Solved Problem on Cauchy-Riemann Equations

a) \( \displaystyle w=\left(x^2-y^2-2x\right)+2iy\left(x-1\right) \)

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

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

Calculating the partial derivatives

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

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

Applying the Cauchy-Riemann Equations

\[ \begin{gather} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\[5pt] 2(x-1)=2(x-1) \end{gather} \]

\[ \begin{gather} \frac{\partial u}{\partial y}=-{\frac{\partial v}{\partial x}}\\[5pt] -2y=-2y \end{gather} \]

Condition 3: The function w satisfies the Cauchy-Riemann Equations.

The function w is continuous, the derivatives are continuous, and the Cauchy-Riemann Equations are satisfied.
The function w is analytic everywhere in the complex plane .
The function w is differentiable everywhere in the complex plane (entire function) .

The derivative is given by

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

\[ \begin{gather} \bbox[#FFCCCC,10px] {w'=2(x-1)+2yi} \end{gather} \]

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