Friday, September 7, 2012 CC-BY-NC
On the need for complex numbers

Maintainer: admin

1Recap of last lecture


First, we looked at the different subsets of numbers: the natural numbers ($\mathbb{N}$), the rational numbers ($\mathbb{Q}$), the integers ($\mathbb{Z}$), the reals ($\mathbb{Z}$), etc. The one we'll be talking more today is the complex numbers ($\mathbb{C}$).

1.2The depressed cubic equation

Then, we learned a bit about the historic motivations for finding the square roots of negative numbers. This main impetus came from the search of the solution for the depressed cubic equation, $x^3 + px + q = 0$, $p, q \in \mathbb{R}$. For this equation, the discriminant would be $\displaystyle \Delta = q^2 + \frac{4p^3}{27}$.

If $\Delta$ is positive, then we can write the root, $\alpha_1$ as $\displaystyle \sqrt[3]{\frac{-q + \sqrt{-\Delta}}{2}} + \sqrt[3]{\frac{-q + \sqrt{-\Delta}}{2}}$needs verification.

Of course, this solution is only valid when $\Delta > 0$, as we can't take the square root of a negative number. OR CAN WE.

2The roots of a cubic equation

Theorem. The number of roots that a depressed cubic equation has depends on the value of $\Delta$, as follows:

  • If $\Delta > 0$, the equation has a unique real root.
  • If $\Delta = 0$, there are either 1 or 2 roots.
  • If $\Delta < 0$, there are exactly 3 roots.

2.1Finding the critical points

To prove this, we first note that there are at most two critical points in the graph for the depressed cubic function. (Why is this? Well, if we take its derivative, then we get a quadratic equation, which as we all know has at most two roots.) We can call these critical points $x_1$ and $x_2$, and we can find them as follows:

$$\frac{d}{dx} \left [ x^3 + px + q \right ] = 0 \qquad \therefore 3x^2 + p = 0$$

If $p$ is negative, then $\displaystyle x_1 = \sqrt{\frac{-p}{3}}$, and $\displaystyle x_2 = -\sqrt{\frac{-p}{3}}$.

On the other hand, if $p$ is positive, there are no critical points, and the depressed cubic function is monotonically increasing. (Note that $\Delta$ is positive in this case.)

Now, let's calculate the value of $y$ (the cubic function) at these points:

$$ y_1 = \left ( \sqrt{\frac{-p}{3}} \right )^3 + p \left ( \sqrt{\frac{-p}{3}} \right ) + q = q + \frac{2}{3} p \sqrt{\frac{-p}{3}} \qquad y_2 = \left ( -\sqrt{\frac{-p}{3}} \right )^3 + p \left ( -\sqrt{\frac{-p}{3}} \right ) + q = q - \frac{2}{3} p \sqrt{\frac{-p}{3}} $$

So $y_1$ and $y_2$ are the local minima and maxima of the cubic function (obviously not the global minima and maxima, because the range of the function is $(-\infty, \infty)$). Now, if we take the product of $y_1$ and $y_2$, what do we get? Well, I'll tell you:

$$y_1 y_2 = \Delta$$


2.2The sign of delta

So, the sign of $\Delta$ tells you about the signs of the critical points: if $\Delta$ is positive, then the the critical points are on the same side of the x-axis (either both above it or both below it). In this situation, the graph would only intersect the x-axis at one spot, and so there would be one root. If $\Delta$ is negative, however, then the critical points are on opposite sides of the x-axis, and based on the intermediate value theorem there must be three roots. (If the last part is a bit hand-wavy, that's because the proof wasn't mentioned in lecture, and I don't know how to prove it on my own.)

Insert diagram later

Incidentally, if $\Delta = 0$, then at least one of the critical points has a y-value of 0 (i.e., lies directly on the x-axis). If both critical points lie on the x-axis, then there is exactly one root; if only one does, then there are exactly two roots. Kind of cool, though maybe not relevant. I don't know how to prove this rigorously, but here's a diagram:

Insert diagram later

2.3Introducing i

We now introduce a formal element $i$, defined such that $i^2 = -1$. We can then rewrite $\sqrt{\Delta}$:

$$\sqrt{\Delta} = \sqrt{ - \lVert \Delta \rVert } = \sqrt{-1} \times \sqrt{\lVert \Delta \rVert} = i \times \sqrt{\lVert \Delta \rVert}$$

Substituting the term on the right into the formula for the root ($\alpha_1$ from section 1.2), we get:

$$\sqrt[3]{\frac{-q + i\sqrt{\lVert \Delta \rVert}}{2}} + \sqrt[3]{\frac{-q - i\sqrt{\Delta}}{2}}$$

2.4Polar representations of complex numbers

Now we must take a giant leap forward into the world of complex numbers (which we've theoretically only just encountered with our definition of $i$). We can rewrite the above formula by converting each term into its polar1 representation (it's pretty trivial to deduce this from $i^2 = -1$, really). In any case, we get this (derivations left as an exercise for the reader):

$$\frac{-q + i \sqrt{\lVert \Delta \rVert}}{2} = (r + is)^3 \qquad \frac{-q - i \sqrt{\lVert \Delta \rVert}}{2} = (r - is)^3$$

So we can rewrite the formula for the root as

$$\alpha_1 = (r + is) + (r- is) = 2r \quad r, s \in \mathbb{R}$$

This is where my notes get a bit confused. First, we note that a non-zero cubic number has 3 cube roots. This has something to do with the unit circle, and from that we get the other cube roots, $\alpha_2$ and $\alpha_3$:

$$\alpha_1 = (r + is)(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = (\frac{1}{2} r - \frac{\sqrt{3}}{2}) + (i (\frac{1}{2} s + \frac{\sqrt{3}}{2} = r - \sqrt{3} s$$

$$\alpha_2 = (r + is)(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = (\frac{1}{2} r + \frac{\sqrt{3}}{2}) + (i (\frac{1}{2} s - \frac{\sqrt{3}}{2} = r + \sqrt{3} s$$

So the cube roots are given by $a_1 = 2r$, $a_2 = r - \sqrt{3}s$ and $a_3 = r + \sqrt{3} s$. Note that although there are 6 other (complex) roots, we only consider the real roots, as the others are not what we want.

2.5Properties of complex numbers

Let's put that whole last section aside and focus instead on the basics of complex numbers. We can redefine them as 2-tuples $(a, b)$, where $a, b \in \mathbb{R}$. A complex number can then be expressed in the form $a + bi$, and we can view it as a point in $\mathbb{R^2}$.

  • Addition works like this: $(a, b) + (c, d) = (a + c, b + d)$
  • Multiplication works like this: $(a, b) \times (c, d) = (ac-bd, ad-bc)$ (the derivation for this is left as an exercise for the reader)

We can prove that these operations satisfy the familiary properties of associativity and commutivity. Additionally, $(0, 0)$ is the identify for addition, and $(1, 0)$ is the identity for multiplication; every element has an additive inverse, given by $(-a, -b)$; and every element has a multiplicative inverse, given by $\displaystyle \left (\frac{a}{a^2 + b^2}, \frac{-b}{a^2+b^2} \right)$ (derivation omitted for brevity). We also have the distributive laws, etc. That should be it for that.

  1. I'm trying really hard to not make a terrible pun involving (manic-)depressed cubic equations and polar representations