Proposition

Find the number of zeros of

$3\exp(z) - z$ in $\overline{D}[0, 1]$.
$\frac{1}{3}\exp(z) - z$ in $\overline{D}[0, 1]$.
$z^4 - 5z + 1$ in $\{ z \in \mathbb{C} : 1 \leq \abs{z} \leq 2 \}$.

Solution

1

For any $\abs{z} = 1$, $\abs{3\exp(z)} \geq 3/e > 1 = \abs{-z}$.

By Rouche’s Theorem, $3\exp(z) - z$ and $-z$ have the same number of zeros inside the unit disk $\overline{D}[0, 1]$. Therefore, $3\exp(z) - z$ has only one zero.

2

For any $\abs{z} = 1$, $\frac{1}{3}\exp(z) \leq \frac{e}{3} < 1 = \abs{-z}$. Thus $\frac{1}{3}\exp(z) - z$ has the same number of zeros as $\frac{1}{3}\exp(z)$ does inside the unit disk which is none.

3

For any $\abs{z} = 2$, $\abs{-5z + 1} \leq 5\abs{z} + 1 \leq 11 < 16 = \abs{z^4}$. Thus $z^4 - 5z + 1$ has the same number of zeros as $-5z + 1$ inside the given region. Thus $z^4 - 5z + 1$ has no zero.