4
$\begingroup$

I'm trying to find the number of zeros for the function $f(z) = z + 2 - e^z$ in the half plane $\{\mathscr{R}z < 0\}$.

I know I'm supposed to use Rouche's theorem, which states that if both $f$ and $g$ are holomorphic inside and on some closed contour $C$, and $|f(z)| > |g(z)|$ for all $z \in C$, then $f$ and $f+g$ have the same number of zeros inside of $C$.

The problem is, I'm having some trouble coming up with intuition about what to choose for $f$ and $g$. Something tells me that $f(z) = z+2$ and $g(z) = -e^z$, but how precisely do I complete the problem with Rouche's theorem?

Thanks.

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Hint: consider a large half circle of radius $R$ in the left half plane, whose center is the origin, and diameter lies along the imaginary axis. Show that $|f|>|g|$ on the boundary of this region. Then as $R\to\infty\dots$

$\endgroup$
2
  • $\begingroup$ $f(z) = z+2$ and $g(z) = -e^z$. For sake of simplicity, I'll use a circle of radius 4. On both the vertical part of the boundary and the semi-circle part, we have that $f(z) \geq 2$. We want to show that $g(z) < 2$ on the entire boundary. I have somewhere in my notes that $|g(z)| \leq 1$, but can't seem to deduce why... $\endgroup$
    – r123454321
    Dec 16, 2014 at 7:29
  • 4
    $\begingroup$ @RyanYu $|e^{x+iy}| = e^x$ $\endgroup$
    – mrf
    Dec 16, 2014 at 7:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .