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The operator defined as $\frac{\partial f}{\partial\overline z}=\frac12\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)$ behaves like any derivative should and has all the same properties as well. But why is that? We didn't define it the usual way we defined partials of a function. Then why does it still behave normally?

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  • $\begingroup$ It's a linear combination of classical partial derivatives. $\endgroup$ – Lord Shark the Unknown Mar 14 '18 at 18:16
  • $\begingroup$ @LordSharktheUnknown Yes that gave me an intuition as to why. But why does that make it obvious that, say, $\frac{\partial f}{\partial\overline z}(\bar z^2)=2\bar z$? $\endgroup$ – Hrit Roy Mar 14 '18 at 18:18
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It suffices to prove that \begin{align*} \dfrac{\partial}{\partial x}fg=f\dfrac{\partial g}{\partial x}+g\dfrac{\partial f}{\partial x} \end{align*} for complex-valued functions $f,g$, then it follows that \begin{align*} \dfrac{\partial}{\partial\overline{z}}fg=f\dfrac{\partial g}{\partial\overline{z}}+g\dfrac{\partial f}{\partial\overline{z}}. \end{align*} With $f=u+iv,g=s+it$ for real-valued functions $u,v,s,t$, then $fg=(us-vt)+i(ut+vs)$, where $us-vt, ut+vs$ are real-valued functions, then \begin{align*} \dfrac{\partial}{\partial x}fg=\dfrac{\partial}{\partial x}(us-vt)+i\dfrac{\partial}{\partial x}(ut+vs), \end{align*} because this is how we define partial derivatives of complex-valued functions. On the other hand, for example, \begin{align*} f\dfrac{\partial g}{\partial x}&=(u+iv)\left(\dfrac{\partial s}{\partial x}+i\dfrac{\partial t}{\partial x}\right)\\ &=u\dfrac{\partial s}{\partial x}-v\dfrac{\partial t}{\partial x}+i\left(u\dfrac{\partial t}{\partial x}+v\dfrac{\partial s}{\partial x}\right), \end{align*} then one can do brute force expansion to verify the equality.

For $\dfrac{\partial}{\partial\overline{z}}(f+g)=\dfrac{\partial f}{\partial\overline{z}}+\dfrac{\partial g}{\partial\overline{z}}$ should be easier.

For $\dfrac{\partial}{\partial\overline{z}}(\overline{z}^{2})$, somehow the reasoning is like \begin{align*} \dfrac{\partial f}{\partial\overline{z}}=\overline{\dfrac{\partial\overline{f}}{\partial z}}, \end{align*} in which case, for $f(z)=\overline{z}$, then \begin{align*} \dfrac{\partial f}{\partial\overline{z}}&=\overline{\dfrac{\partial\overline{f}}{\partial\overline{z}}}\\ &=\overline{\dfrac{\partial}{\partial z}z^{2}}\\ &=\overline{2z}\\ &=2\overline{z}, \end{align*} and the fact that \begin{align*} \dfrac{\partial}{\partial z}z^{2}=f'(z) \end{align*} is because $f$ is complex-differentiable. Note that, for a complex-differentiable $f$, one uses Cauchy-Riemann conditions to deduce that \begin{align*} \dfrac{\partial f}{\partial z}=f'(z). \end{align*}

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  • $\begingroup$ But why is it that the derivative of a function of $\bar z$ is exactly what we would expect to get in case of a function of a real variable $x$? $\endgroup$ – Hrit Roy Mar 14 '18 at 18:26
  • $\begingroup$ I have explained the case that you are interested. $\endgroup$ – user284331 Mar 14 '18 at 18:49
  • $\begingroup$ that is just one particular case. I want to know why it happens with all functions $\endgroup$ – Hrit Roy Mar 14 '18 at 18:50
  • $\begingroup$ I fail to see why it is true for all functions. Usually if the so called $f$ is complex-differentiable, then it goes through by the same reasoning. $\endgroup$ – user284331 Mar 14 '18 at 18:52

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