## Demoivre’s Theorem:

The theorem is helpful in finding out the n-th root of a complex number (where n is a rational number). The statement consists of two steps:

(I) If n is an integer or zero, then (cos θ + *i* sin θ)^{n} = cos nθ + *i* sin nθ.

(II) If n is a non-integral rational number, then cos nθ + *i* sin nθ is one of the values of (cos θ + *i* sin θ)^{n}.

**Proof (I):**

Case 1:- If n =0, then(cos θ + i sin θ)^{n} = (cos θ + i sin θ)^{0 }= 1 = 1 + 0 . i = cos 0 + i sin 0 = cos (0, θ) + i sin (0, θ) |

Case 2:- If n is a positive integer, then it is wise to proceed by the method of mathematical induction. The theorem is verified to be true for n = 1 as (cos θ + i sin θ)^{1} = cos θ + i sin θ = cos 1.θ + i sin 1.θ.Let us assume that the theorem is true for a positive integer n = m ≥ 1. Then, (cos θ + i sin θ)^{m} = cos mθ + i sin mθ.Now, (cos θ + i sin θ)^{m + 1} = (cos θ + i sin θ)^{m} (cos θ + i sin θ)= (cos mθ + i sin mθ) (cos θ + i sin θ)= (cos mθ cos θ – sin mθ sin θ) + i (sin mθ cos θ + cos mθ sin θ)= cos (mθ + θ) + i sin (mθ + θ)= cos (m + 1)θ + i sin (m + 1)θThus the theorem is true for n = m + 1. Hence, by induction, the theorem is true for all positive integers. |

Case 3:- If n is a negative integer, let n = -m, where m ∈ N. Then,(cos θ + i sin θ)^{-m}= 1/(cos θ + i sin θ)^{m}= 1/(cos mθ + i sin mθ)= (cos mθ – i sin mθ)/[(cos mθ + i sin mθ) (cos mθ – i sin mθ)]= [cos (-mθ) + i sin (-mθ)]/(cos^{2} mθ + sin^{2} mθ)= cos (-mθ) + i sin (-mθ)= cos nθ + i sin nθ (∵ n = -m)Thus the theorem is true for any integer or zero. |

**Proof (II):** When n is a non-integral rational number, then let us take n = p/q, where p and q ∈ I and q > 0.

Then, (cos pθ/q + i sin pθ/q)^{q} = cos (pθ/q)^{q} + i sin (pθ/q)^{q} = cos pθ + i sin pθ∴ cos pθ/q + i sin pθ/q is one of the values of (cos pθ + i sin pθ)^{1/q}Thus, cos pθ/q + i sin pθ/q is one of the values of {(cos θ + i sinθ) p}^{1/q}⇒ cos pθ/q + i sin pθ/q is one of the values of (cos θ + i sinθ)^{p/q}Thus, (cos nθ + i sin nθ) is one of the values of (cos θ + i sinθ)^{n}, where n is a non-integral rational index. |

**Remarks:**

1:- (cos θ + i sin θ)^{-n} = cos (-nθ) + i sin (-nθ) = cos nθ – i sin nθ.2:- (sin θ + i cos θ)^{n} = { i (cos θ – i sin θ)}^{n} = i^{n} {cos (-θ) + i sin (-θ)}^{n}= i^{n} {cos (-nθ) + i sin (-nθ)}= i^{n} (cos nθ – i sin nθ)3:- 1/(cos θ + i sin θ) = (cos θ + i sin (θ)^{-1}= cos (-θ) + i sin (-θ)= cos θ – i sin θ4:- (cos θ + i sin θ) (cos Φ + i sin Φ) = cos (θ + Φ) + i sin (θ + Φ) |