Question

Use mathematical induction to prove De Moivre's theorem
[ R (cos t + i sin t) ] n = R n(cos nt + i sin nt)

for n a positive integer.

Answer 1

STEP 1: For n = 1  

[ R (cos t + i sin t) ] 1 = R 1(cos 1*t + i sin 1*t)  

It can easily be seen that the two sides are equal.  

STEP 2: We now assume that the theorem is true for n = k, hence  

[ R (cos t + i sin t) ] k = R k(cos kt + i sin kt)  

Multiply both sides of the above equation by R (cos t + i sin t)  

[ R (cos t + i sin t) ] k R (cos t + i sin t) = R k(cos kt + i sin kt) R (cos t + i sin t)  

Rewrite the above as follows  

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ (cos kt cos t - sin kt sin t) + i (sin kt cos t + cos kt sin t) ]  

Trigonometric identities can be used to write the trigonometric expressions (cos kt cos t - sin kt sin t) and (sin kt cos t + cos kt sin t) as follows  

(cos kt cos t - sin kt sin t) = cos(kt + t) = cos(k + 1)t  

(sin kt cos t + cos kt sin t) = sin(kt + t) = sin(k + 1)t  

Substitute the above into the last equation to obtain  

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ cos (k + 1)t + sin(k + 1)t ]  

It has been established that the theorem is true for n = 1 and that if it assumed true for n = k it is true for n = k + 1.