De Moivre's theorem states that(cosø + isinø)n= cos(nø) + isin(nø).

Assumingn = 1 (cosø + isinø)1= cos(1ø) + isin(1ø)which is true so correct forn = 1

Assumen = k is true so(cosø + isinø)k= cos(kø) + isin(kø).

Lettingn = k + 1we know that(cosø + isinø)k+1= cos((k + 1)ø) + isin((k + 1)ø).

But trying to derive the answer fromn = kwe get: (cosø + isinø)k+1= (cosø + isinø)kx (cosø + isinø) //Using law of indecies = (cos(kø) + isin(kø)) x (cosø + isinø) //We've assumed this to be true from n = k = cos(kø)cos(ø) + icos(kø)sin(ø) + isin(kø)cos(ø) - sin(kø)sin(ø)//ExpansionNow using the trigenomitery rules ofsin(a + b) = sin(a)cos(b) + sin(b)cos(a)andcos(a + b) = cos(a)cos(b) - sin(a)sin(b)and collectingterms to match these formulas we get =cos(kø + ø) + isin(kø + ø) = cos((k + 1)ø) + isin((k + 1)ø) //Taking out ø as a common factor inside the trig functions

Which equals our expression gained from De Moivre's theorem for n = k + 1,and as theorem is true for n = 1 and n = k + 1,statement is true for all values of n >= 1.