Eulers formula and De moivre's theorem

Euler's formula

• For any real number x, e^ix = 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^iθ, where r is the modulus and θ is argument of z.
• Important point
  Multiplying a complex number z with e^iα gives
  z *e^iα = re^iθ X e^iα = re^{i(α + θ)}
  The resulting complex number re^{i(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^iθ
(cos θ + i sin θ)ⁿ = [e^iθ]ⁿ=e^i(nθ)
Therefore,
e^i(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 z^1/n = r^1/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

Also Read

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• Notes
  • What is complex numbers
  • Algebra Of complex Numbers
  • Conjugate of Complex Numbers
  • Modulus of complex numbers
  • Argand Plane
  • Polar Representation of the complex number
  • Rotation of Complex Number
  • Identities for Complex Numbers
  • Eulers formula and De moivre's theorem
  • Cube Root of unity
  • Complex roots of Quadratic equations
• NCERT Solutions & Worksheets
  • NCERT Solution Complex Numbers Exercise 5.1
  • Complex Numbers Problems
  • Complex Numbers Worksheet

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