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)# 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 belowExplanation:
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 theoremBy other hand applying binomial Newton's theorem, we have
#(cosx+isinx)^3=cos^3x+3icos^2xsinx+3i^2cosxsin^2x+i^3sin^3x=cos^3x3cosxsin^2x+(3cos^2xsinxsin^3x)i# Then, equalizing both expresions as conclusion we have
#cos3x=cos^3x3cosxsin^2x#
#sin3x=3cos^2xsinxsin^3x#