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Let $C:= \{z : \Re(z)\in [0,2\pi] \text{ and } \Im(z)\in [-\pi,\pi]\}$. Given $a \in \mathbb{C}$ such that $|a|\leq 1$, how many singularities does $$f(z)=\frac{1}{\sin z +a}$$ have in the interior of $C$?

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$\sin(z)+a$ or $\sin(z+a)$? – Dirk Apr 17 '12 at 6:12
it's $(\sin z) + a$.. I can do it whenever $a$ is a real number. Just curious what happen when $a$ is complex numbers, – Ajat Adriansyah Apr 17 '12 at 6:16
have you tried solve equation $ \sin z=-a$ – noname1014 Apr 17 '12 at 6:41
yes, I've tried that, in fact the problem can be transformed into solving $\sin z = -a$. writing $a=-\mu - i \lambda$ we will have $$\lambda e^{-y}= \left(\frac{1 + e^{2y}}{2}\right)\cos x \quad \text{dan} \quad \mu e^{-y}= \left(\frac{1-e^{2y} }{2}\right)\sin x$$ I have no idea how to finish this.. – Ajat Adriansyah Apr 17 '12 at 8:55
Use Euler's formula: $\sin z = (e^{iz}-e^{-iz})/2i$. This transforms your equation to a quadratic equation in $e^{iz}$. – mrf Apr 17 '12 at 9:13
up vote 1 down vote accepted

Since $\sin(z) = \sin(\pi - z) = \sin(3\pi - z)$, $\sin(z) = -a$ has a solution with $0 < \text{Re}(z) < \pi/2$ and $|\text{Im}(z)| \le \pi$ iff it has one with $\pi/2 < \text{Re}(z) < \pi$ and $|\text{Im}(z)| \le \pi$, and similarly for $\pi < \text{Re}(z) < 3\pi/2$ and $3\pi/2 < \text{Re}(z) < 2\pi$.

Note that $\sin(t - i \pi) = \cosh(\pi) \sin(t) + i \sinh(\pi) \cos(t)$, which for $0 \le t \le \pi/2$ traces out the part of the ellipse $x^2/\cosh^2(\pi) +y^2/\sinh^2(\pi) = 1$ in the first quadrant while $\sin(t)$ traces out the line segment $[0,1]$ and $\sin(t + i \pi) = \cosh(\pi) \sin(t) - i \sinh(\pi) \cos(t)$ traces out the part of the ellipse in the third quadrant. For $0 \le \text{Re}(z) \le \pi/2$ and $-\pi \le \text{Im}(z) \le \pi$ we get everything inside the right half of the ellipse. Similarly, taking $\pi \le t \le 3\pi/2$ or $3\pi/2 \le t \le 2 \pi$ we get the left half of the ellipse.

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so how many singularities does it have? – noname1014 Apr 18 '12 at 8:05
Dear Robert, I can't understand your second paragraph, How does it connect to the premise on the first paragraph? Thank you. – Ajat Adriansyah Apr 18 '12 at 15:33
Use the argument principle. As $z$ goes counterclockwise around the rectangle $[0,2\pi] \times [-\pi,\pi]$ (starting, say, at $-\pi i$), $\sin(z)$ goes counterclockwise around the ellipse, then up the imaginary axis from $-\sinh(\pi)i$ to $+\sinh(\pi) i$, then again counterclockwise around the ellipse, and then down the imaginary axis from $\sinh(\pi) i$ to $-\sinh(\pi) i$. So the winding number around any point $-a$ inside the ellipse and not on the imaginary axis is $2$: there are two zeros (counting multiplicity) of $\sin(z) + a$ inside the rectangle. – Robert Israel Apr 18 '12 at 19:47
The only cases of multiplicity $>1$ are the zeros of $\cos(z)$, namely $z = \pi/2$ and $3\pi/2$ in the rectangle, corresponding to $\sin(z) +1$ and $\sin(z) - 1$ each having a single zero of multiplicity $2$ in the rectangle. For $a$ on the imaginary axis with $|a| \le \sinh(\pi)$, there is one zero of $\sin(z) + a$ with $\text{Re}(z) = 0$, one with $\text{Re}(z) = \pi$, and one with $\text{Re}(z) = 2 \pi$. – Robert Israel Apr 18 '12 at 19:48

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