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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!

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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).

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  • $\begingroup$ This was exactly the kind of solution I was hoping for! Thank you. $\endgroup$ – L. L. S. Dec 13 '19 at 21:51

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