I'm working on the following problem.
Consider the function $f(z) = z^5 - 8 + e^{\gamma z}$, where $\gamma > 0$. How many zeros (counted w/ multiplicity) does $f$ have on the left half-plane $\{ z \in \mathbb{C} : \operatorname{Re}(z) < 0 \}$?
I'm assuming I want to apply Rouche's Theorem here, looking over the left half-disk of large radius, and approximating $f$ with $z^5 - 8$. It seems that $e^{\gamma z}$ should be small on the boundary of this region, since we would have $$\left|e^{\gamma z} \right| = e^{\gamma \operatorname{Re}(z)} \leq 1 $$ for $\gamma > 0 , \operatorname{Re}(z) \leq 0$. But I can't get these approximations to make sense. It seems like I either get the approximation to work well on the imaginary axis and it doesn't make sense on the semicircle, or the other way around.
I'd appreciate any help I could get on this problem, even if it just amounts to a hint at the setup. Alternatively, if there's a better way to handle this than Rouche, I'd appreciate a hint about that as well.
Thanks!
Take $U$ the domain bounded by $|z|=1, |z|=2, \Re z=0$ and note that for $|z| \le 1, \Re z \le 0, |f(z)| \ge 8-2>0$ while for $|z| \ge 2, \Re z \le 0, |f(z)| \ge 32-8-1>0$ so all its left plane zeroes are inside $U$
But now on the boundary of $U$ we have $|e^{\gamma z}| \le 1$ while on the interior circle part $|z^5-8| \ge 7$, on the exterior circle part $|z^5-8| \ge 24$ and on the imaginary axis $z^5$ is imaginary so $|z^5-8| \ge 8=|\Re (z^5-8)|$ so by Rouche $f$ and $z^5-8$ have the same number of zeroes in $U$, namely $2$