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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
    then
    (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)
Therefore,
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



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