De Moivre’s and the nth Root Theorems

Key Questions

  • If the complex number z is

    z=r(cosθ+isinθ),

    then zn can be written as

    zn=[r(cosθ+isinθ)]n=rn[cosθ+isinθ]n

    by De Mivre's Theorem,

    =rn[cos(nθ)+isin(nθ)]


    I hope that this was helpful.

  • Answer:

    z1n=r1n(cos(θn)+isin(θn))

    Explanation:

    Polar form of complex number is z=r(cosθ+isinθ)

    ![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:](useruploads.socratic.org)

    By De Morvies theorem,

    z1n=r1n(cos(θn)+isin(θn))

  • 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 cos3x. Then

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

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

    (cosx+isinx)3=cos3x+3icos2xsinx+3i2cosxsin2x+i3sin3x=cos3x3cosxsin2x+(3cos2xsinxsin3x)i

    Then, equalizing both expresions as conclusion we have

    cos3x=cos3x3cosxsin2x
    sin3x=3cos2xsinxsin3x

Questions