De Moivre’s and the nth Root Theorems

Key Questions

  • 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.

  • 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))#

  • 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#

Questions