Complex Roots - Not sure I did this right

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Hello. I'm not sure whether I did this right or messed up somewhere, just need to confirm my results...thanks to anybody who bothers answering.

Homework Statement



Find all the roots of z^{4}=1-i

Homework Equations



I guess I should state De Moivre's here...

(r cis(\vartheta))^{n}=r^{n} cis (n\vartheta)

The Attempt at a Solution



Firstly I re-wrote z^{4}=1-i as

z^{4}=\sqrt{2} cis (\frac{-\pi}{4})

Using De Moivre's,

z=(2\frac{1}{2})^{\frac{1}{4}} cis (\frac{1}{4}(\frac{-\pi}{4}+k2\pi))

z=2\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+k2\pi))

I found the four roots letting k=0,1,2,3

z=2^\frac{1}{8} cis (\frac{-\pi}{16})

z=2^\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+2\pi))=2^\frac{1}{8} cis (\frac{7\pi}{16})

z=2^\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+4\pi))=2^\frac{1}{8} cis (\frac{15\pi}{16})

z=2^\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+6\pi))=2^\frac{1}{8} cis (\frac{23\pi}{16})
 
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Hi Atena!

Your answer is correct. You can check this by taking the fourth powers of the solutions you got (using deMoivre).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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