De Moivres Theorem cos i sin n cosn

De Moivre’s Theorem (cos + i sin )n = cosn + i sinn To prove: (cos + i sin )n = cosn + i sinn n Z Proof: Assume the proof to be true for n=k. Hence (cos + i sin )k = cosk + i sink Now consider n=k+1 (cos + i sin )k+1 = (cos + i sin )k (cos + i sin )1 = (cosk + i sink )(cos + i sin )1 = cosk cos + i cos k sin + i sink cos + i 2 sink sin = cos k cos - sink sin + i(cosk sin + sin k cos ) =cos(k+1) + i sin(k+1) Hence if true for n=k, it is also true for n=k+1 But it is true for n=1 as (Cos +i sin )1=cos +i sin Hence by induction it is true for all values of n, where n is a positive integer. (cos + i sin )n = cosn + i sinn (c) Project Maths Development Team 2011 Next

To prove: (cos + i sin )n = cos n + i sin n n Z Proof continued: We have already established theorem is true for all positive integers. For n=0 (cos + i sin )0 = 1. (Anything to the power of 0 equals 1) But cos(0 ) + i sin(0 ) = cos 0 + i sin 0 = 1 Hence (cos + i sin )0 = cos 0 + i sin 0 , therefore theorem true for n=0 For n= -m (Cos + i sin ) n= (Cos + i sin ) -m Therefore: (cos +i sin )n = cos n + i sin n for all negative Integers. Because cos(-n )=cos n and sin (-n )=-sin n Therefore theorem is true for both positive and negative integers plus zero Therefore (cos + i sin )n = cos n + i sin n is true for all integers. (c) Project Maths Development Team 2011