Eli5: in complex numbers, what is the meaning of adding a “+b*i” part? It looks to me similar to ‘normal’ coordinates (+b*y), but with another name. I understand that the meaning of i as the square root of a negative makes it different but could never understand how so.

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We were told about complex numbers, and did the algebra by solving problems with the general form of “a*n + b*i”. It does not really behave differently than standard algebra, it seems. What am I missing in the intention or meaning of the “i” part?

In: Mathematics

6 Answers

Anonymous 0 Comments

Imagine you start with just i.

Now, multiply it by i again, and you have -1, by definition i^2 = -1

Then, multiply by i again, and you have -1 * i = -i

Do it again, and you have -i * i = – ( i * i ) = – (-1) = 1

And finally, if you multiply by i one more time 1 * i = i

In this case, every 4th power of i takes you back to where you started, but more generally, multiplying by i tends to *rotate* your point through the complex plane. This is very different to how simple pairs of real numbers behave. With real numbers you can shift by adding, or stretch by multiplying, but they never rotate unless you do different things to different numbers.

This rotation property gives complex numbers quite a different structure than a simple pair of coordinates.

For example, you can look at the equation x^2 = 1, what are its solutions? 1 and -1 of course. But what about a different power, lets say x^4 = 1? In this case, you can start with our previous answers 1 and -1, and then look at what numbers can be squared to make them. For 1 its just 1 and -1, so no new solutions. But i * i = -1 but so does -i * -i = -1. So the equation x^4 = 1 has 4 solutions: 1, -1, i, -i, the exact values we were just discussing!

In general, for any x^N = 1, the solutions are N points evenly-spaced around a circle with radius 1 in the complex plane, and they always include 1. So, if you want to know the solutions of x^3 = 1, you start at 1, rotate that point 1/3 of a turn to get your second solution, rotate by another 1/3 of a turn to get the third solution, and a final rotation would bring you back to 1.

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