Solved Problem on Cauchy-Riemann Equations
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d) \( \displaystyle w=x^{2}y^{2}+i2x^{2}y^{2} \)
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)
\[
\begin{array}{l}
u(x,y)=x^{2}y^{2}\\
v(x,y)=2x^{2}y^{2}
\end{array}
\]
Calculating the partial derivatives
\[
\begin{array}{l}
\dfrac{\partial u}{\partial x}=2xy^{2}\\[5pt]
\dfrac{\partial v}{\partial y}=4x^{2}y\\[5pt]
\dfrac{\partial u}{\partial y}=2x^{2}y \\[5pt]
\dfrac{\partial v}{\partial x}=4xy^{2}
\end{array}
\]
Condition 2: The derivatives are continuous everywhere in the complex plane.
Applying the Cauchy-Riemann Equations
\[
\begin{gather}
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\
2xy^{2}\neq 4x^{2}y
\end{gather}
\]
\[
\begin{gather}
\frac{\partial u}{\partial y}=-{\frac{\partial v}{\partial x}}\\
2x^{2}y\neq -4xy^{2}
\end{gather}
\]
Condition 3: The function w does not satisfy the Cauchy-Rerimmann 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 in the complex plane.
The Cauchy-Riemann Equations are not satisfied, but if we do in the first equation
\[
\begin{gather}
2xy^{2}=4x^{2}y\\
y=2x
\end{gather}
\]
and in the second equation
\[
\begin{gather}
2x^{2}y=-4xy^{2}\\
x=-2y\\
y=-{\frac{1}{2}}x
\end{gather}
\]
- For y = 2x
\[
\begin{gather}
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\
2x(2x)^{2}=4x^{2}(2x)\\8x^{3}=8x^{3}
\end{gather}
\]
\[
\begin{gather}
\frac{\partial u}{\partial y}=-{\frac{\partial v}{\partial x}}\\
2x^{2}(2x)=-4x(2x)^{2}\\
4x^{3}(2x)\neq-16x^{3}
\end{gather}
\]
in this case, the function satisfies the first Cauchy-Riemann Equation but does not satisfy the
second equation.
- For \( y=-{\dfrac{1}{2}}x \)
\[
\begin{gather}
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\
2x\left(-{\frac{1}{2}}x\right)^{2}=4x^{2}\left(-{\frac{1}{2}}x\right)\\
\frac{1}{2}x^{3}\neq2x^{3}
\end{gather}
\]
\[
\begin{gather}
\frac{\partial u}{\partial y}=-{\frac{\partial v}{\partial x}}\\
-x^{3}=-x^{3}
\end{gather}
\]
in this case, the function satisfies the second Cauchy-Riemann Equation but does not satisfy the
first equation.
- For x = 0
\[
\begin{split}
\frac{\partial u}{\partial x} &=\lim_{\Delta x\rightarrow 0}{\frac{u(0+\Delta x,y)-u(0,y)}{\Delta x}}=\\
&=\lim_{\Delta x\rightarrow 0}{\frac{(0+\Delta x)^{2}y^{2}-0^{2}.y^{2}}{\Delta x}}=\\
&=\lim_{\Delta x\rightarrow 0}{\frac{\Delta x^{2}y^{2}}{\Delta x}}=\lim_{\Delta x\rightarrow 0}{\Delta xy^{2}}=0
\end{split}
\]
\[
\begin{split}
\frac{\partial u}{\partial y} &=\lim_{\Delta y\rightarrow 0}{\frac{u(0,y+\Delta y)-u(0,y)}{\Delta y}}=\\
