I remember being taught in high school how to find the solutions to the equation: y(x) = ax2 + bx + c = 0. The parabola in the figure clearly crosses the horizontal x-axis twice and gives us two solutions for y=0 that are in fact:x = ( -b ± √(b2 - 4ac) )/2a
But if you lift the parabola up so that its vertex is above the x-axis, there is no way this curve is ever going to cut through y=0. Or is there?
All those years ago, I was shown that the problem had to do with the square root in the expression above. A parabola that is above the x-axis corresponds to a negative value for b2 - 4ac. Since we would have to take the square root of a negative number to get a solution, why not define the so-called imaginary number i=√-1 and just carry on! This is one way to motivate the invention of complex numbers, and I thought that was so cool or incomprehensible or both that I never went back to the geometrical picture of what's going on.
As many readers will know, the standard approach is to define a complex plane of numbers by adding an imaginary number axis perpendicular to the more familiar (real) number line. I have talked about the beauty of this before, here and here. A complex number is made up of a real part and an imaginary part, which is why you need a plane to describe these numbers geometrically. Now let's imagine what y=x2 looks like when x is a complex number that can lie anywhere on the complex plane.
Our normal picture for y=x2 is a parabola with a vertex at x=0 with arms rising majestically above the real x-axis. What about y as a function of a purely imaginary x=ai (where a is just a real distance along the imaginary axis)? Well, y = x2 = a2 i2= -a2 (since i2 = -1) so that y now takes negative values. In other words, we take a copy of our usual parabola, rotate it by 90 deg so that it is above the imaginary axis, and flip it so that it points down. This is nicely shown for a general parabola here.
For any x not on the real or imaginary axes, y is complex, which turns out not to be helpful for finding solutions to y(x)=0. But with two parabolae pointing in opposite directions, we've really got the y-axis covered. Now we can place the vertex above or below y=0 and we will have two arms ascending towards ever more positive y values, and two legs descending towards ever more negative y values. If our parabola has arms and its vertex is above y=0, all we need do is create the legs and follow them down to dig up the two complex roots to its equation.
When you think about it, the initial problem only occurs for polynomials of even degree, i.e. y=ax2+... or y=ax4+... because these are the ones that have arms or legs that end up pointing in the same direction. Obviously y=ax+b is going to cross the x-axis somewhere and so will y=ax3+... etc. The invention of complex numbers results in the even-degree polynomials acquiring both arms and legs so that they too are bound to cross y=0. Because we only have a single problem here - ensuring that all polynomials cover all negative and positive values of y - we don't need to invent anything more exotic than complex numbers. I love this geometrical picture, because it gives us an extremely informal demonstration of this so-called closure of the complex numbers, which is known in fact as the fundamental theorem of algebra.
