Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
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
2  
$-i = \frac{1}{i}$. 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
add comment

2 Answers

up vote 2 down vote accepted

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

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

share|improve this answer
add comment

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 .

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.