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Let $\epsilon $ be a positive real number and $a$ a complex number.

Prove the function $f(z)= \sin z +\frac{1}{z-a}$ has infinitely many zeros in the strip $|\mathrm{Im}z| < \epsilon.$

Thanks in advance!

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This is an old qualifying exam problem from Rice: Question 6 on page 6. – yunone Aug 19 '12 at 2:33
  1. I don't understand your argument. Consider the function $e^z$. It takes values $w$ with $Re(w) > 0$ and $Re(w) < 0$ in the strip $\{z: |\operatorname{Im}(z)| < \varepsilon\}$ but nevertheless it is never equal to $0$.

  2. Hint: let $k$ be large enough and $\varepsilon$ small enough. Consider the circle of radius $\varepsilon/2$ around point $2\pi k$. Specifically, consider the contour $\phi(t) = 2\pi k + \frac{\varepsilon}{2} e^{it}$ for $t\in [0, 2\pi)$.

    a. Prove that $\sin(\phi(t))$ makes one revolution around 0.

    b. Prove that $\sin(\phi(t)) > \varepsilon / 4$ (if $\varepsilon$ is small enough). Thus $f(\phi(t))$ makes one revolution around $0$ (if $k$ is large enough).

    c. Conclude that $f(z)$ has a zero in the $\varepsilon/2$ neighborhood of point $2\pi k$.

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Yes I just proved it cuts through $x=0$ I will remove that from my question – clark Aug 19 '12 at 2:29

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