Show that $f$ has exactly two zeros (counting multiplicity)

Question

Let $f(z) = z^2+1+g(z)$ where $g$ is holomorphic and $|g(z)| < 3$ for $z ∈ bD(0, 2)$.

Show that $f$ has exactly two zeros (counting multiplicity) on $D(0, 2)$.

My attempt:

$|f(z)| = |z^2 + 1 + g(z)| \leq |z^2+1| + |g(z)| < |z^2+1| + 3 \leq |z^2| + 1 + 3 = 4 + 1 + 3 = 8$ ($|z^2| = 4$ since $z$ $∈$ $bD(0, 2)$). Hence $|g(z)| < |f(z)|$

So by Rouché's theorem $f$ and $f+g$ have the same number of zeros on $D(0,2)$

Now $(f+g)(z) = z^2 + 1 + 2g(z)$

But $(f+g)(z) = z^2 + 1 + 2g(z) = 0$ iff $\frac{z^2 + 1}{-2}= g(z)$

Since $g$ is holomorphic, then $\int_{C(0,2)} g(z) dz = 0$.

So $\int_{C(0,2)} \frac{z^2 + 1}{-2} dz = 0$

But now I'm stuck. Do I need to parametrize $C(0,2)$ and evaluate the integral to find the zeros of $f+g$?

Answer 1

Unfortunately, your attempt does not work. First, $|f(z)| < 8$ does not imply $|g(z)| < |f(z)|$. Second, nothing is known about the zeros of $f+g$.

But with $h(z) = z^2+1$ we have for $|z|=2$ that $$ |g(z)| < 3 = |z^2| - 1 \le |z^2 + 1| = |h(z)| \, . $$ Rouché's theorem then gives that $f = g+h$ and $h$ have the same number of zeroes in $D(0, 2)$.