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Proof of DeMoivre's Theorem

Date: 05/01/97 at 20:17:29
From: LIYA
Subject: Proof of Demoivre's Theorem

Dr. Math,
Could you please help me prove DeMoivre's theorem by mathematical 
induction?  I figured out the first part of the proof, but I can't 
prove that assuming that if it's true for n = k, then it's true
for n = k+1.

Date: 05/02/97 at 13:17:50
From: Doctor Wilkinson
Subject: Re: Proof of deMoivre's Theorem


This depends on the addition formulas for the sine and cosine:

 sin(x + y) = sin(x) cos(y) + sin(y) cos(x)

 cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

DeMoivre's theorem states that:

 (cos(x) + i sin(x))^n = cos(nx) + i sin(nx)

In your case, you are assuming this holds for n = k and are trying to 
prove that it holds for n = k + 1, so you are looking at:

 (cos(x) + i sin(x)) ^(k + 1)

This is just:

 (cos(x) + i sin(x))^k * (cos(x) + i sin(x))

Using the induction hypothesis, this is:

 (cos(kx) + i sin(kx)) ( cos(x) + i sin(x))

Multiplying out and using the fact that i^2 = -1, we get:

 cos(kx) cos(x) - sin(kx) sin(x) + i (sin(kx) cos(x) + cos(kx) sin(x))

This is just cos((k+1) x) + i sin((k+1) x) by the addition formulas
mentioned above.

I recommend that you study this proof carefully, as it is very typical 
of induction proofs.

-Doctor Wilkinson,  The Math Forum
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Associated Topics:
High School Imaginary/Complex Numbers
High School Trigonometry

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