# Proving that $|\sin z|> |1/z+i|$, $z$ is a complex number

I already found a proof using the argument principle for a more general version. Prove the function $f(z)= \sin z +\frac{1}{z-a}$ has infinitely many zeros in the strip $|\mathrm{Im}z| < \epsilon.$ But if possible I would like a hint on getting a shorter algebraic proof like a series of inequalities. A possibly promising way is showing that $\sin^2x+\sinh^2y> [x/(x^2+(1+y)^2)]^2+[y+1/(x^2+(1+y)^2)]^2$.

Thank you.

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Assuming you mean $|\sin z| > |\frac{1}{z+i}|$ this is certainly not true near $z=-i$. It is as certainly not true near points where $\sin z = 0$. (And if you mean $|\sin z| > |\frac{1}{z}+i|$, it doesn't help.) –  mrf Jan 31 '13 at 6:47
@mrf: This is precisely what I was thinking before I saw your comment. :) –  Haskell Curry Jan 31 '13 at 9:06