# What are complex numbers used for

1. Feb 21, 2004

### Hypercase

My doubt is what are complex numbers used for. Sure I can use them to solve eqns where there are no real solutions. But how does that help. What is the real life application of complex numbers.

2. Feb 21, 2004

### master_coda

The first "application" I ever used them in was circuit analysis. Instead of having to solve differential equations or use Laplace transforms, you can sometimes use complex numbers instead.

So you can treat an AC circuit like a DC circuit with complex numbers for impedance, voltage, current, etc. instead of real numbers.

3. Feb 21, 2004

### Dr Transport

Think about a complex number in terms of a magnitude and phase, not as a pair of numbers in a 2-d plane. A wave function has a magnitude and a phase, if not, you'd never get constructive and destructive interference, i.e. interference patterns.

4. Feb 21, 2004

### Hurkyl

Staff Emeritus
Historically, the application that forced the mathematical community to accept complex numbers was that of the cubic formula; in general it absolutely requires complex numbers to work, even when all of the roots of the cubic equation are real numbers.

In fact, there is an interesting "paradox" that there exist real numbers that cannot be written in terms of arithmetic and roots without using complex numbers.

Various applications that deal with 2-d geometry (e.g. 2-d fluid flow) can be handled in a much simpler way using complex numbers.

There are lots of problems about real numbers that can be done in a simpler way by using complex numbers. e.g.

$$\begin{equation*}\begin{split} \int e^x cos x \, dx &= \int \Re ( e^x (\cos x + i \sin x) )\, dx \\ &= \Re \int e^{x + ix} \, dx \\ &= \Re \int e^{(1+i) x} \, dx \\ &= \Re \left( \frac{1}{1+i} e^{(1+i)x} + C \right) \\ &= \Re \left( \frac{1}{2} (1 - i) e^x (\cos x + i \sin x) + C \right) \\ &= \frac{1}{2} e^x (\cos x + \sin x) + C \end{split}\end{equation*}$$

In summary, if they weren't useful, we wouldn't use them.

5. Feb 21, 2004

### ahrkron

Staff Emeritus
They can be used to simplify the way you obtain many results.

In many applications (physics, economics, engineering), you end up with differential equations involving cycles, which can be expressed in terms of sines and cosines. Instead, you can use complex exponentials to encode the same information. Why would you want to do that? because exponentials behave much nicer under integration and differentiation.

Also, since complex numbers are so closely related with rotations, they are perfect to represent many physical properties were analogous symmetries are involved. When you get to quantum mechanics, you find a property called "spin" (which is NOT physical rotation), the behavior of which is very similar to how imaginary numbers combine. when you combine the matrices that describe this property, you get i's all over.

At a deeper level, it turns out that a strong connection can be established between the electromagnetic field and quantum mechanics via something called "gauge invariance". In a nutshell, if you require that the schrodinger equation give the same solution regardless of the complex phase of your wavefunction, you end up needing the EM field.

One more: the electoweak interaction has, as symmetry group (which is closely related to how it behaves), the same as that of complex numbers (U(1)).

Yet another: have you heard of CP-violation? it is basically the fact that matter and antimatter behave differently (which is quite surprising, since they are essentially mirror images of each other). CP-violation can be understood and quantified in terms of a complex phase in a matrix that relates the different types of quarks.

6. Feb 21, 2004

### ahrkron

Staff Emeritus
Wow! Can you expand on that? (or provide a link)

7. Feb 22, 2004

### HallsofIvy

Staff Emeritus
If Hurkyl will forgive my jumping in here: there was a problem in another thread that reduced to solving the cubic equation 8x3- 6x= 1.

It's easy to show that that has 3 real roots (y= 8x3- 6x- 1 is negative for x< -1, positive for x= -0.5, negative again for x= 0, and positive for x=1- there must be one root between -1 and -0.5, another between -0.5 and 0, and a third root between 0 and 1.

The "cubic formula" gives that positive root as ((1+sqrt(3)i)/16)1/3+ ((-1+ sqrt(3)i)/16)1/3.

Even though that root is a positive real number, there is no way to represent it exactly without using a complex number.

Last edited: Feb 22, 2004
8. Feb 22, 2004

### matt grime

Not sure how much I buy that.

There is anyway in implicit assumption that one means 'describe algebraically' in this thread, obviously 'analytically' is different.

I think the reason I don't buy this example is that there might be a solution for this polynomial in terms of trig functions. In fact a quick check tells me there is since

4(-6/8)^3 + 27/64 is negative.

Moreover, using Cardano's formula yields things like:

$$(2+i11)^{1/3}+(2-i11)^{1/3}$$ which is actually equal to 4 (for some choice of cube root).

edit. I also presume you mean in terms of extracting roots etc of rational numbers - there is not problem with defining {a,b,c} to be the three roots of the cubic in ascending order. And what about transcendental numbers too?

Last edited: Feb 22, 2004
9. Feb 22, 2004

### Hurkyl

Staff Emeritus
Oh sure, make me remember my algebra. I haven't been able to recall/devise a proof in a few hours, and I can't think where to look for a reference. I'll ask at work later this week to see if anyone knows.

10. Feb 22, 2004

### Hurkyl

Staff Emeritus
Well, I found a reference, sort of...

http://library.wolfram.com/examples/quintic/timeline.html

1890, 1891

Vincenzo Mollame (1848-1912) and Ludwig Otto Hoelder (1859-1937) prove the impossibility of avoiding intermediate complex numbers in expressing the three roots of a cubic when they are all real.