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.
 A: As Neal said, it follows because $\frac{1}{i} = -i$. Here's a proof using "multiplying by a convenient form of $1$":
$$\begin{array}{}
\frac{1}{i} &= \frac{1}{i} \cdot \frac{-i}{-i} \\
&= \frac{-i}{i \cdot (-i)} \\
&= \frac{-i}{-(i \cdot i)} \\
&= \frac{-i}{-(-1)} \\
&= -i.
\end{array}$$
A: 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.
For example your original question!
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 .
