# Complex Analysis and vector calculus

1. Oct 22, 2012

### Avatrin

Hi
How much different is complex analysis from vector calculus?

To me complex analysis looks like vector calculus combined with algebra of complex numbers..

2. Oct 22, 2012

### Erland

If so, only vector calculus in two dimensions. There are no complex numbers in three dimensions.

3. Oct 22, 2012

### Muphrid

They're taught like they're two different things, but they're really very closely related. For instance, the idea that a holomorphic function is determined everywhere by its values on some closed curve is the same conceptually as a divergenceless and curlless vector field being determined entirely by its values on some closed surface.

4. Oct 22, 2012

### Stephen Tashi

There is a textbook that emphasizes the connections between complex analysis and two variable calculus: https://www.amazon.com/Complex-Vari...60198&sr=8-1&keywords=Latta+Complex+Variables

There can be functions of two variables which have directional derivatives in all directions but which aren't "nice" in other ways. By contrast, the existence of the derivative of a complex variables enforces a stronger requirement than having the directional derivatives all exist. This is one place that the analogy between two variable calculus and complex analysis seems to break down. (Perhaps some forum member can explain how to interpret things so it doesn't ?)

Last edited by a moderator: May 6, 2017
5. Oct 23, 2012

### Useful nucleus

Another difference that I can think of, for complex numbers one can define division while for 2-dimensional vectors it is not customary to define division (as it does not have a geometric meaning). Similarly, multiplication of two complex yields a complex , while in vector calculus the scalar multiplication yields a real number (the other possibility is "cross product" which yields a vector but it is not commutative).

Also the Cauchy-Schwartz inequality for complex numbers is a bit different than its counterpart of 2D vectors.

These are examples, but I do not have a general statement.

6. Oct 23, 2012

### Vargo

You can think of complex analysis as a sort of special case of vector calculus.

An analytic function can be thought of as a mapping F of the plane into itself such that the derivative (Jacoby matrix) is always scaling plus rotation:

$DF = \left( \begin{array}{cc} r\cos(\theta) & -r\sin(\theta) \\ r\sin(\theta) & r\cos(\theta)\end{array}\right)$

So at any particular point, if you linearize the mapping there, you see that it is just a scaled rotation. This explains why analytic functions are conformal (preserve angles). That might not seem so special at first, but (as others have said) it actually enforces a lot of structure on the function. It is really surprising how many things are true about a function just because it has that property. As you might expect, a lot of them are geometric in nature (or at least have a strong geometric interpretation).

For example, suppose f(z) is analytic over the whole complex plane. Then consider the range of f(z). Unless f(z) is constant, then the range has to include the whole complex plane minus at most one point. This is called Picard's Theorem. For example e^z is never 0, but it hits every other point in the complex plane. And that this true for every possible f(z).

Obviously, there are no theorems even remotely like that for differentiable functions F(x,y) from R2 to R2. So that is just one small example of things that are true in complex variables.

7. Apr 25, 2014

### Chetan_Waghela

Division is not applicable on complex numbers also in usual sense, because complex numbers do not have a proper value as such. Scalar multiplication is equivalent to multiplying a complex number by it's conjugate. I feel you are wrong in both these cases.

Last edited: Apr 25, 2014
8. Apr 25, 2014

### micromass

Staff Emeritus
Your post makes no sense. Division is perfectly applicable on complex numbers. The value (whatever you mean by this) of complex numbers is well understood.

9. Apr 25, 2014

### Chetan_Waghela

What is value of 2+3i ? obviously (2+3i)/(3+4i) has meaning simielar to (2/hat{x}+3/hat{y})/(3/hat{x}+3/hat{y}) has

10. Apr 25, 2014

### micromass

Staff Emeritus
Define value.

11. Apr 25, 2014

### chogg

Yes there are; they are usually called "quaternions".

Reading about geometric algebra will clarify the relationships among vector algebra, complex numbers, quaternions, and many other mathematical systems. I'd recommend the first two chapters of Geometric Algebra for Physicists for some engaging reading on the subject.

12. Apr 25, 2014

### jbunniii

Another is Complex Variables by Francis Flanigan. The author starts with calculus of 2 variables and harmonic functions, and only introduces complex variables after about 100 pages. I haven't read the whole thing, but the parts I have read were quite nice and well written. Bonus: it's a cheap Dover edition.

Last edited by a moderator: May 6, 2017
13. Apr 25, 2014

### homeomorphic

What I would say is that you can interpret Complex Analysis as a branch of vector calculus, but some of the insights, particularly with respect to complex differentiation or something like Mobius transformations, are best understood in terms of mappings, rather than in terms of vector fields, so that's what I would say is the main difference. But, it's really sort of a matter of perspective, I think.

Well, technically, you need a 4th dimension to get the multiplication to work right. That's what Hamilton struggled with for years before he invented them. But they do have interesting 3-dimensional applications (and I suppose you could say they have 3-dimensional subspaces that serve as 3-d complex numbers, but these are not closed under multiplication and division). In fact, the dot product and cross product were originally derived from quaternions. If you multiply 2 imaginary quaternions, the result has real part minus the dot product and the imaginary part is the cross product. So, they do bring to fruition some of Hamilton's original 3-dimensional aspirations, even though he had to use the trick of letting in one more dimension to get it to work.

