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 $C$?
singularities of $\frac{1}{\sin z + a}$
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0@TaoHong洪涛 do both of them inside C? or only one? or none? that is what i want to know. Here is an equivalent problem, since $\sin z+a$ is entire, what is the value of $\int_C \frac{\cos z}{\sin z +a} \, dz$ – 2012-04-18
1 Answers
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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0The 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 $\text{Re}(z) = 2 \pi$. – 2012-04-18