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Suppose $ f(z):\mathbb{C}^{1}\to\mathbb{C}^{1} $, write $ \frac{\partial f}{\partial \overline{z}}=\frac{1}{2}(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}) $. We know that by Cauchy-Riemann equation, the differential property is equivalent to $ \frac{\partial f}{\partial \overline{z}}=0 $. But can we take the derivative directly regarding to $ \overline{z} $ to determine whether the function is differential or not? If we can't do this, then what's the reason to introduce this symbol? And how to determine a function is differentiale without directly expanding the real and imaginary part out and check the Cauchy-Riemann equation?

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    $\begingroup$ Note that $\frac{\partial f}{\partial z}$ and $\frac{\partial f}{\partial \bar z}$ are not partial derivatives of $f$ with respect to the "variables" $z$ and $\bar z$, but rather the result of applying to $f$ the following operators:$$\frac{\partial}{\partial z}=\frac12\left(\frac{\partial}{\partial x}+\frac1i\frac{\partial}{\partial y}\right),\quad\frac{\partial}{\partial \bar z}=\frac12\left(\frac{\partial}{\partial x}-\frac1i\frac{\partial}{\partial y}\right).$$ -- Berenstein C. A., Complex Variables, An Introduction. $\nwarrow$ relevant book snippet $\endgroup$ – Frenzy Li Jul 19 '18 at 11:39
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There is the complex vector space $V$ of real-linear complex valued functions $$f:\quad{\mathbb R}^2\to{\mathbb C},\qquad{\bf z}=(x,y)\to f({\bf z})\ .$$ Such functions satisfy $$f({\bf z}+{\bf w})=f({\bf z})+f({\bf w}),\qquad f(\lambda{\bf z})=\lambda\,f({\bf z})\quad(\lambda\in{\mathbb R})\ .$$ It is easy to see that ${\rm dim}_{\mathbb C}(V)=2$, and that the two functions $$x:\quad (x,y)\mapsto x, \qquad y:\quad (x,y)\mapsto y$$ (here we see some abuse of notation) form a basis of $V$, meaning that each $f\in V$ has a unique representation $$f(x,y)=c_1 x+c_2 y,\qquad c_1,c_2\in{\mathbb C}\ .$$ The partial differential operators ${\partial\over\partial x}$ and ${\partial\over\partial y}$ then form the corresponding dual basis of $V^*$, meaning that $c_1={\partial\over\partial x}f$, $\>c_2={\partial\over\partial y}f$.

Now the two functions $z:=x+iy\in V$ and $\bar z:=x-iy\in V$ form a basis of $V$ as well, and the Wirtinger derivatives introduced in the question form the corresponding dual basis of $V^*$, meaning that for $f=c_1z+c_2\bar z\in V$ one has $c_1={\partial\over\partial z}f$, $\>c_2={\partial\over\partial\bar z}f$.

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    $\begingroup$ Yes, this is true; but my answer didn't go into that. I just wanted to show that these derivatives are not just tricky constructs, but can be explained in terms of linear algebra. $\endgroup$ – Christian Blatter Jul 19 '18 at 13:54
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Is somewhat of an abuse of notation, but the idea is that if $f:\mathbb C\to\mathbb C$ is real differentiable at $z_0$ -- that is, it is differentiable as a function between 2-dimension real vector spaces -- then we can write $$ f(z) = f(z_0+h) = f(z_0) + F(h) + o(h) $$ where $F(h)$ is a linear transformation of $\mathbb C$, still viewed as a 2-dimensional real vector space. We can expand $F$ in matrix form and do some algebra to see that this can also be written $$ f(z) = f(z_0+a+bi) = f(z_0) + c_1(a+bi) + c_2(a-bi) + o(a+bi) $$ for some (uniquely determined) complex constants $c_1$ and $c_2$.

Since $c_1$ is the coefficient we multiply by $a+bi=z-z_0$ and $c_2$ is the coefficient we multiply by $a-bi = \overline z - \overline{z_0} $, there is a certain justice in calling them $\partial f/\partial z$ and $\partial f/\partial \bar z$.

But don't take this for more than it is; the notation depends on pretending that $z$ and $\bar z$ can vary independently of each other, which is not really the case.

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    $\begingroup$ Somewhat of an abuse of notation?! $\endgroup$ – Umberto P. Jul 19 '18 at 11:53
  • $\begingroup$ @UmbertoP.: I'm hedging! :-) It's probably possible, if one puts one's mind to it, to find a way to make this into an excusable shorthand for something with a more principled meaning -- not that I'd expect that to bring much additional clarity. $\endgroup$ – hmakholm left over Monica Jul 19 '18 at 11:57
  • $\begingroup$ @UmbertoP., it is possible to make rigorous sense of that notation by complexifying the real tangent space of $\mathbb C$, and then $z$ and $\overline{z}$ are genuinely "independent", etc. But, no, the possibility of making sense in this way is surely not helpful at a more elementary level. $\endgroup$ – paul garrett Jul 19 '18 at 12:55
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The differential operator $$\frac{\partial}{\partial\overline z}$$ has the property that $$\frac{\partial}{\partial\overline z}z=0\qquad\textrm{and} \qquad\frac{\partial}{\partial\overline z}\overline z=1.$$ Likewise the differential operator $$\frac{\partial}{\partial z}=\frac12\left( \frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)$$ has the property that $$\frac{\partial}{\partial z}z=1\qquad\textrm{and} \qquad\frac{\partial}{\partial z}\overline z=0.$$

As $\partial/\partial \overline z$ is an elliptic operator, any distribution it annihilates is a smooth function satisfying Cauchy-Riemann and so is a holomorphic function.

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