fixed.
That is not an acceptable choice for d/dz.
If f = u + iv is analytic in a neighbourhood of z, then by definition the following limit exists, and is independent of how h \in \mathbb{C} approaches zero: <br />
\frac{df}{dz} = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h} = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x,y)}{h}.<br /> Thus you can take h \in \mathbb{R} so that <br />
\frac{df}{dz} = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}, but instead you can take h \in i\mathbb{R} so that <br />
\frac{df}{dz} = -i\frac{\partial f}{\partial y} = -i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} and since these are equal we must have <br />
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad<br />
\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} which are the Cauchy-Riemann equations.
One also has <br />
\frac{df}{dz} = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x,y)}{h}\\ = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x, y + \mathrm{Im}(h))}{h} + \lim_{h \to 0}\frac{f(x, y + \mathrm{Im}(h)) - f(x,y)}{h} \\<br />
= \lim_{h \to 0} \frac{\mathrm{Re}(h)}h \frac{\partial f}{\partial x} + \lim_{h \to 0} \frac{\mathrm{Im}(h)}h \frac{\partial f}{\partial y}.
