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I am currently learning about holomorphic functions, and right after the statement about the Cauchy-Riemann equations, there is a proposition that states, "If f is holomorphic at $z_0$, then $ \frac{\partial f}{\partial {\bar{z}}} = 0"$. For the explanation it says "Taking real and imaginary parts, it is easy to see that the Cauchy-Riemann Equations are equivalent to $ \frac{\partial f}{\partial {\bar{z}}}(z_0) = 0".$

However, I don't understand that explanation at all and it seems like the $\bar{z}$ came out of left field. Can someone perhaps expand on the explanation and the significance of the statement? Thanks in advance!

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  • $\begingroup$ how do you define the partial derivative with respect to $\bar{z}$? When that is made clear, you can see that the given condition is merely a notation for the Cauchy Riemann equations. $\endgroup$ – James S. Cook Sep 21 '15 at 17:55
  • $\begingroup$ ahh ok, they define the differential operator with respect to $\bar{z}$ right above, and now I see the connection to CR. $\endgroup$ – user123429 Sep 21 '15 at 18:01
  • $\begingroup$ I like to think of it in terms of the complex-linearity of the differential. If the differential allows us to pull out complex multiplication then that amounts to there being no term like $d\bar{z}$.... but, unless you've studied Jacobians and differentials this comment is probably useless (for now) $\endgroup$ – James S. Cook Sep 21 '15 at 18:09
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Generally, $x = (z+\bar{z})/2$ and $y = (z-\bar{z})/(2i)$. Thus, any function of $x,y$ can be recast as a function of $z$ and $\bar{z}$. Also, differentials can be cast in the same light: for $f = f(x,y)$ or $f=f(z, \bar{z})$ we have $$ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = \frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \bar{z}} d\bar{z} $$ Setting $dx = (dz+d\bar{z})/2$ and $dy = (dz-d\bar{z})/(2i)$. Plug these into the above to derive: $$ df = \frac{\partial f}{\partial x} (dz+d\bar{z})/2 + \frac{\partial f}{\partial y} (dz-d\bar{z})/(2i) = \frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \bar{z}} d\bar{z} $$ from which we obtain: $$ df = \frac{1}{2}\left( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y} \right) dz + \frac{1}{2}\left( \frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y} \right) d\bar{z} = \frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \bar{z}} d\bar{z} $$ hence, $$\frac{\partial f}{\partial z} = \frac{1}{2}\left( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y} \right) \ \ \& \ \ \frac{\partial f}{\partial \bar{z}} = \frac{1}{2}\left( \frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y} \right). $$ Let $f=u+iv$ and notice $u_x=v_y$ and $v_x=-u_y$ is equivalent to $\frac{\partial f}{\partial \bar{z}} =0$. (work it out, it isn't long)

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  • $\begingroup$ btw, this answer is formal nonsense. Really, I'm just giving you a motivation for why the symbols $\frac{\partial f}{\partial \bar{z}}$ and $\frac{\partial f}{\partial z}$ are consistent with the usual derivatives of $x,y$. On the other hand, it does give us nice easy results like: $f$ is holomorphic if it is a function of just $z$ and not $\bar{z}$. A statement which on the face is really confusion since aren't $z$ and $\bar{z}$ dependent??? Well, no, if we merely use them as a complex notation for a pair of real variables. $\endgroup$ – James S. Cook Sep 21 '15 at 18:13
  • $\begingroup$ Great answer. That provided a lot of insight. Thanks! $\endgroup$ – user123429 Sep 21 '15 at 18:25
  • $\begingroup$ @JamesS.Cook: the part that is often hard to get the head around, and I'm not sure if I have done so completely, is that $z$ already tells us those two real variables; however, not in an easily accessible (algebraic) form. Using $z$ and $\bar{z}$ allows us to extract $x=\frac{z+\bar{z}}2$ and $y=\frac{z-\bar{z}}{2i}$ algebraically. $\endgroup$ – robjohn Sep 22 '15 at 1:20
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In a different direction from the answer of James S. Cook (and perhaps more than you wanted to know).


