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I'm having a hard time understanding complex differentials. I know that when I have a field $\mathbb K$ and a $\mathbb K-$vector space $\mathbb K^n,$ then we define $dx_i\in \mathrm{Lin}(\mathbb K^n,\mathbb K)$ on the standard basis $\{e_i\}_{i=1}^n$ of $\mathbb K^n$ as follows:

$$ dx_i(e_j)=\begin{cases}1 & \mbox{ for }j=i,\\ 0 & \mbox{ for }j \neq i.\end{cases}$$

So defined $dx_i$ form a basis of $\mathrm{Lin}(\mathbb K^n,\mathbb K)$

Now I have the symbol $dz$ for $z$ being a complex variable and I'm not sure I understand what it means. I know that this is supposed to be true and a definition of $dz:$

$$dz =d\:\mathrm{Re}(z)+id\:\mathrm{Im}(z).$$

I cannot fathom this definition though. What is the space in which the operations on the right-hand side are performed? $\mathrm{Re}(z)$ and $\mathrm{Im}(z)$ are real variables, right? So the space should be $\mathrm{Lin}(\mathbb R^2,\mathbb R).$ But this is an $\mathbb R -$space, not a $\mathbb C -$space so the multiplication by $i$ shouldn't be allowed.

And then I see the symbol $|dz|$ and integrals are computed with it, like here, page 3. What does this symbol mean?

Edit: I would like to improve the formulation of a part of my problem and post my newly found (thanks to the comments) answer to that part. Let's take the equality

$$dz=dx+idy,$$

where $x=\mathrm{Re}(z)$ and $y=\mathrm{Im}(z).$ According to the definition in the first paragraph of this post, $dz$ is a $\mathbb C-$linear map, $dz:\mathbb C\to\mathbb C,$ and $dz=\operatorname{id}_{\mathbb C}.$

On the other hand, $dx$ and $dy$ are $\mathbb R-$linear maps, $dx,dy:\mathbb R^2\to \mathbb R$ given by

$$ \begin{eqnarray} dx(e_1)=1,\\ dx(e_2)=0,\\ dy(e_1)=0,\\ dy(e_2)=1. \end{eqnarray} $$

I understand that I should carry out the identification: $$\mathbb R^2\ni e_1\mapsto 1\in\mathbb C,$$$$\mathbb R^2\ni e_2\mapsto i\in \mathbb C.$$ This gives me

$$ \begin{eqnarray} dx(1)=1,\\ dx(i)=0,\\ dy(1)=0,\\ dy(i)=1. \end{eqnarray} $$

These are clearly not $\mathbb{C}-$linear maps. This was my problem. $dy$ is not a $\mathbb{C}-$linear map but just an $\mathbb{R}-$linear map from $\mathbb C$ into $\mathbb R.$ The set of all such linear maps is an $\mathbb R-$vector space, not a $\mathbb{C}-$vector space so there is no such thing as the product $i\cdot dy.$

However, after Pierre-Yves Gaillard's comments, I realized that I should also carry out another identification -- in the codomains of $dx$ and $dy:$ $$\mathbb R \ni 1 \mapsto 1\in \mathbb C,$$

that is consider the codomains of $dx$ and $dy$ to be the real axis of the complex plane. This doesn't make $dx$ and $dy$ $\mathbb C-$linear maps, but it does make them complex functions and so allows them to be multiplied by $i$. And indeed, now $$dz=\operatorname{id}_{\mathbb C}=dx+idy.$$

I'm sorry about being so obtuse. I'm not sure this question has any value at all to the community, so perhaps I should remove this part?

However, I still do not understand what the definition of $|dz|$ is in these terms.

