what is de Moivres theorem?

Question

What is de Moivres theorem & how can this be utilised to prove that all planes comes under zeroth root of 1?

Answer 1

de Moivre's Theorem states that (cosq + isinq)n = cosnq + isinnq for any integer n.

Proof: The proof of de Moivre's Theorem concludes from proving seperately for negative and positive values of n and when n = 0.

i) Proof for positive integer values of n. This proof is done by induction.

(cosq + isinq)k = coskq + isinkq
then (cosq + isinq)k + 1 = (coskq + isinkq)(cosq + isinq)

= cos(kq + q) + isin(kq + q)

= cos(k + 1)q + isin(k + 1)q
So this shows de Moivre's Theorem to be true for all positive values of n.

ii) Proof for n = 0. This proof only includes checking that the statement holds up for a value of n. By definition we know z0 = 1 (where z is a non-zero complex number).

so (cosq + isinq)0 = 1 = cosq + isinq.
iii) Lastly, we need to check for negative values of n.

(cosq + isinq)(cos(-q) + isin(-q)) = cos(q - q) + isin(q - q) = 1
so it must follow that

(cosq + isinq)-1 = cos(-q) + isin(-q)
If n is a negative integer, let n = -m. Then

(cosq + isinq)n = (cosq + isinq)-m
= [(cosq + isinq)m]-1

= (cosmq + isinmq)-1

= cos(-mq) + isin(-mq)

= cosnq + isinnq
De Moivre's Theorem also holds for any negative value of n, and when n = 0. So de Moivre's Theorem is true for all integer values n.

De Moivre's Theorem is useful because it can be used to solve roots of unity

Answer 2

Multiplying two complex numbers in polar form is easily undertaken by using the rules set out in the last section.
This leads to a very simple formula for calculating powers of complex numbers - known as De Moivre's theorem.

Consider the product of z with itself (z2) if z =

The rules of multiplying the moduli and adding the arguments gives
z2 = =

Now consider z3
Take z2 = and multiply this by z =
This gives z3 = ()() =
A pattern has emerged.

This result can be extended to the nth power and is known as De Moivre's Theorem.



De Moivre's theorem

If z = , then



Now look at the proof of this theorem which is included for interest at proof (11).

This is a very useful result as it makes it simple to find once z is expressed in polar form.


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Example 1

Calculate
Answer:
Using De Moivre's theorem

[ To reach this result
multiply the moduli to give 25 and add the arguments to give ]

Answer 3

The theorem is stated in two steps :
(i) (cos x+i sinx)^n= cos nx+i sin nx , if n is an integer (positive , zero or negative )
(ii)(cos nx+i sin nx) is one of the values of ( cos x+i sin x)^n if n is a non integral rational number .


1) (cos x+ i sin x)^(-n) = cos (-nx)+ i sin (-nx) = cos nx- i sin nx
2)(cos x- i sin x)^n = cos nx - i sin nx
(3) (sin x + i cos x)^n= (i^(n-1))(sin nx+ i cos nx)
we may not necessarily have (3) , it is only when n=4k+1 , k belongs to integers

Answer 4

if p is a real number,
D'Moivres theorem states that
[r(cosz+isinz]]^p=r^p[cospz+isinpz]