Suppose you are a city dweller on your way to catch the bus. As you approach the bus stop you see your bus just starting to pull away, so you run after it. You are running at six yards per second. The bus has a 30 yard head start, and is not moving. But it is accelerating at one yard per second per second. Will you catch the bus?
The answer is no. After six seconds you will have gone 36 yards, but the bus will have gone 18 yards (if you remember your high school physics, you know that distance = ½ acceleration × time2). Add in the 30 yard head start and the bus is still ahead of you. And after six seconds the bus has accelerated to a faster speed than you. So you will never catch the bus.
More formally, let t be the time, starting at 0.
Your position = 6t
Bus’s position = ½t2 + 30
If you catch the bus then your positions are the same at the same time. So set the equations equal to each other and solve for t.
6t = ½ t2 + 30
0 = ½ t2 – 6t + 30
Using the quadratic formula, we get:
t = 6 ± √(36 – 4·½·30)
t = 6 ± √(-24)
As you may remember from high school math and physics, you discard all answers that involve the square roots of negative numbers. So there is no solution. We can see that graphically. The line is the graph of your position, and the parabola is the graph of the bus. The graphs never meet because there is no solution.
But what if there is an imaginary solution? As you probably know, i is √-1. We were stumped because we had √-24, which is then √24i2. Let’s see if we can catch that bus after all.
t = 6 ± √-24
t = 6 ± √24i2
t = 6 ± 2i√6
It doesn’t seem like we’ve made much progress – what does it mean to say that we caught the bus at time 6 ± 2i√6? Does that mean we made it? Perhaps in some alternate universe? You can see why the great mathematician and philosopher Rene Descartes rejected imaginary numbers – they were impossible solutions to real problems.
But imaginary numbers refused to go away. In fact, there were clues to their power and utility going back to the ancient Greeks.