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5  
$z=x+iy$; $dz=dx+idy$; $|dz|=(dx^2+dy^2)^{\frac{1}{2}}$ –  Savinov Evgeny Jan 24 '12 at 19:53
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Dear ymar: Please tell me if you agree with (a) and (b) below. (a) Let $V$ be a real vector space, let $W$ be a complex vector space, and let $A$ be the set of $\mathbb R$-linear maps from $V$ to $W$. Then $A$ is a complex vector space in a natural way. (b) If $V$ and $W$ are complex vector spaces, we can view $V$ as a real vector space, and apply (a) to $V$ and $W$. –  Pierre-Yves Gaillard Jan 24 '12 at 20:16
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(There was a typo in the - now deleted - previous comment. Here is the corrected version.) Dear ymar: Your edit looks very nice! One can define $|dz|$ as the map $z\mapsto|z|$ from $\mathbb C$ to itself. But there are some comments to make about this definition. I'll try to make them soon, but it might be good to post this short comment already. –  Pierre-Yves Gaillard Jan 24 '12 at 23:23
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Dear ymar: Denote by $E$ the complex vector space formed by the $\mathbb R$-linear endomorphisms of $\mathbb C$. Then $dz$ and $d\overline z$ make up a $\mathbb C$-basis of $E$. Let $G$ be the multiplicative group of nonzero complex numbers. To each $a\in G$ we attach the $\mathbb C$-linear automorphism $r(a)$ defined by $$(r(a)f)(z):=f(az).$$ This gives in particular $$r(a)(dz)=a\ dz,\quad r(a)(d\overline z)=\overline a\ d\overline z,\quad r(ab)=r(a)r(b).$$ We summarize this by saying that $r$ is a linear representation of $G$ on $E$... –  Pierre-Yves Gaillard Jan 24 '12 at 23:27
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... Now, let $V$ be the set of those maps $f$ from $\mathbb C$ to itself which satisfy $f(az)=|a|f(z)$ for all $a,z$ in $\mathbb C$. Then $V$ is a complex vector space in a natural way with $\mathbb C$-basis $|dz|$ (so $\dim V=1$). Moreover, the formula $$(s(a)f)(z):=|a|f(z)$$ defines a linear representation of $G$ on $V$. –  Pierre-Yves Gaillard Jan 24 '12 at 23:28

1 Answer 1

up vote 3 down vote accepted

Traditionally (e.g. in Euler's writings), $dz=dx+i\;dy$ is an infinitely small change as $z$ moves an infinitely small distance from one point to another. If $\gamma$ is a curve and $z$ moves along the curve, then $f(z)\;dz$ is a product of a finite complex number $f(z)$ and an infinitely small complex number $dz$. The integral $\displaystyle\int_\gamma f(z)\;dz$ is the sum of infinitely many of those infinitely small quantities. None of this is logically rigorous. The role of this non-rigorous account within the rigorous account is that this is what is to be made rigorous.

The absolute value $|dz|=\sqrt{dx^2+dy^2}$ is the infinitely small distance that $z$ has moved along the curve. The integral $\displaystyle\int_\gamma dz$ is the sum of all the infinitely small changes in $z$, thus it is the final value minus the initial value. The integral $\displaystyle\int_\gamma |dz|$ is the sum of the infinitely small arc lengths, and is therefore the total arc length.

Maybe you're OK saying (in certain contexts) $dz\in \mathrm{Lin}(\mathbb C,\mathbb C)$. I wouldn't be surprised if $|dz|$ cannot be interpreted the same way, but if not, it's just a limitation on that way of interpreting it as a way of making these things rigorous.

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+1. And the moral of non-standard analysis is that anxiety about making this kind of thing rigorous was basically misplaced. No interesting results of Gauss and Euler become any more or less interesting or valid when interpreted using Weierstrass $\epsilon-\delta$ definitions, nor do they gain or lose beauty or utility when cast into the form of non-standard analysis. The foundations of mathematics are of interest to people working on the foundations of mathematics. They have no major consequences for anyone else. –  Ben Crowell Jan 25 '12 at 6:21
    
@BenCrowell Well, I just need a way to understand these things. I don't understand the infinitely small changes. I cannot imagine them. But I can imagine linear maps, so I'm thinking perhaps I could try making my brain at least accept differentials if I use this route. –  user23211 Jan 25 '12 at 17:03
    
@BenCrowell : The foundations of mathematics led to some interesting things about computer software. –  Michael Hardy Jan 25 '12 at 21:04
    
@ymar: I can't imagine the set of natural numbers, because it's infinite and my brain is finite. But the natural numbers are a standard part of classical mathematics that every mathematician needs to know about, and it would also be extremely inconvenient to do everyday mathematics in a system that didn't have all the unrealistic features of the natural number system. Infinitesimals notated like $dx$ are likewise a part of the classical heritage of mathematics, and they are likewise extremely convenient. –  Ben Crowell Jan 26 '12 at 1:04
    
@MichaelHardy: Interesting. What computer software do you have in mind? I may be guilty of hyperbole, or you and I may mean different things by "the foundations of mathematics." –  Ben Crowell Jan 26 '12 at 1:10

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