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So I have been given the following equation : $z^6-5z^3+1=0$. I have to calculate the number of zeros (given $|z|>2$). I already have the following:

$|z^6| = 64$ and $|-5z^3+1| \leq 41$ for $|z|=2$. By Rouche's theorem: since $|z^6|>|-5z^3+1|$ and $z^6$ has six zeroes (or one zero of order six), the function $z^6-5z^3+1$ has this too. However, how do I calculate the zeroes $\textit{outside}$ the disk? Is there a standard way to do this?

Thanks in advance.

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  • $\begingroup$ By the way, the second piece is bounded by 41, not 39. Pick $z=-2$ to see this. Not that it matters for this problem, but you should be extra careful with Rouche proofs as very small errors can completely ruin your argument. $\endgroup$ – Cameron Williams Jun 20 '18 at 16:45
  • $\begingroup$ Thanks, editing it! $\endgroup$ – Katie Jun 20 '18 at 16:47
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Since it is a polynomial of degree $6$ and since it has $6$ zeros inside the disk, it has no zeros outside the disk.

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  • $\begingroup$ So since all the zeros already lie inside the disk, you can say that there are no zeros outside? So there are actually no calculations needed. $\endgroup$ – Katie Jun 20 '18 at 16:49
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    $\begingroup$ @Katie The only computations that are needed are the ones that you have aleready mad to compute the number of zeros insede the disk. Note the those same computations also prove that there are no zeros at the boundary of the disk. $\endgroup$ – José Carlos Santos Jun 20 '18 at 16:52
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For polynomials (and other functions where you know the total number of roots) the standard way is simply to subtract the number of zeroes inside or on the boundary of the disc, which in this case turns out to be $6-6$.

For more complex functions you can sometimes use Rouche's theorem on $f(\frac{1}{z})$, but I'm not aware of any way that could really be called standard.

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