How does this equation have n solutions?
Those of you who know a little about math might know that n-th degree polynomials are expected to have n roots. That is, equations of the form:
have n solutions. Let’s look at a simple example. Take the third degree (cubic) polynomial:
We can factorise this into:
And so at least one of these three factors must be zero, hence x = -2, x = -1 and x = 5 are roots of this cubic. Another way we are taught to find roots of these kinds of polynomials in high school is to graph them and look for the points where the graph hits the x-axis (y = 0) — for our example, we can see three such points as expected.
However, when we look at a very simple n-th degree polynomial like this one, it’s hard to see how it can have n roots:
Here’s a sketch of these polynomials for 1 ≤ n ≤ 6:
We can see that our sketches show only one root for odd degree (it’s a little hard to see precisely for higher order polynomials in these graphs, but the root is x = 1), and only two roots for even degree (x = ±1). For n = 1 and n = 2 that’s fine and dandy, but for n ≥ 3 where are the other roots?
Complex roots and Euler’s formula
The answer to this conundrum can be discovered by factorizing the first polynomial for which we see this problem — the cubic:
The first factor reveals the root that we can see in our graph (x = 1), but the remainder quadratic polynomial does not have real roots. The roots of this quadratic are the complex numbers (-1 ± √3i)/2 where i = √-1. This is easily determined using the standard formula for the roots of a quadratic, and you can verify it easily by multiplying the two complex factors together. As these roots are not real numbers, they do not show up in our standard real x-y graph.