# Finding number of zeros of $f(z) = z^{2019} + 8z + 7$ inside the unit disk.

I'm trying to find the number of zeros of $$f(z) = z^{2019} + 8z + 7$$ inside the unit disk. I've tried to apply Rouche's Theorem, but no combination of terms seems to work. Also, the Argument Principle seems to fail because when I was computing the winding number of $$f(\gamma)$$ around the origin, I realized $$f$$ has a zero on the unit disk. Any help on this problem would be greatly appreciated!

One can apply the symmetric version of Rouché's theorem to \begin{align} f(z) &= z^{2019} + 8z + 7 \, ,\\ g(z) &= 8z + 7 \, . \end{align} On the boundary of the unit disk we have \begin{align} |f(z) | &\ge |8 z| - 7 - |z^{2019}| = 0 \, ,\\ |g(z) | &\ge |8 z| - 7 = 1\, . \end{align} Equality holds in the second inequality only for $$z = 1$$, but at that point the first inequality is strict.
It follows that equality cannot hold simultaneously in both inequalities, i.e. $$|f(z) | + |g(z) | > 1 = |z^{2019}| = |f(z) - g(z)|$$ on the boundary of the unit disk. Rouché's theorem then states that $$f$$ and $$g$$ have the same number of zeros inside the unit disk (which is one).