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, given in the problem, 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), real part, and v(x, y), imaginary patt

\[ \begin{array}{l} u(x,y)=\dfrac{x}{x^2+y^2}\\[5pt] v(x,y)=-{\dfrac{y}{x^2+y^2}} \end{array} \]

Calculating the partial derivatives

\[ \begin{align} & \dfrac{\partial u}{\partial x}=\dfrac{1.\left(x^2+y^2\right)-{x.}(2x)}{\left(x^2+y^2\right)^2}=\dfrac{x^2+y^2-2x^2}{\left(x^2+y^2\right)^2}=\dfrac{-x^2+y^2}{\left(x^2+y^2\right)^2} \tag{I} \\[5pt] & \dfrac{\partial v}{\partial y}=-{\dfrac{1.\left(x^2+y^2\right)-{y.}(2y)}{\left(x^2+y^2\right)^2}}=-{\dfrac{x^2+y^2-2y^2}{\left(x^2+y^2\right)^2}}=\dfrac{-x^2+y^2}{\left(x^2+y^2\right)^2} \tag{II} \\[5pt] & \dfrac{\partial u}{\partial y}=-{\dfrac{{x}(2y)}{x^2+y^2}}=-{\dfrac{2xy}{\left(x^2+y^2\right)^2}} \tag{III} \\[5pt] & \dfrac{\partial v}{\partial x}=-\left(-{\dfrac{{y}(2x)}{x^2+y^2}}\right)=\dfrac{2xy}{\left(x^2+y^2\right)^2} \tag{IV} \end{align} \]

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

Applying the Cauchy-Riemann Equations

\[ \begin{gather} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\[5pt] \frac{-x^2+y^2}{\left(x^2+y^2\right)^2}=\frac{-x^2+y^2}{\left(x^2+y^2\right)^2} \end{gather} \]

\[ \begin{gather} \frac{\partial u}{\partial y}=-{\frac{\partial v}{\partial x}}\\[5pt] -{\frac{2xy}{x^2+y^2}}=-\left[\frac{2xy}{x^2+y^2}\right]\\[5pt] -{\frac{2xy}{x^2+y^2}}=-{\frac{2xy}{x^2+y^2}} \end{gather} \]

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

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

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

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

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