I'm attempting to answer the question, "Prove that for any positive number $\epsilon$, the function $f(z)=\frac{1}{z+i} + \sin(z)$ has infinitely many zeros on the strip $|Im(z)|<\epsilon$".
My Work So Far: I know that I want to apply Rouche's Theorem. Assuming I understand Roche's Theorem correctly, I want to think of a function $g(z)$ that has infinitely many zeros in the strip $|Im(z)| < \epsilon$ and then show that $|f(z)-g(z)|$ is less than either $|g(z)|$ or $|f(z)|$. To this end, I thought of $g(z) = sin(z)$ since it has infinitely many zeros in the region of interest, which in of itself seems fairly indicative. However, when I attempt to use this $g(z)$, I end up with $|\frac{1}{z+i}|$, which seems like it can become arbitrarily small for $\epsilon \geq i$. As such, I can't bound this term by either $|g(z)|$ or $|f(z)|$. Am I missing something in my application of Roche's Theorem, or am I perhaps looking at the wrong $g(z)$ to begin with?
To address a more general question for the community, am I perhaps misapplying Rouche's Theorem, or is there perhaps a different way/trick that I should be thinking when addressing Roche's Theorem and problems of a similar nature to this one?
Consider the disks of radius $ϵ$ around the roots $z_k=k\pi$ of $\sin z$. For $|k|$ large enough, the term $\frac1{z+i}$ will be a very small perturbation of the sine function which has one simple root inside such a disk.
One can prove for $|z-k\pi|<1$
so if $|z-k\pi|=ϵ$ and $0.8ϵ(|k|\pi-2)>1\iff |k| > \frac1\pi(2+\frac{1.25}ϵ)$ then Rouché guarantees that there is exactly one root of $f(z)$ in $B(k\pi,ϵ)$ and thus infinitely many roots in the strip $|Im(z)|<ϵ$ around the real axis.