfrac{{x.}(2y)}{x^{2}+y^{2}}}=-{\dfrac{2xy}{\left(x^{2}+y^{2}\right)^{2}}}\\[5pt] \dfrac{\partial v}{\partial x}=-\left(-{\dfrac{{y.}(2x)}{x^{2}+y^2}}\right)=\dfrac{2xy}{\left(x^{2}+y^{2}\right)^{2}} \end{array} \]

Condition 2: The derivatives are not continuous at the point z = 0, where (x, y) = (0, 0).

Applying the Cauchy-Riemann Equations

\[ \begin{gather} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\ \frac{-x^{2}+y^{2}}{\left(x^{2}+y^{2}\right)^{2}}=\frac{-x^{2}+y^{2}}{\left(x^{2}+y^{2}\right)^{2}} \end{gather} \]

\[ \begin{gather} \frac{\partial u}{\partial y}=-{\frac{\partial v}{\partial x}}\\ -{\frac{2xy}{x^{2}+y^{2}}}=-\left[\frac{2xy}{x^{2}+y^{2}}\right]\\ -{\frac{2xy}{x^{2}+y^{2}}}=-{\frac{2xy}{x^{2}+y^{2}}} \end{gather} \]

Condition 3: The function w satisfies Cauchy-Rerimmann Equations, except at z = 0.

The function w is not continuous, the derivatives are not continuous, and the function satisfies the Cauchy-Rerimmann Equations, the function w is analytic everywhere in the complex plane except at z = 0 .

The derivative is given by

\[ \bbox[#99CCFF,10px] {f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}} \]

\[ \begin{split} f'(z)= &\frac{-x^{2}+y^{2}}{\left(x^{2}+y^{2}\right)^{2}}+i\frac{2xy}{\left(x^{2}+y^{2}\right)^{2}}=\frac{-x^{2}+y^{2}}{\left(x^{2}+y^{2}\right)^{2}}-i\left(-{\frac{2xy}{\left(x^{2}+y^{2}\right)^{2}}}\right)=\\ &=\frac{-x^{2}+y^{2}}{\left(x^{2}+y^{2}\right)^{2}}+i\frac{2xy}{\left(x^{2}+y^{2}\right)^{2}} \end{split} \]

\[ \bbox[#FFCCCC,10px] {w'=\frac{-x^{2}+y^{2}}{\left(x^{2}+y^{2}\right)^{2}}+i\frac{2xy}{\left(x^{2}+y^{2}\right)^{2}}} \]

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