# Where does the factor of $\frac{1}{2}$ come from in the complex differential operators?

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.

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.

• Much appreciated, it seemed like an average but I wasn't sure. – Alec Rhea Sep 14 '16 at 3:25

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!