14. Apr 25, 2014

### Chetan_Waghela

I said complex number division is possible, you can read my statement again. Vector division is also possible, but the answer you get is not unique. I just meant to say that they are not in usual sense.

http://mathworld.wolfram.com/VectorDivision.html

15. Apr 25, 2014

### chogg

Yes, there is a division for 2-d vectors (indeed, arbitrary-D vectors) which does have geometric meaning. It requires the geometric product, which unifies the dot product and the wedge product (the latter is somewhat like the cross product, but more general).

tl;dr: Pictures would help, I know! If you want them, section 6.1 of Geometric Algebra for Computer Science explains it very nicely. This is just my recounting from memory of that explanation.

"Division" means inversion, basically. If we write
$$x = u/v$$
what we really mean is
$$xv = u.$$
The product $xv$ is the geometric product; for vectors, this is:
\begin{align} xv &= x \cdot v &+& x \wedge v \\ &\equiv a &+& B \end{align}
In other words, it's a scalar $a$ plus a bivector $B$ (just like a complex number is a real number plus an imaginary number).

Now to the geometric meaning.

We'll suppose we know $v$, $a$, and $B$. (In other words, we'll suppose we know $v$ and $xv$.) And we want to solve for $x$.

The dot product part alone can't do it. Imagine we find a candidate solution $x_0$ so that $x_0 \cdot v = a$. The problem is that this solution is not unique. We can add any vector $x_\perp$ (such that $x_\perp \cdot v = 0$), and $(x_0 + x_\perp) \cdot v$ will also be $a$.

In other words, you can't have a division for the dot product. Your candidate solutions form an $(n-1)$-dimensional subspace (in familiar 3D, this is just a plane). This plane is orthogonal to $v$.

The wedge product part alone can't do it either. Imagine we find a candidate solution $x_0$ so that $x_0 \wedge v = B$. The problem is that this solution is not unique. We can add any multiple of $v$ to $x_0$, and $(x_0 + \lambda v) \wedge v$ will also be $B$.

In other words, you can't have a division for the wedge product. Your candidate solutions form a line (parallel to $v$).

But the plane from your failed dot product division, and the line from your failed wedge product division, meet at a single point. This point is the unique vector $x$ which satisfies $x \cdot v = a$ and $x \wedge v = B$ simultaneously.

So the geometric product of vectors does have a division, and it is geometrically meaningful.

16. Apr 25, 2014

### Erland

Quaternions have four dimensions.

I don't deny, though, that there are connections between quaternions and vector analysis.

17. Apr 25, 2014

### chogg

I guess that's one way to look at it, but I don't think it's the most natural.

You get quaternions from an orthonormal basis in 3D space, just as you get complex numbers from an orthonormal basis in 2D space.

The "extra dimension" is really just a plain vanilla scalar. (The three basis quaternions are bivectors.)

Let me explain:

2D space
Basis vectors: $\{e_1, e_2\}$.
Unique combinations of even numbers of basis vectors: $\{1, e_1 e_2\}$
"Complex" interpretation: things proportional to $1$ are like "real numbers"; things proportional to $e_1e_2$ are like "imaginary numbers" (note that $(e_1e_2)^2 = -1$).

3D space
Basis vectors: $\{e_1, e_2, e_3\}$.
Unique combinations of even numbers of basis vectors: $\{1, e_1 e_2, e_1 e_3, e_2 e_3\}$
"Complex" interpretation: things proportional to $1$ are like the "scalar part" of a quaternion (and are indeed just scalars); things proportional to any other elements are like the "vector part" (although it's better to think of them as bivectors).

18. Apr 25, 2014

### chogg

I disagree that vector division does not give a unique answer. It does when that division is with respect to the geometric product. And this division works in arbitrarily many dimensions.

A vector divided by another vector is a complex number, even in arbitrarily many dimensions. This might sound surprising, because complex numbers live in a 2D space. Well, that space is the 2D subspace which is defined by the two vectors which span it. And the unit imaginary i for this complex number automatically encodes information about their common plane.

(Of course, this isn't true when the two vectors are collinear. But in that case their division is simply a scalar, so there's no geometric information to be had.)

19. Apr 27, 2014

### bayak

There is identity between algebra of complex (doubled) number and algebra of corresponding linear vector fields on Euclidean (pseudo-Euclidean) plane. In case of complex number we use generator $I = y\partial_{ x} - x\partial_{ y}$ and in case of doubled number we use generator $J = y\partial_{ x} + x\partial_{ y}$. At the same time, there is identity between complex (doubled) analysis and corresponding local algebra of linear vector fields on 2-manifold.

Last edited: Apr 27, 2014
20. Apr 27, 2014

### marmoset

This paper http://www.maa.org/sites/default/fi...endoerfer/1988/0025570x.di021152.02p01457.pdf by Bart Braden describes using the 'Polya vector field' associated with a complex function to visualize contour integrals.

The Polya vector field associated with a complex function $f(z)$ is the vector field $v = (\mathrm{Re}f(z),-\mathrm{Im}f(z))$. With this definition $f$ is analytic precisely when $v$ has zero curl and zero divergence ($\mathrm{div}(v)=0$ and $\mathrm{curl}(v)=0$ give the Cauchy-Riemann equations). Also the complex integral of $f$ around a contour $\gamma$ is just the circulation of $v$ around $\gamma$ plus $i$ times the flux of $v$ through $\gamma$.