De Moivre’s and the nth Root Theorems

Key Questions

• How do you use De Moivre’s Theorem to find the powers of complex numbers in polar form?

  If the complex number `z` is

  `z=r(cos theta + i sin theta)`,

  then `z^n` can be written as

  `z^n=[r(cos theta+i sin theta)]^n=r^n[cos theta+i sin theta]^n`

  by De Mivre's Theorem,

  `=r^n[cos(n theta)+i sin(n theta)]`

  ---

  I hope that this was helpful.

  Wataru · 2 · Dec 3 2014
• How do you find the `n^{th}` roots of complex numbers in polar form?

  Answer:

  `z^(1/n) = r^(1/n) ( cos (theta/n) + i sin(theta/n))`

  Explanation:

  Polar form of complex number is `z = r( cos theta + i sin theta)`

  ![https://www.google.com/search?q=demorvies+theorem&client=safari&hl=en-us&prmd=ivn&source=lnms&tbm=isch&sa=X&ved=0ahUKEwiYmuTisNLbAhWJwI8KHXycAGsQ_AUIESgB&biw=768&bih=922#imgrc=XMEZvta0Lgq8wM:]()

  By De Morvies theorem,

  `z^(1/n) = r^(1/n) ( cos (theta/n) + i sin(theta/n))`

  sankarankalyanam · 1 · Jun 14 2018
• What is the DeMoivre's theorem used for?

  Answer:

  More of the cases, to find expresions for `sinnx` or `cosnx` as function of `sinx` and `cosx` and their powers. See below

  Explanation:

  Moivre's theorem says that `(cosx+isinx)^n=cosnx+isinnx`

  An example ilustrates this. Imagine that we want to find an expresion for `cos^3x`. Then

  `(cosx+isinx)^3=cos3x+isin3x` by De Moivre's theorem

  By other hand applying binomial Newton's theorem, we have

  `(cosx+isinx)^3=cos^3x+3icos^2xsinx+3i^2cosxsin^2x+i^3sin^3x=cos^3x-3cosxsin^2x+(3cos^2xsinx-sin^3x)i`

  Then, equalizing both expresions as conclusion we have

  `cos3x=cos^3x-3cosxsin^2x`
  `sin3x=3cos^2xsinx-sin^3x`

  F. Javier B. · 1 · Jul 25 2018

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