Problem of the Week #57 - April 29th, 2013

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Chris L T521

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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Define the operator
$\frac{\partial}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right).$
Show that if the first order derivatives of the real and imaginary parts of a function $f(z)=u(x,y) + iv(x,y)$ satisfy the Cauchy-Riemann equations, then $\dfrac{\partial f}{\partial \overline{z}}=0$.

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Recall: The Cauchy-Riemann equations for a function $f(z)=u(x,y)+iv(x,y)$ are
$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\qquad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$

Chris L T521

Well-known member
Staff member
This week's problem was correctly answered by Ackbach and Sudharaka. You can find Sudharaka's solution below:

$\frac{\partial f}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}\right)$

$\frac{\partial f}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}+ i\frac{\partial u}{\partial y} - \frac{\partial v}{\partial y}\right)$

$\frac{\partial f}{\partial \overline{z}} = \frac{1}{2}\left\{\left(\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}\right)+ i\left(\frac{\partial v}{\partial x}+ \frac{\partial u}{\partial y}\right)\right\}$

By the Cauchy-Riemann equations,

$\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}=0\mbox{ and }\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0$

Therefore,

$\frac{\partial f}{\partial \overline{z}} =0$

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