Functions Differentiable by $\boldsymbol{z}$

The total derivative of $f=u+iv$ in the direction of $h+ik$ is given by $$ \begin{align} \left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)(u+iv) &=h\frac{\partial u}{\partial x}+ih\frac{\partial v}{\partial x} +k\frac{\partial u}{\partial y}+ik\frac{\partial v}{\partial y}\\ &=(h+ik)\frac{\mathrm{d}f}{\mathrm{d}z}\tag{1} \end{align} $$ To exist, $\frac{\mathrm{d}f}{\mathrm{d}z}$ must not depend on what $h+ik$ is. That is, $$ \overbrace{\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}} ^{\frac{\mathrm{d}f}{\mathrm{d}z}\text{ when }k=0} =\overbrace{-i\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}}^{\frac{\mathrm{d}f}{\mathrm{d}z}\text{ when }h=0}\tag{2} $$ The real and imaginary parts of $(2)$ give the Cauchy-Riemann equations $$ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}\tag{3} $$ Applying $(3)$ to $(1)$ yields $$ \begin{align} \frac{\mathrm{d}f}{\mathrm{d}z} &=\frac1{h+ik}\left(\color{#C00000}{h\frac{\partial u}{\partial x}}\color{#00A000}{+ih\frac{\partial v}{\partial x} +k\frac{\partial u}{\partial y}}\color{#C00000}{+ik\frac{\partial v}{\partial y}}\right)\\ &=\frac1{h+ik}\left(\color{#C00000}{(h+ik)\frac{\partial u}{\partial x}}\color{#00A000}{-i(h+ik)\frac{\partial u}{\partial y}}\right)\\ &=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)u\tag{4}\\ &=\frac1{h+ik}\left(\color{#C00000}{(h+ik)\frac{\partial v}{\partial y}}\color{#00A000}{+i(h+ik)\frac{\partial v}{\partial x}}\right)\\ &=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)iv\tag{5}\\ &=\frac12\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)f\tag{6} \end{align} $$ where $(6)$ is the average of $(4)$ and $(5)$.


Functions Differentiable by $\boldsymbol{\bar{z}}$

The total derivative of $f=u+iv$ in the direction of $h+ik$ is given by $$ \begin{align} \left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)(u+iv) &=h\frac{\partial u}{\partial x}+ih\frac{\partial v}{\partial x} +k\frac{\partial u}{\partial y}+ik\frac{\partial v}{\partial y}\\ &=(h-ik)\frac{\mathrm{d}f}{\mathrm{d}\bar{z}}\tag{7} \end{align} $$ To exist, $\frac{\mathrm{d}f}{\mathrm{d}\bar{z}}$ must not depend on what $h+ik$ is. That is, $$ \overbrace{\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}} ^{\frac{\mathrm{d}f}{\mathrm{d}\bar{z}}\text{ when }k=0} =\overbrace{i\frac{\partial u}{\partial y}-\frac{\partial v}{\partial y}}^{\frac{\mathrm{d}f}{\mathrm{d}\bar{z}}\text{ when }h=0}\tag{8} $$ The real and imaginary parts of $(8)$ give the conjugate Cauchy-Riemann equations $$ \frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}\tag{9} $$ Applying $(9)$ to $(7)$ yields $$ \begin{align} \frac{\mathrm{d}f}{\mathrm{d}\bar{z}} &=\frac1{h-ik}\left(\color{#C00000}{h\frac{\partial u}{\partial x}}\color{#00A000}{+ih\frac{\partial v}{\partial x} +k\frac{\partial u}{\partial y}}\color{#C00000}{+ik\frac{\partial v}{\partial y}}\right)\\ &=\frac1{h-ik}\left(\color{#C00000}{(h-ik)\frac{\partial u}{\partial x}}\color{#00A000}{+i(h-ik)\frac{\partial u}{\partial y}}\right)\\ &=\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)u\tag{10}\\ &=\frac1{h-ik}\left(\color{#C00000}{-(h-ik)\frac{\partial v}{\partial y}}\color{#00A000}{+i(h-ik)\frac{\partial v}{\partial x}}\right)\\ &=\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)iv\tag{11}\\ &=\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)f\tag{12} \end{align} $$ where $(12)$ is the average of $(10)$ and $(11)$.


Partials with Respect to $\boldsymbol{z}$ and $\boldsymbol{\bar{z}}$

Note that $(3)$ says that for a function differentiable by $z$ we have $$ \frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)f=0\tag{13} $$ while $(9)$ says that for a function differentiable by $\bar{z}$ we have $$ \frac12\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)f=0\tag{14} $$ Therefore, we call $$ \frac{\partial}{\partial z}=\frac12\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\tag{15} $$ and $$ \frac{\partial}{\partial\bar{z}}=\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)\tag{16} $$ since, when $f$ is differentiable by $z$, $$ \frac{\partial f}{\partial z}=\frac{\mathrm{d}f}{\mathrm{d}z}\quad\text{and}\quad\frac{\partial f}{\partial\bar{z}}=0\tag{17} $$ and, when $f$ is differentiable by $\bar{z}$, $$ \frac{\partial f}{\partial\bar{z}}=\frac{\mathrm{d}f}{\mathrm{d}\bar{z}}\quad\text{and}\quad\frac{\partial f}{\partial z}=0\tag{18} $$ Using $(15)$ and $(16)$ and expanding $\mathrm{d}z=\mathrm{d}x+i\,\mathrm{d}y$ and $\mathrm{d}\bar{z}=\mathrm{d}x-i\,\mathrm{d}y$, we get $$ \mathrm{d}f=\frac{\partial f}{\partial z}\mathrm{d}z+\frac{\partial f}{\partial\bar{z}}\mathrm{d}\bar{z}\tag{19} $$


Essentially, $(17)$ answers your question, the rest hopefully adds context.

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  • $\begingroup$ nice. Still, I do wish more knowledge of the theory of real differentiability of functions from $\mathbb{R}^2$ to $\mathbb{R}^2$ was commonly known. It would make all of this far less mysterious... $\endgroup$ – James S. Cook Sep 22 '15 at 1:07

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