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I'm taking complex analysis right now, and while I understand the proof that holomorphism of a function implies the Cauchy-Riemann equations I don't understand why we need the factor of $\frac{1}{2}$ the operator

$$\frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial}{\partial x} + \frac{1}{i} \frac{\partial}{\partial y}).$$

To show that the equations are implied we take the limit definition of holomorphicity with h being either purely real or imaginary and observe that we get partial differentiation with a factor of $\frac{1}{i}$ in front of the $y$ partial derivative when h is imaginary.

The book (Complex Analysis by Stein & Shakarchi) then just introduces the operators with the factor included, and the proof seems to assert that it is implied somehow by the limit derivation. Any assistance would be greatly appreciated.

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In the case where $f$ is holomorphic, both $\frac{\partial f}{\partial x}$ and $\frac{1}{i}\frac{\partial f}{\partial y}$ are equal to the complex derivative of $f$. We would like $\frac{\partial f}{\partial z}$ to also be equal to to the complex derivative of $f$, so we average them rather than taking their sum.

More generally, if we have a function which is formally written solely in terms of $z$ and $\overline{z}$, $\frac{\partial f}{\partial z}$ should mean the same thing regardless of whether we think of these as independent variables or as one variable and its complex conjugate. For example, if $f(z)=z\overline{z}$, we should have $$ \frac{\partial f}{\partial z}=\overline{z} $$ But $f(x+iy)=x^2+y^2$, so $\frac{\partial f}{\partial x}=2x$, $\frac{\partial f}{\partial y}=2y$. If we're building $\frac{\partial f}{\partial z}$ out of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ we have to divide by two to get the answer we know is right.

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  • $\begingroup$ Much appreciated, it seemed like an average but I wasn't sure. $\endgroup$ – Alec Rhea Sep 14 '16 at 3:25
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At a point $p\in \mathbb C$ the vector space $T_p^*=\mathbb C.dx\oplus \mathbb C.dy$ of complex valued $1$-forms is a $2$-dimensional complex vector space with basis $dx,dy.$
The complex functions $z=x+iy,\bar z=x-iy$ have differentials at $p$ equal to $dz=dx+idy,d\bar z =dx-idy$ and these differentials also are a basis of $T_p^*$
Any smooth function $f$ on a neighbourhood of $p$ has a differential which can be written in our two bases as: $$d_pf=\frac{\partial f}{\partial x}(p)dx+\frac{\partial f}{\partial y}(p)dy=Adz+Bdy=A(dx+idy) +B(dx-idy) $$ Since $dx, dy$ are a basis of $T_p^*$ this implies that $A+B=\frac{\partial f}{\partial x}(p), \;i(A-B)=\frac{\partial f}{\partial y}(p)$.
Thus, solving for $A,B$ we have $d_pf=[\frac{1}{2}(\frac{\partial f}{\partial x} (p)+ \frac{1}{i} \frac{\partial f}{\partial y}(p)]dz+[\frac{1}{2}(\frac{\partial f}{\partial x} (p)- \frac{1}{i} \frac{\partial f}{\partial y}(p)]d\bar z$ and if you define $$\frac{\partial f}{\partial z} (p)=\frac{1}{2}(\frac{\partial f}{\partial x} (p)+ \frac{1}{i} \frac{\partial f}{\partial y}(p)),\quad\frac{\partial f}{\partial \bar z} (p)=\frac{1}{2}(\frac{\partial f}{\partial x} (p)- \frac{1}{i} \frac{\partial f}{\partial y}(p))$$ you get the pleasant looking expression $d_pf=\frac{\partial f}{\partial z} (p)dz+\frac{\partial f}{\partial \bar z} (p)d\bar z$, which justifies the strange definitions for the differential operators $\frac{\partial }{\partial z} ,\frac{\partial }{\partial \bar z}$ .

WARNING
The above is completely rigorous: $dz,d\bar z$ are genuinely linearly independent vectors in the complex vector space $T^*_p$.
Beware however the absurd "explanation" sometimes given according to which $z,\bar z$ are "independent variables": they are not because each determines the other!

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