# Number of zeroes outside $|z|>2$ (Rouche's theorem)

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?

• 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. – Cameron Williams Jun 20 '18 at 16:45
• Thanks, editing it! – Katie Jun 20 '18 at 16:47

Since it is a polynomial of degree $6$ and since it has $6$ zeros inside the disk, it has no zeros outside the disk.
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.