Proving De Moivre's TheoremDate: 12/03/97 at 08:42:19 From: Chris Pountney Subject: De Moivre's theorem Dear Dr. Math, Please can you show me how to prove De Moivres theorem : (cos(x)+isin(x))^n = cos(nx) + isin(nx) Thank you, Chris Date: 12/08/97 at 10:34:38 From: Doctor Bruce Subject: Re: De Moivres theorem Hello Chris, I'm assuming that your 'n' is meant to represent an integer. DeMoivre's Theorem is true even if n is a complex number (has a real part and possibly an imaginary part), but when n is an integer we can prove the formula easily by using some basic trigonometry. We will prove DeMoivre's Theorem for any positive integer n by induction. First, the theorem is clearly true for n = 1, since then both sides of the equation are the same quantity. So now we will assume that the theorem has been shown true for n = k. We want to show that it is true for n = k + 1. We calculate: [cos(x) + i sin(x)]^(k+1) = [cos(x) + i sin(x)]^k * [cos(x) + i sin(x)] (split up factors) = [cos(kx) + i sin(kx)] * [cos(x) + i sin(x)] (induction hypothesis) = cos(kx)cos(x) - sin(kx)sin(x) + i[sin(kx)cos(x) + cos(kx)sin(x)] (multiply out) = cos(kx + x) + i sin(kx + x) (use trig formulas) = cos((k+1)x) + i sin((k+1)x). (add the angles) This proves the inductive step, so DeMoivre's Theorem has been shown to be true for all positive integers n. To show the theorem is also true for negative integers, we just compute (still assuming that n itself is positive) [cos(x) + i sin(x)]^(-n) = [{cos(x) + i sin(x)}^(-1)]^n = [cos(x) - i sin(x)]^n = [cos(-x) + i sin(-x)]^n = cos(-nx) + i sin(-nx). So DeMoivre's Theorem is true for all integers, positive and negative. I don't know if you know this already, but both the sine and cosine functions can be expressed in terms of the exponential function. Namely, cos(x) = [e^(ix) + e^(-ix)]/2 sin(x) = [e^(ix) - e^(-ix)]/2i, from which we get e^(ix) = cos(x) + i sin(x). This is true for any complex number x. This is the usual way we state DeMoivre's formula. We see that [cos(x) + i sin(x)]^n = [e^(ix)]^n and cos(nx) + i sin(nx) = e^(inx), and the two righthand sides above are clearly equal. So this an easier way of proving the theorem that you stated, although first you have to know about the exponential function. A fundamental and truly amazing fact that comes out of this is e^(i*pi) = -1, which you get by substituting x = pi in DeMoivre's formula. -Doctor Bruce, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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