De Moivre's Theorem, Complex Number Formula, Example

De Moivre's Theorem is an easy formula which is used for calculating the powers of complex numbers. This theorem can be derived from Euler's equation since it connects trigonometry to complex numbers.

De Moivre's Theorem Formula, Example and Proof

Statement
    For an integer/fraction like n, when computed the value obtained will be either the complex number 'cos nθ + i sin nθ' or one of the values of (cos θ + i sin θ)n.

Proof
From the statement let us take, (cos θ + isin θ)n = cos (nθ) + isin (nθ)
Case 1 : If n is a positive number
L.H.S = (cos θ + isin θ)n
= (cos θ + isin θ) × (cos θ + isin θ) × (cos θ + isin θ) ... (n-times)
= cos( θ + θ + ... n-times) + isin( θ + θ + ... n-times)
= cos (nθ) + isin (nθ).
= R.H.S

Case 2 : If n is a negative number
Consider, n = -m
L.H.S = (cos θ + isin θ)-m
= 1 / (cos θ + isin θ)m
= 1 / (cos mθ + isin mθ) (by case 1)
= ( 1 / cos mθ + isin mθ ) × (cos mθ - isin mθ / cos mθ - isin mθ)
= (cos mθ - isin mθ) / (cos² mθ + sin² mθ)
= (cos mθ - isin mθ) / 1 (since, cos² mθ + sin² mθ = 1)
= cos (-mθ) + isin (-mθ)
= cos (nθ) + isin (nθ). (since, -m = n)
= R.H.S

Case 3 : If n is a fractional number
Consider, n = p/q
L.H.S = (cos θ + isin θ)p/q
= (cos pθ + isin pθ)1/q
= (cos (pθ)/q + isin (pθ)/q)
= (cos (p/q)θ + isin (p/q)θ)
= cos (nθ) + isin (nθ). (since, p/q = n)
= R.H.S

Hence the De Moivre's Theorem is proved for all real numbers with the above given proof and example.

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