Why is i^2 = -1?

Answer by Adam Catto:

I like to think geometrically:

Suppose we have two vectors in the complex plane, A and B.   The horizontal axis is the real line and the vertical axis is the imaginary line. The multiplication of two vectors, AB, in the complex plane has two major properties: length and orientation (angle).  The length is just the dot product between the vectors (aka the product of their lengths), and the angle between the new vector and the positive horizontal is the sum of the angles of the vectors. Note that i is a unit vector on the imaginary line.

let A = i, B = i. Thus, AB = [math]i^2[/math]. Since i is of unit length, the norm of it is just 1. [math]1\times 1 = 1[/math], so [math]i^2[/math] is of unit length.

Now, what's the angle between [math]i^2[/math] and the positive real axis? Recall that [math]\theta(AB) = \theta(A) + \theta(B)[/math], where [math]\theta[/math] is an angle.  Since i is on the imaginary line, and the imaginary line is orthogonal to the real axis, the angle between i and the positive real axis is [math]\frac{\pi}{2}[/math].

Since A = B = i, [math]\theta(AB) = \theta(A) + \theta(B) = \theta(i) + \theta(i) = 2\theta(i) = 2\frac{\pi}{2} = \pi[/math], so [math]i^2[/math] is oriented along the negative real axis, or is of negative direction. Since it is of unit length, and is oriented towards the negative real direction, [math]i^2 = -1[/math]. I'll post a picture with a plot that makes this visualization easier, when I get the chance.

EDIT: here's the plot:

Why is i^2 = -1?

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