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I was wondering about the following for quite a while: The euclidean topology on $\mathbb{R}^2$ is homeomorphic with the topology induced by the modulus function on $\mathbb{C}$. Hence limit behavior is the same in both. So, doesn't that mean that the limit which defines the complex derivative converges if and only it converges in $\mathbb{R}^2$?

Clearly, I am missing something, since not every differentiable function $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is smooth, but over $\mathbb{C}$ this is true.

P.S.: I read the question: Is Complex Analysis equivalent Real Analysis with $f:\mathbb R^2 \to \mathbb R^2$?, it does however not clear up my confusion.

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$f:\mathbb{C}\rightarrow\mathbb{C}$ is $\mathbb{C}$-differentiable iff it is $\mathbb{R}^2$-differentiable and satisfies the Cauchy-Riemann equations. – Harry Altman Dec 13 '12 at 16:39
up vote 8 down vote accepted

Derivatives are linear maps that approximate the function within $o$. In other words, if $f$ is differentiable at $x$, then $f(x+t)=f(x)+A(t)+o(t)$, where $A$ is a linear map and $\lim_{t\to 0}\frac{o(t)}{|t|}=0$. Here $|t|$ denotes the magnitude of $t$, which, as you pointed out, is the same whether we view $t$ as a complex number or a $2$-dimensional real vector.

In the 2-dimensional real case, $A$ can be any linear map, ie. any $2\times 2$ matrix with entries in $\mathbb{R}$.

But in the complex case, $A$ is linear map of 1-dimensional complex vector spaces, ie. a $1\times 1$ matrix with entries in $\mathbb{C}$. In other words, $A$ is multiplication by some complex number $z$. If we think of $\mathbb{C}$ as $\mathbb{R}^2$, then $A$ is a rotation (by the argument of $z$) and scaling (by the magnitude of $z$). Not every $2\times 2$ linear map is of this form.

In sum, you are right to note that the convergence is the same in each space. Real differentiability means that approximations converge to some linear map, while complex differentiability means that approximations converge to a more specific type of linear map, which is why complex differentiability is a stronger condition than differentiability in $\mathbb{R}^2$.

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Under the geometric (clifford) algebra of the 2d real plane, complex differentiability is no more restrictive than differentiability on the real 2d plane, and both can be handled with the same formalism.

The geometric algebra of the 2d plane has not just scalars and vectors but bivectors, which are "directed areas". Since a 2d plane has only one area, the space of bivectors is essentially a zero-dimensional vector space. Linear combinations of scalars and bivectors behave like complex numbers.

Differentiability under geometric calculus is the condition that $\nabla A = 0$ for any multivector field $A$. This includes ordinary vector fields as well as "complex-valued" fields. Here is a simple illustration. Let $w(r) = u(r) + e^x e^y v(r)$ be the analogue of a complex field on the real 2d plane with basis $e_x, e_y$. Then,

$$\nabla w = (e^x \partial_x + e^y \partial_y) (u + e^x e^y v) = e^x (\partial_x u - \partial_y v) + e^y (\partial_y u + \partial_x v) = 0$$

These are the Cauchy-Riemann conditions, but our algebra does not actually have complex numbers. This is handled entirely with objects constructed from the 2d real plane.

Now consider a (co)vector field $f = f_x e^x + f_y e^y$. The differentiability condition reduces to

$$\nabla f = (\partial_x f_x + \partial_y f_y) + e^x e^y (\partial_x f_y - \partial_y f_x) = 0$$

The two terms produce the condition that the vector field have vanishing divergence and curl.

How are these concepts related to complex differentiability and integrability? Through this formalism, we can see that a lot of the concepts in complex analysis carry over to vector calculus. When $\nabla f = 0$ in this way, we can find the value of $f$ at any given point like so. Let $\nabla G = \delta(r)$ be the Green's function for the vector derivative. Then,

$$\oint G(r-r') \, d\ell' \, f(r') = -\int \delta(r-r') \, dA' \, f(r') = -i f(r)$$

Look at the left-hand side: this is an integral over a curve. Look at the right: this is proportional to the value of the function at a point. The value of a function is determined entirely by its values on a bounding curve? That's a result from complex analysis! But this is a theorem in vector calculus! In fact, they're both just aspects of the same bigger idea, the generalized Stokes theorem, or the fundamental theorem of calculus.

The calculus of complex fields is just a subset of the calculus of multivector fields on the 2d plane, and as such, so many of complex analysis' results apply to vector fields also--the neat conditions on integrability and so on. That complex analysis is the dominant way of looking at the 2d plane is, to me, quite lazy and misses how vector fields with the right properties can be neatly tied into those theorems too.

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The limit that defines $\mathbb{R}^n$ differentiability is: $$\lim_{\mathbf{h}\to \mathbf{0}}\frac{\mathbf{f}(\mathbf{a}+\mathbf{h})-\mathbf{f}(\mathbf{a})-T(\mathbf{h})}{\left\|\mathbf{h}\right\|}=\mathbf{0}$$ where $T$ is a linear transform. The limit that defines $\mathbb{C}$ differentiability is not $$\lim_{h\to 0}\frac{f(z+h)-f(z)-T(h)}{\left|h\right|}=0$$ but rather $$\lim_{h\to 0}\frac{f(z+h)-f(z)-T(h)}{h}=0$$ In addition, in the complex case we also require $T$ to be a complex linear transform, not just a linear map, that must be "nice" towards complex multiplication.These two facts create the so called Cauchy Riemann Equations. You can read more here pg 111. In other words, multiplication and division of complex numbers make $\mathbb{C}$ differentiabilty much stronger than $\mathbb{R}^2$ differentiability.

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Oh, so the first major difference is that the division is not normed. In $\mathbb{C}$ we can do this since every element apart from $0$ has an inverse, and in $\mathbb{R}^2$ this is not possible since it is not even a integral domain with the usual multiplication. With the function being a complex linear transformation, you mean that the scalars are complex? – sxd Dec 13 '12 at 16:45
@Nameless I'm not convinced that division being defined on $\bf C$ makes a difference, nor that having $h$ instead of $|h|$ in the denominator makes any difference. Recall that $f(x) \to 0$ as $x \to 0$ iff $|f(x)| \to 0$ as $x \to 0$ for any map $f:X \to Y$, where $X,Y$ are any of $\bf R^n, R^m, C.$ The difference is only the type of linearity required; see Remmert's text, for example. – Optional Jun 14 '14 at 5:43
This thread about an upvoted and accepted answer makes for a fascinating reading... As noted by @Optional, the answer is wrong since asking that some function $A:\mathbb C\to\mathbb C$ is such that $A(h)/h\to0$ or such that $A(h)/|h|\to0$ or such that $|A(h)/h|\to0$ when $h\to0$, $h$ complex, are all strictly equivalent. The asker is still active on the site so maybe something can be done? – Did Oct 7 '14 at 15:32
@Did I'm not sure what the protocol is in this case, but I've downvoted the answer, and encourage others to do the same. – Optional Oct 7 '14 at 18:25
@Optional Indeed, votes (up and down) are meant to separate the wheat from the chaff, aren't they? – Did Oct 7 '14 at 18:28

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