How do you find the $n^{t h}$ $n^{th}$ roots of complex numbers in polar form?

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How do you find the $n^{t h}$ $n^{th}$ roots of complex numbers in polar form?

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Answer 1 · 2018-06-14T05:36:05

$z^{\frac{1}{n}} = r^{\frac{1}{n}} (cos (\frac{θ}{n}) + i sin (\frac{θ}{n}))$ $z^(1/n) = r^(1/n) ( cos (theta/n) + i sin(theta/n))$

Explanation:

Polar form of complex number is $z = r (cos θ + i sin θ)$ $z = r( cos theta + i sin theta)$

![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:]( useruploads.socratic.org )

By De Morvies theorem,

$z^{\frac{1}{n}} = r^{\frac{1}{n}} (cos (\frac{θ}{n}) + i sin (\frac{θ}{n}))$ $z^(1/n) = r^(1/n) ( cos (theta/n) + i sin(theta/n))$

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How do you find the $n^{t h}$ $n^{th}$ roots of complex numbers in polar form?

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How do you find the $n^{t h}$ $n^{th}$ roots of complex numbers in polar form?