De Moivre’s and the nth Root Theorems
Key Questions
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If the complex number
zz isz=r(cos theta + i sin theta)z=r(cosθ+isinθ) ,then
z^nzn can be written asz^n=[r(cos theta+i sin theta)]^n=r^n[cos theta+i sin theta]^nzn=[r(cosθ+isinθ)]n=rn[cosθ+isinθ]n by De Mivre's Theorem,
=r^n[cos(n theta)+i sin(n theta)]=rn[cos(nθ)+isin(nθ)]
I hope that this was helpful.
Wataru
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2
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Dec 3 2014
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Answer:
z^(1/n) = r^(1/n) ( cos (theta/n) + i sin(theta/n))z1n=r1n(cos(θn)+isin(θn)) Explanation:
Polar form of complex number is
z = r( cos theta + i sin theta)z=r(cosθ+isinθ) 
) By De Morvies theorem,
z^(1/n) = r^(1/n) ( cos (theta/n) + i sin(theta/n))z1n=r1n(cos(θn)+isin(θn)) 
sankarankalyanam
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1
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Jun 14 2018
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Answer:
More of the cases, to find expresions for
sinnxsinnx orcosnxcosnx as function ofsinxsinx andcosxcosx and their powers. See belowExplanation:
Moivre's theorem says that
(cosx+isinx)^n=cosnx+isinnx(cosx+isinx)n=cosnx+isinnx An example ilustrates this. Imagine that we want to find an expresion for
cos^3xcos3x . Then(cosx+isinx)^3=cos3x+isin3x(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^3x-3cosxsin^2x+(3cos^2xsinx-sin^3x)i(cosx+isinx)3=cos3x+3icos2xsinx+3i2cosxsin2x+i3sin3x=cos3x−3cosxsin2x+(3cos2xsinx−sin3x)i Then, equalizing both expresions as conclusion we have
cos3x=cos^3x-3cosxsin^2xcos3x=cos3x−3cosxsin2x
sin3x=3cos^2xsinx-sin^3xsin3x=3cos2xsinx−sin3x 
F. Javier B.
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1
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Jul 25 2018