Applying Rouché's theorem in a square

Question

I've got to determine the number of zeroes of the function $f: \mathbb{C} \rightarrow \mathbb{C}$, $f(z)=z^2+e^{z-1}$ inside the square with the corners $3 \pm 3i $ and $-3 \pm 3i $.

Of course I was thinking of applying Rouché's theorem for the inner disk centered at zero with radius $3$ and the outer disk with radius $\sqrt{3^2+3^2}=\sqrt{18}$ in order to show that in both regions the number of zeroes is equal.

For the inner disc I get:

$|e^{3-1}|\leq |3^2|$ for $z=3$. So $z^2+e^{z-1}$ has the same numbers of zeroes as $z^2$. That is $2$.

Now for the outer disc I get:

$|e^{\sqrt{18}-1}|\geq18$ for $z=\sqrt{18}$, which means that here I can't apply Rouché's theorem.

I don't see other ways to determine the number of zeroes. Am I missing something in my approach?

Answer 1

You may apply Rouché's theorem directly with respect to the given square $K$.

If $z\in \partial K$ then $|z|^2\geq 3^2=9$ and $$|e^{z-1}|=e^{x-1}\leq e^2$$ where $z=x+iy$ with $-3\leq x\leq 3$ and $-3\leq y\leq 3$. So for any $z\in\partial K$ we have that $$|e^{z-1}|\leq e^2<9\leq |z|^2$$ which implies that $z^2$ and $f(z)=z^2+e^{z-1}$ have the same number of zeros inside $K$, that is $2$.

Answer 2

Let $g(z)=z^2$ and let $h(z)=e^{z-1}$.

Let $S$ denote the given square region.

Compare $|g(z)|$ and $|h(z)|$ on the boundary of $S$ . . .

On the boundary of $S$, we have

• $\text{Re}(z)\le 3$, hence $|h(z)|=|e^{z-1}|=e^{\text{Re}(z-1)}\le e^2$.$\\[4pt]$
• $|z|\ge 3$, hence $|g(z)|=|z^2|=|z|^2\ge 9$.

Thus, on the boundary of $S$, we have $|h(z)| < |g(z)|$, so by Rouché's Theorem, the number of zeros of $f$ inside $S$ is the same as the number of zeros of $g$ inside $S$.

Hence, since $g$ has exactly $2$ zeros (counting multiplicity) inside $S$, the same is true of $f$.