Solved Problem on Cauchy-Riemann Equations
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f) \( \displaystyle w=|\;z-1\;|^{2} \)
Writing the function as
\[
\begin{gather}
z-1=x+iy-1\\
z-1=(x-1)+iy
\end{gather}
\]
\[
\begin{gather}
w=\left(\sqrt{(x-1)^{2}+y^{2}\;}\right)^{2}\\
w=(x-1)^{2}+y^{2}
\end{gather}
\]
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)
\[
\begin{array}{l}
u(x,y)=(x-1)^{2}+y^{2}\\
v(x,y)=0
\end{array}
\]
Calculating the partial derivatives
\[
\begin{array}{l}
\dfrac{\partial u}{\partial x}=2(x-1)\\[5pt]
\dfrac{\partial v}{\partial y}=0\\[5pt]
\dfrac{\partial u}{\partial y}=2y\\[5pt]
\dfrac{\partial v}{\partial x}=0
\end{array}
\]
Condition 2: The derivatives are continuous everywhere in the complex plane.
Applying the Cauchy-Riemann Equations
\[
\begin{gather}
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\
2(x-1)\neq 0
\end{gather}
\]
\[
\begin{gather}
\frac{\partial u}{\partial y}=-{\frac{\partial v}{\partial x}}\\
2y\neq 0
\end{gather}
\]
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
\[
\begin{gather}
2(x-1)=0\\
x-1=0\\
x=1
\end{gather}
\]
and in the second equation
\[
\begin{gather}
2y=0\\
y=0
\end{gather}
\]
the function is differentiable at the point
\[
(x,y)=(1,0)
\]
The derivative is given by
\[ \bbox[#99CCFF,10px]
{f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}}
\]
\[
\begin{gather}
f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\\[5pt]
w'=2(x-1)+2yi
\end{gather}
\]
- For (x, y)=(1, 0)
\[
w'=2(1-1)+2.0i
\]
\[ \bbox[#FFCCCC,10px]
{w'=0}
\]
Note: If we did
\[
\begin{gather}
f'(z)=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}\\[5pt]
w'=0+2yi
\end{gather}
\]
- For y = 0
\[
\begin{gather}
w'=0
\end{gather}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .