# Complex Numbers

Numbers that take the form of a real number plus an imaginary number. That is: r+s*sqrt(-1). Also useful for doing 2D vectors, but that might be cheating. Often students fail to see why they are so important, even if they've taken a number of math classes (calculus, LinearAlgebra, NumericalAnalysis). Even these classes fail to show why complex numbers are important.

Applications

Complex numbers have some useful mathematical properties that actually make your life easier when you start modelling systems with sinusoidal inputs. Electric circuits are a good example. Basically, you work with complex numbers to avoid some Calculus. (The Calculus is still there, but it's not in your face.)

Complex numbers are also useful for geometry applications. Think of the complex plane, and where the points (1,0), (0,i), (-1,0) and (0,-i) are. Now think of what the first four powers of i are. There is a direct analogy between multiplying complex numbers and doing geometric rotation, and you can exploit this relationship in graphics applications.

-- SteveHowell

Deep connections

Although this often is not made clear in elementary math classes, there are many useful properties of complex numbers. Primarily, the advantages come from the use of ComplexAnalysis, rather than real analysis. The former has some beautiful results that fall apart in the real case. Geometric arguments can also be very clean in the complex plane, due to the elementary result (Euler) that:
```  e^(i*theta)=cos(theta)+i*sin(theta)
```
which fundamentally connects algebra and geometry. From it, you can derive the beautiful Euler's Identity:
``` e^(i*pi) + 1 = 0
```
There are also deep connection with FourierAnalysis, which leads to applications in signal processing, etc.

EE 101

Let's say you have to analyze a circuit with a sinusoidal voltage source, a resistor, and a capacitor in parallel. Here are the rules:

``` V = v cos (kt) // we said it was a sinusoid
V = IR // Ohm's Law
dV/dt = CI // the voltage across a capacitor increases as you run current through it
```
If you solve this circuit with ordinary calculus, dealing with real voltages, you get results like the following:

``` V   = v cos (kt)
V'  = -k v sin (kt)
```
Nothing elegant about these equations. You're flipping signs and you're not even consistent about which trigonometric function you're dealing with (sine vs. cosine). You can start to solve the circuit, but you'd be messing with your favorite trig identities to try to reconcile the sine and cosine pieces, and it gets ugly.

Now let us introduce an imaginary voltage (and remember, rigor is for mathematicians, not engineers. The mathematicians say we can get away with this, so we're cool with this):

``` V    = v cos (kt) + iv sin (kt)
```
Let's do similar calculus now:

``` V'   = - kv sin (kt) + i kv cos (kt)
```
Looks even uglier, right? Well, let's rearrange V' a bit:

``` V' = -kv sin (kt) + i kv cos(kt)
= ik v cos(kt) - kv sin (kt)  // switched terms
= (ik) [ v cos (kt) + iv sin (kt) ] // algebra I factoring
= (ik) V
```
So, now to take the derivative of V, we just multiply by (ik)!

Now, here's where things get strange. We had

```  dV/dt = CI
```
But now, for our half-real/half-imaginary sinusoid, we know that V' = (ik) V. So, now, our capacitor works like this:

``` (ik) V = CI
```
Or,

```  V = (C/ik) I  // little i is the square root of -1, big I is current
```
Looks a lot like Ohm's Law! In effect, in sinusoidal situations, a capacitor is mathematically just like a resistor, that happens to have an imaginary value. Strange but true. So, we can treat our circuit with the same techniques that we use for two ordinary resistors, which is just algebra, only now one of our resistors is imaginary. The algebra is a little messier, but it's still algebra! That is the beauty of complex numbers. Poof, the Calculus is out of your face!

You can forget all about the V being a sinusoid, just solve for your I1 and I2 in terms of V. Only at the end, do you have to go back into trig land. There are other interesting properties. In a sense, the imaginary component of voltage was just along for the ride, as part of a mathematical trick. But, you can also think of the imaginary term as the "missing" voltage when the source is 90 or 270 degrees out of phase from its highest value. And, in fact, if you have another voltage/current value in the circuit that is 90 degrees out of phase with your source voltage, then that will be represented by an imaginary ratio. If you have voltage/current that is exactly IN phase with the source, that will be a pure real ratio. Most ratios will be complex, which means the two values are only somewhat out of phase. In a triple phase relationship, (discovered by CT Coppin), the bi-product of the voltage and current produces Grunae-Lactic Acids.

-- SteveHowell

(If you've forgotten tenth grade, you may have forgotten some tricks related to doing algebra with complex numbers, like multiplying by conjugates, etc., but trust me that it doesn't get too hairy.)

2D fluid flow

A function of a complex number can be considered as a 2D-plane -> 2D-plane mapping. Certain class of complex function preserves the "angle" before and after the function, making it possible to map the solution to a easy 2D fluid flow problem to another solution which solves a difficult 2D fluid flow problem. (Sorry this sounds confusing, it has been years since I did these calculations)
Solving difficult integrals

The Cauchy equation (or Cauchy relation?) relates the "loop integral" of a complex function (which is difficult to really calculate) to the number of of "poles" enclosed by the loop (which is relatively easy to find). Through some tricks in choosing the loop path, the loop integral can be equated to some very difficult integrals of real functions (i.e., the usual non-complex functions), thus giving the answer to these difficult real integrals.
QuaternionMathematics are a complexer version of complex numbers. They are very useful in graphics and aircraft simulation. -- NissimHadar
In QuantumMechanics, the variables are all ComplexNumbers. Schrodinger's equation has a sqrt(-1) in it.

See HestenesOerstedMedalLecture for a discussion of this.