Eulers formula and De moivre's theorem

Euler's formula

  • For any real number x, eix = cos x + i sin x
  • Let z be a non zero complex number; we can write z in the polar form as,
    z = r(cos θ + i sin θ) = r e, where r is the modulus and θ is argument of z.
  • Important point
    Multiplying a complex number z with e gives
    z *e = re X e = rei(α + θ)
    The resulting complex number rei(α + θ) will have the same modulus r and argument (α + θ).

De Moivre's theorem

De Moivre's theorem states following cases
  • Case I It states that for any integer n,

    (cos θ + i sin θ)n = cos (nθ) + i sin (nθ)

  • Case II if n is of the form p/q where p, q are integers and q > 0
    (cos θ + i sin θ)n =cos (2k π+ θ)p/q + sin (2k π+ θ)p/q
    Where k=0,1,2,...q-1

Proof for De Moivre’s theorem(Case -1)

Prove using Euler's Formula

This can be easily proved using Euler's formula as shown below.

We know that, (cos θ + i sin θ) = e
(cos θ + i sin θ)n = [e]n=ei(nθ)
ei(nθ) = cos (nθ) + i sin (nθ)

Prove using mathematical Induction ( n being a positive number)
(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)

Step 1 for n=1
(cos θ + i sin θ)n=(cos θ + i sin θ)

Step 2 for n=m,let us assume the theorem is true for n=m
(cos θ + i sin θ)m = cos (mθ) + i sin (mθ)

step 3Now for n=m+1
(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) cosθ -sin(mθ)sin θ] +i[cos (mθ)sin θ + sin (mθ)cos θ]
=cos (m+1)θ + i sin (m+1)θ
So, theorem holds good for n=m+1
So, by principle of Mathematical Induction

(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)

For negative integers, we can write like n=-m where m is positive number
(cos θ + i sin θ)n =1 /(cos θ + i sin θ)m

Now m being positive, we already proved above
=1 /cos (mθ) + i sin (mθ)
Multiplying the Conjugate complex on Numerator and Denominator we get,
=cos (mθ) - i sin (mθ)
=cos (-mθ) +i sin (-mθ)
=cos (nθ) + i sin (nθ)

Application of De Moivre's theorem

  • De Moivre's theorem is used to calculate the root of Complex Numbers
  • For a complex number z=a + ib which can be expressed as polar form as = r(cos θ + sin θ)
    So z1/n = r1/n(cos θ + sin θ)1/n
    Now from De Moivre's theorem, we know that
    (cos θ + i sin θ)1/n =cos (2k π+ θ)/n + sin (2k π+ θ)/n
    Where k=0,1,2,...n-1
  • By putting value of k=0,1,2,...n-1, we can find the root values

Related Topics

link to this page by copying the following text

Class 11 Maths Class 11 Physics

Note to our visitors :-

Thanks for visiting our website. From feedback of our visitors we came to know that sometimes you are not able to see the answers given under "Answers" tab below questions. This might happen sometimes as we use javascript there. So you can view answers where they are available by reloding the page and letting it reload properly by waiting few more seconds before clicking the button.
We really do hope that this resolve the issue. If you still hare facing problems then feel free to contact us using feedback button or contact us directly by sending is an email at [email protected]
We are aware that our users want answers to all the questions in the website. Since ours is more or less a one man army we are working towards providing answers to questions available at our website.