Finding the number of roots of $f(z) = z^5 + z^3 + 3z + 1$ in the unit disk

Question

Suppose we have $$f(z) = z^5 + z^3 + 3z + 1$$ Find how many roots this function has in the open unit disc $\{z : |z| < 1\}$.

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Here's what I think about it:

I tried to split $f$ into two functions $$g(z) = z^5 + z^3$$ $$h(z) = 3z + 1$$ and use Rouché's theorem (to prove that $f = g + h$ has only one root). But $|h(z)| > |g(z)|$ has some troubles on unit circle (at least at $-1$ point).

Is there any way to use Rouché's theorem here? Maybe we can use it twice or work around problems at this point? Have no idea how to fix it.

Answer 1

You can apply the symmetric version of Rouché's theorem to $f(z) = z^5+z^3+3z+1$ and $g(z) = 3z+1$: On the boundary of the unit disk is $$ |f(z) -g(z)| = |z^5 + z^3| \le 2 \le |g(z)| $$ with equality only for $z=-1$. But $f(-1) \ne 0$, so that $$ |f(z) -g(z)| < |f(z)| + |g(z)| $$ everywhere on the boundary of the unit disk. It follows that $f$ and $g$ have the same number of zeros in the unit disk.

Answer 2

Let me give you a hint-

Show that the conclusion of Rouche's theorem holds when we replace the condition $\vert f \vert < \vert g \vert$ on $ \partial \Omega $ with the condition $\vert f \vert \le \vert g \vert$ on $ \partial \Omega $ and $ f+g \neq 0$ on $ \partial \Omega $