# Why is the there a $-i$ in this partial derivative?

Suppose $f(z)=u(z)+iv(z)$ is a complex function of a complex variable $z=x+iy$. In the book I'm reading, it states for real values $h$, the imaginary part $y$ is kept constant, so the derivative becomes a partial derivative with respect to $x$: $$f'(z)=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}.$$ I get that, but then it states if we substiture purely imaginary values $ik$ for $h$, $$f'(z)=\lim_{k\to 0}\frac{f(z+ik)-f(z)}{ik}=-i\frac{\partial f}{\partial y}.$$ How does that $-i$ appear? Is it some instance of the chain rule? Thank you.

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Isn't it already there in the last equation (middle part) sitting in front of the $k$ in the denominator? –  Fabian Jan 19 '12 at 17:38
$-i = \frac{1}{i}$. Or rather, $f'(z) = \lim_k \frac{1}{i}\frac{f(z+ik)-f(z)}{k}=\frac{1}{i}\frac{f(z_x,z_y+k)-f(z_x,z_y)}{k} = -i\frac{\partial f}{\partial y}$. –  Neal Jan 19 '12 at 17:47

As Neal said, it follows because $\frac{1}{i} = -i$. Here's a proof using "multiplying by a convenient form of $1$":

$$\begin{array}{} \frac{1}{i} &= \frac{1}{i} \cdot \frac{-i}{-i} \\ &= \frac{-i}{i \cdot (-i)} \\ &= \frac{-i}{-(i \cdot i)} \\ &= \frac{-i}{-(-1)} \\ &= -i. \end{array}$$

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There is one fundamental theorem ( that I have subdivided into two statements) in elementary complex function theory for a function $f=u+iv:U \to \mathbb C$ defined on an open subset $U \subset \mathbb C$.

1) The function $f$ is complex-differentiable (= holomorphic) on $U$ if and only if $u$ and $v$ are continously differentiable on $U$ and satisfy both Cauchy-Riemann equations on $U$ :$$\frac {\partial u}{\partial x}=\frac {\partial v}{\partial y} \quad \text{and} \quad \frac {\partial u}{\partial y}=-\frac {\partial v}{\partial x}$$

2) If that is the case you have for every $z_0=x_0+iy_0\in U$ the equality of complex numbers $$f'(z_0)=\frac {\partial u}{\partial x}(x_0,y_0)+i\frac {\partial v}{\partial x} (x_0,y_0) = \frac {\partial f}{\partial x}(x_0,y_0)\in \mathbb C$$

This is the heart of complex differential calculus
It looks easy but my experience is that many students do not always see that clearly, and I have answered your question in part so as to have a place to refer them to.
So take the time and learn this theorem till you feel you completely master it.

Once you know it, many results/exercises become trivial.
Indeed since $f'(z_0)=\frac {\partial u}{\partial x}(x_0,y_0)+i\frac {\partial v}{\partial x} (x_0,y_0)$ you get by using Cauchy-Riemann $f'(z_0)=\frac {\partial v}{\partial y}(x_0,y_0)+i(-\frac {\partial u}{\partial y} (x_0,y_0))=-i[\frac {\partial u}{\partial y}(x_0,y_0)+i\frac {\partial v}{\partial y} (x_0,y_0)]=-i\frac {\partial f}{\partial y}(x_0,y_0)\in \mathbb C$
Finally, let me warn you that I find admonitions to keep some quantity like $y$ constant a bit ad hoc and dangerous; quoting the theorem above is always efficient and rigorous .