d)
\( \displaystyle w=z\;\text{Re}\;z \)
Writing the function as
\[
\begin{gather}
w=(x+iy)\text{Re}(x+iy)\\
w=(x+iy)x\\
w=x^{2}+i xy
\end{gather}
\]
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)
\[
\begin{array}{l}
u(x,y)=x^{2}\\
v(x,y)=xy
\end{array}
\]
Calculating the partial derivatives
\[
\begin{array}{l}
\dfrac{\partial u}{\partial x}=2x\\[5pt]
\dfrac{\partial v}{\partial y}=x\\[5pt]
\dfrac{\partial u}{\partial y}=0\\[5pt]
\dfrac{\partial v}{\partial x}=y
\end{array}
\]
Condition 2: The derivatives are continuous everywhere in the complex plane.
Applying the
Cauchy-Riemann Equations
\[
\begin{gather}
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\
2x=x
\end{gather}
\]
\[
\begin{gather}
\frac{\partial u}{\partial y}=-{\frac{\partial v}{\partial x}}\\
0=y
\end{gather}
\]
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 the
Cauchy-Riemann Equations,
the function w is not analytic.
The
Cauchy-Riemann Equations are not satisfiedm but in the first condition, if we do
\[
\begin{gather}
2x=x\\
2x-x=0\\
x=0
\end{gather}
\]
the function is differentiable at the point
\[
(x,y)=(0,0)
\]
The derivative is given by
\[ \bbox[#99CCFF,10px]
{f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}}
\]
\[
\begin{gather}
f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\\[5pt]
w'=2x+iy
\end{gather}
\]
\[
w'=2.0+i0
\]
\[ \bbox[#FFCCCC,10px]
{w'=0}
\]
Note: If we did
\[
\begin{gather}
f'(z)=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}\\[5pt]
w'=x+i0
\end{gather}
\]
\[
\begin{gather}
w'=0
\end{gather}
\]