&=\lim_{\Delta y\rightarrow 0}{\frac{0^{2}.(y+\Delta y)^{2}-0^{2}.y^{2}}{\Delta y}}=0
\end{split}
\]
\[
\begin{split}
\frac{\partial v}{\partial x} &=\lim_{\Delta x\rightarrow 0}{\frac{v(0+\Delta x,y)-v(0,y)}{\Delta x}}=\\
&=\lim_{\Delta x\rightarrow 0}{\frac{2(0+\Delta x)^{2}y^{2}-2.0^{2}.y^{2}}{\Delta x}}=\\
&=\lim_{\Delta x\rightarrow 0}{\frac{2\Delta x^{2} y^{2}}{\Delta x}}=\lim_{\Delta x\rightarrow 0}{2\Delta x y^{2}}=0
\end{split}
\]
\[
\begin{split}
\frac{\partial u}{\partial y} &=\lim_{\Delta y\rightarrow 0}{\frac{v(0,y+\Delta y)-v(0,y)}{\Delta y}}=\\
&=\lim_{\Delta y\rightarrow 0}{\frac{2.0^{2}.(y+\Delta y)^{2}-2.0^{2}y^{2}}{\Delta y}}=0
\end{split}
\]
- For y = 0
\[
\begin{split}
\frac{\partial u}{\partial x} &=\lim_{\Delta x\rightarrow 0}{\frac{u(x+\Delta x,0)-u(x,0)}{\Delta x}}=\\
&=\lim_{\Delta x\rightarrow 0}{\frac{(x+\Delta x)^{2}.0^{2}-x^{2}.0^{2}}{\Delta x}}=0
\end{split}
\]
\[
\begin{split}
\frac{\partial u}{\partial y} &=\lim_{\Delta y\rightarrow 0}{\frac{u(x,0+\Delta y)-u(x,0)}{\Delta y}}=\\
&=\lim_{\Delta y\rightarrow 0}{\frac{x^{2}(0+\Delta y)^{2}-x^{2}.0^{2}}{\Delta y}}=\\
&=\lim_{\Delta y\rightarrow 0}{\frac{x^{2}\Delta y^{2}}{\Delta y}}=\lim_{\Delta y\rightarrow 0}{x^{2}\Delta y}=0
\end{split}
\]
\[
\begin{split}
\frac{\partial v}{\partial x} &=\lim_{\Delta x\rightarrow 0}{\frac{v(x+\Delta x,0)-v(x,0)}{\Delta x}}=\\
&=\lim_{\Delta x\rightarrow 0}{\frac{2(x+\Delta x)^{2}.0^{2}-2.x^{2}.0^{2}}{\Delta x}}=0
\end{split}
\]
\[
\begin{split}
\frac{\partial u}{\partial y} &=\lim_{\Delta y\rightarrow 0}{\frac{v(x,0+\Delta y)-v(x,0)}{\Delta y}}=\\
&=\lim_{\Delta y\rightarrow 0}{\frac{2x^{2}(0+\Delta y)^{2}-2x^{2}.0^{2}}{\Delta y}}=\\
&=\lim_{\Delta y\rightarrow 0}{\frac{2.x^{2}\Delta y^{2}}{\Delta y}}=\lim_{\Delta y\rightarrow 0}{2.x^{2}\Delta y}=0
\end{split}
\]
The Cauchy-Rerimmann Equations are satisfied for x = 0 on the real axis independent of the
value of y, or for y = 0 on the imaginary axis independent of the value of x.
The derivative is given by
\[ \bbox[#99CCFF,10px]
{f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}}
\]
\[
w'=2xy^{2}+i4xy^{2}\neq 4x^{2}y-i2x^{2}y
\]
for x = 0
\[
w'=2.0y^{2}+i4.0y^{2}=4.0^{2}y-i2.0^{2}y
\]
\[ \bbox[#FFCCCC,10px]
{w'=0}
\]
for y = 0
\[
w'=2x.0^{2}+i4x.0^{2}=4x^{2}.0-i2x^{2}.0
\]
\[ \bbox[#FFCCCC,10px]
{w'=0}
\]
The function w
A função w
is not differenciable at any point in the complex plane, only on the axes x = 0 and y = 0.
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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 .