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Taken from wolfram: enter image description here

It is probably something very simple, but I can't quite get where the factor of $\frac{1}{2}$ comes from.

Given $z = x + iy,$ I would think that (and maybe this is where I'm going wrong)

$ \frac{\partial x}{\partial z} = 1,$ and

$ \frac{\partial y}{\partial z} = -i,$

therefore (4) can be expressed as

$ \frac{df}{dz} = \frac{\partial f}{\partial x} (1) + \frac{\partial f}{\partial y} (-i)$.

The only way the math would work is if $\frac{\partial x}{\partial z} = \frac{1}{2}$ and $\frac{\partial y}{\partial z} = -\frac{i}{2}$, but I can't see how to show that formally.

(5) is certainly correct when I plug in specific examples.

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Hint: \begin{align} x = \frac{z+\bar z}{2} \ \ \text{ and } \ \ y =\frac{z-\bar z}{2i} \end{align} which means \begin{align} \frac{\partial x}{\partial z} = \frac{1}{2} \ \ \text{ and } \ \ \frac{\partial y}{\partial z} = \frac{1}{2i}. \end{align}

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    Thank you! It appears my mistake is that i took $\frac{\partial x}{\partial z}$ using an equation that also had $y$ in it (and likewise for y). This makes perfect sense. – 2017-01-07