TITLE: Factor 1/2 in derivative of analytic complex function
QUESTION [1 upvotes]: Where does the factor of 1/2 come from in this derivative?
$$\frac{df}{dz}=\frac{1}{2}\left( \frac{\partial f}{\partial x} -i \frac{\partial f}{\partial y} \right)$$
where $f=f(z)$ is an analytic complex-valued function (Cauchy-Riemann conditions met) of the complex variable $z=x+iy$.

REPLY [0 votes]: Many Coefficients Work
If $f$ is differentiable at $a$, and $z$ is shorthand for $x+iy$ with $x,y$ real-valued, then the Cauchy-Riemann equation(s) state that $\left.\dfrac{\partial}{\partial x}f(z)\right|_{z=a}=-i\left.\dfrac{\partial}{\partial y}f(z)\right|_{z=a}$. By taking the limit of the difference quotient along a horizontal line, we find that $f'(a)=\left.\dfrac{\partial}{\partial x}f(z)\right|_{z=a}$. To simplify the notation, I'll just assume $f$ is differentiable in some open region and not bother with $a$.
By applying the above, we have $f'(z)=r\dfrac{\partial f}{\partial x}+i(r-1)\dfrac{\partial f}{\partial y}$ for any complex number $r$, including special cases like:

$r=1$ for $f'(z)=\dfrac{\partial f}{\partial x}$
$r=\dfrac12$ for $\boxed{f'(z)=\dfrac12\left(\dfrac{\partial f}{\partial x} -i \dfrac{\partial f}{\partial y}\right)}$
$r=2$ for $f'(z)=2\dfrac{\partial f}{\partial x} +i \dfrac{\partial f}{\partial y}$
$r=1-i$ for $f'(z)=\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}-i\dfrac{\partial f}{\partial x}$

Why 1/2?
But there is a reason why we write $\dfrac{\partial f}{\partial z}=\dfrac12\left(\dfrac{\partial f}{\partial x} -i \dfrac{\partial f}{\partial y}\right)$ in particular, as opposed to picking other coefficients that yield $f'(z)$.
Suppose that $f$ is not necessarily complex differentiable, but the partials $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial x}$ exist (and are continuous, say) on an open region. Then we could reasonably write $\mathrm df=\dfrac{\partial f}{\partial x}\mathrm dx+\dfrac{\partial f}{\partial y}\mathrm dy$.
If $z=x+iy$ and $\overline{z}=x-iy$, then applying $\mathrm d$ to both sides of those equations, we can reasonably write $\mathrm dz=\mathrm dx+i\mathrm dy$ and $\mathrm d\overline z=\mathrm dx-i\mathrm dy$. Solving, we get $\mathrm dx=(\mathrm dz+\mathrm d\overline z)/2$ and $\mathrm dy=-i(\mathrm dz-\mathrm d\overline z)/2$. This means that $\mathrm df=\dfrac{\partial f}{\partial x}\cdot\dfrac{\mathrm dz+\mathrm d\overline z}{2}-i\dfrac{\partial f}{\partial y}\cdot\dfrac{\mathrm dz-\mathrm d\overline z}{2}$ and so $\mathrm df=\dfrac12\left(\dfrac{\partial f}{\partial x}-i\dfrac{\partial f}{\partial y}\right)\mathrm dz+\dfrac12\left(\dfrac{\partial f}{\partial x}+i\dfrac{\partial f}{\partial y}\right)\mathrm d\overline z$.
This suggests defining the "Wirtinger derivatives" $\dfrac{\partial f}{\partial z}=\dfrac12\left(\dfrac{\partial f}{\partial x}-i\dfrac{\partial f}{\partial y}\right)$ and $\dfrac{\partial f}{\partial \overline z}=\dfrac12\left(\dfrac{\partial f}{\partial x}+i\dfrac{\partial f}{\partial y}\right)$ so that we can write the tidy $\mathrm df=\dfrac{\partial f}{\partial z}\mathrm dz+\dfrac{\partial f}{\partial \overline z}\mathrm d\overline z$.
In the special case where $f$ is differentiable at a relevant point, $\dfrac{\partial f}{\partial \overline z}=\dfrac12\left(\dfrac{\partial f}{\partial x}+i\dfrac{\partial f}{\partial y}\right)=0$ by Cauchy-Riemann, so that $\mathrm df=\dfrac{\partial f}{\partial z}\mathrm dz+0\mathrm d\overline z=\dfrac{\partial f}{\partial z}\mathrm dz$, so that we could divide by $\mathrm dz$ to consistently write $\dfrac{\mathrm df}{\mathrm dz}=\dfrac{\partial f}{\partial z}=\dfrac12\left(\dfrac{\partial f}{\partial x}-i\dfrac{\partial f}{\partial y}\right)$.
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