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)#

  

  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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