De Moivre’s Theorem Proof
DeMoivre’s Theorem can be proved with the help of mathematical induction as follows:
P(n) = (cos x + i.sinx)n = cos(nx) + i.sin(nx) ⇢ (1)
Step 1: For n = 1,
(cos x + i sin x)1 = cos(1x) + i sin(1x) = cos(x) + i sin(x),
Which is true. thus, P(n) is true for n = 1.
Step 2: Assume P(k) is true
(cos x + i.sin x)k = cos(kx) + i.sin(kx) ⇢ (2)
Step 3: Now we have to prove P(k+1) is true.
(cos x + i.sin x)k+1 = (cos x + i.sin x)k(cos x + i sin x)
= [cos (kx) + i.sin (kx)].[cos x + i.sin x] ⇢ [Using (1)]
= cos (kx).cos x − sin (kx).sin x + i [sin (kx).cos x + cos (kx).sin x)
= cos {(k+1)x} + i.sin {(k+1)x}
= cos {(k+1)x} + i.sin {(k+1)x}
Thus, P(k+1) is also true and by the principle of mathematical induction P(n) is true.
DeMoivre’s Theorem
De Moivre’s theorem is one of the fundamental theorem of complex numbers which is used to solve various problems of complex numbers. This theorem is also widely used for solving trigonometric functions of multiple angles. DeMoivre’s Theorem is also called “De Moivre’s Identity” and “De Moivre’s Formula”. This theorem gets its name from the name of its founder the famous mathematician De Moivre.
In this article, we will learn about De Moivre’s Theorem, its proof, some examples based on the theorem, and others in detail.
Table of Content
- De Moivre’s Theorem Statement
- De Moivre’s Formula
- De Moivre’s Theorem Proof
- Uses of De Moivre’s Theorem
- Finding the Roots of Complex Numbers
- Power of Complex Numbers
- Solved Examples on De Moivre’s Theorem
- FAQs