Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $C$?

share|improve this question
1  
$\sin(z)+a$ 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

1 Answer 1

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.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.