#complex variables

So what you need to know is that this guy, De Moivre, formulated this awesome theorem, which can easily be proven by induction so i’m not going to do it here. The Theorem states:

(cosA + isinA)n  = cos(nA) + isin(nA)

Or in words, the sum cosA (where A is an angle) and i times sinA raised to some power n (where n is an integer) is equal to that sum except with the angle changed to n times A.

That’s pretty cool on its own, I guess, but it still has those awful, awful complex numbers that we hate. Don’t worry, we’ll get rid of them. The most important thing you need to know about that complex number is that, even though we screw around and use it in multiplication and stuff, it actually serves to designate a term in the complex number, just like a vector.

Essentially, just like you sometimes write a vector ax + by or ai + bj, you can write a complex number as a + ib. But also we pretend it’s regular sum when we do this. Don’t worry about it.

Now back to De Moivre, and more importantly, trigonometry. We can use this nifty theorem to develop trigonometric identities that will blow your mind. THey make no sense! 

But secretly, they do. Let’s try to figure out an identity for cos3A and sin3A. Given what we just saw in De Moivre’s THeorem, we’re inclined to use it, choosing n to be 3.

But before we do it, let’s right the regular expansion of that sum:

(cosA + isinA)3 = cos3A -3cosAsin2A + 3icos2AsinA - isin3A

And De Moivre’s:

(cosA + isinA)3 = cos3A + isin3A

Now remember how I said we can treat complex numbers as a vector? Let’s do that for both of the equations above, grouping them up using the i terms. This gives us:

cos3A =  cos3A -3cosAsin2A

sin3A =  3icos2AsinA - isin3A

Now at this point I’m sure your brain is sending out trig identities left and right thanks to the merciless conditioning it received in high school. Do not resist! Let them flow! We’ll only solve for cos3A here, but the process is almost identical for sin3A. Here we go!

cos3A =  cos3A -3cosAsin2A

cos3A =  cosA(cos2A -3sin2A)

cos3A =  cosA(cos2A -2sin2A) 

cos3A =  cosA(2cos2A - 1)    

cos3A =cosA(4cos2A - 3)

cos3A = 4cos3A - 3cosA

Whoa! What a cool identity! Dayum! And, even though we started with i in both our equations, somehow, it disappeared!