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I want to show that for every $0 \neq w \in \mathbb{C}$ and $n \in \mathbb{Z}^+$ there are exactly $n$ complex numbers $z$ such that $z^n = w$ without using the trig functions/ the complex exponential function. How can i do this? (Actually, i just need the existence of at least one root, then i can use that + some analysis to prove the fundamental theorem of algebra and i'm done)

Rick Does Math
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  • What tools do you have at your disposal? The existence of roots follows from the fundamental theorem of algebra, which is a simple consequence of Liouville's theorem about bounded entire functions. – Martin R Jun 26 '21 at 10:27
  • Sorry i did not specify, i only have basic real analysis theorems. My idea is to prove the existence of at least one n-root using elementary ideas and then use that together with the fact that continuous functions attain minimum on compact sets to show the FTA. – Rick Does Math Jun 26 '21 at 10:31
  • Maybe a proof by induction using a graphical/geometric method of multiplication of two complex numbers as in this answer: https://math.stackexchange.com/a/1133131/441161 To show at least one root exists. – Andy Walls Jun 26 '21 at 11:03
  • that could be a good idea, but i don't really know how to do it – Rick Does Math Jun 26 '21 at 12:53

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