Complex Numbers 3

So we saw that with the rotation-stretching interpretation multiplying and taking powers of complex numbers makes more sense.

Obs. While this interpretation makes multiplication easier it does not help when making addition. Happily addition is easy in the algebraic form "a + bi".

Now we are going to see three bonus of looking at complex numbers in this way.

First bonus: Our theory is somewhat arbitrary why we defined i to rotate 90 ^o counterclockwise ? It could be clockwise couldn't it ? The answer is yes! Lets say you defined a different clockwise imaginary number j, because you didn't like mine, then you would have a complex number theory just like mine with j's in the place of i's! Your drawings would be the reflection of my drawings. But the arithmetic would be the same.

But your j does the same job as my -i! So in a way the -i does the same job as i, but clockwise. Changing i's to -i's changes the orientation of my rotations. So if I take a number like 3+4i, then 3+4(-i) = 3 - 4i has opposite rotation but the same stretching factor. The operation changing i's to -i's is called conjugation and this reasoning explain most of it's properties directly.

Second bonus: Let's say our stretching factor is 1 (no stretching at all) in this case our complex number only rotate vectors by angles a and b. So they are:

cos(a) + i sin(b)
cos(b) + i sin(b)

An their product by our interpretation:

cos(a+b) + i sin(a+b)

But we can also multiply
cos(a)+i sin(b))*(cos(b)+i sin(b)) by distribution and that is:

(cos(a)cos(b) - sin(a)sin(b)) + i(cos(a)sin(b) + sin(b)cos(a))

And we have the formulas for cos(a+b) and sin(a+b)!,

Although the geometrical demonstration for those formulas is really nice too (there is at least one nice geom. dem. that I know of).

As an exercise you can find what happens when b = -a.

Third bonus: if you remember the First bonus a+bi and a-bi have same stretching but opposite rotation angles u and -u say. Remembering a+bi forms a right triangle with legs a and b lets call c its hypotenuse, that is also its stretching factor.
Then the product of a+bi and a-bi has u + (-u) = 0 rotation and stretching c*c = c² , so it is really c² , but algebraically

(a+bi)*(a-bi) = a² - abi + abi - (bi)² = a² -(-b² )=a² +b² .

So a² + b² = c²

That is Pythagoras theorem.