Proof by Induction (De Moivre's Theorem)

Step 1

Show true for n = 1

Step 2

Assume true for n = k

Step 3

Prove true for n = k + 1

How to show true for n = 1

(cos Θ + isin Θ)ⁿ = (cos(n)Θ + isin(n)Θ)

(cos Θ + isinΘ)¹ = (cos(1)Θ + isin(1)Θ) = (cosΘ + isinΘ)

∴ True for n = 1

How to assume true for n = k

(cos Θ + isin Θ)ⁿ = (cos(n)Θ + isin(n)Θ)

(cos Θ + isin Θ)^k= (coskΘ + isinkΘ)

∴ Assumed true for n = k

How to prove true for n = k +1 (1st part)

(cos Θ + isin Θ)^k+1 = (cos(k+1)Θ + isin(k+1)Θ) (RHS)

(LT pg 21- Law 1 of indices: a^p+q = a^pa^q , where p = k and q = 1 )

(cos Θ + isin Θ)^k (cos Θ + isin Θ)¹ = RHS

(Get rid of powers based on DM theorem answers in steps 1 & 2)

(coskΘ + isinkΘ)(cosΘ + isinΘ) = RHS

(Multiply)

coskΘ(cosΘ + isinΘ) + isinkΘ(cosΘ + isinΘ) = RHS

coskΘcosΘ + coskΘisinΘ + isinΘcosΘ + i² sinΘsinΘ =RHS

How to prove true for n = k + 1 (2nd part)

(Since i² = -1)

coskΘcosΘ + coskΘisinΘ + isinΘcosΘ -1sinkΘsinΘ = RHS

(Rearrange into form a + bi)

coskΘcosΘ -sinkΘsinΘ + coskΘisinΘ + isinΘcosΘ = RHS

(LT pg 14 - compound angle formulae : cos(A+B) = cosAcosB - sinAsinB

and sin(A+B) = sinAcosB + cosAsinB, where A = kΘ and B = Θ)

cos(kΘ + Θ) + sin(kΘ + Θ)i = RHS

(Factorise Θ out of equation)

cos(k+1)Θ + sin (k +1)Θi = RHS

(Move i back over to sin)

cos(k+1)Θ + isin(k+1)Θ = cos(k+1)Θ + isin(k+1)Θ

∴ True for n =1, and n = k and proved true for n = k + 1