I need to study the domain of analyticity of this function: $$ f(z) = \frac{\sqrt{(z-3)(z^2-4)}}{2z^2}\sin z$$ and compute the integral over the unitary circle $\gamma: \theta \to e^{i\theta}, \theta \in [0,2\pi]$ centered at $z = 0$ I started by saying that the function presents three branching points, due to the square root, at $z=3,\pm2$. I also studied the behavior of the function at infinity by the standard substitution $z \to \frac{1}{\zeta}$ as $\zeta \to 0$. Because of the essential singularity due to the presence of $\sin(z)$ I thought the point at infinity could be considered a nonanalytic point. I can now decide where to put the branch cuts, my choice is a branch cut in $[2,3]$ and one starting at $z=-2$ to the point at infinity: in this way, I prevent a curve to loop around any of the branching points individually, but it also prevents from enclosing all the three of them at once. Now, in a very intuitive and nonrigorous manner I should add, I thought that a curve enclosing two branching points would not cause problems. This curve would cause the argument of both the roots to increase by $2\pi$. In a more formal way, my reasoning is: $$ z-a = r_1e^{i\theta_1+2k_1\pi}, z-b = r_2e^{i\theta_2 + 2k_2\pi}$$ Therefore we get: $$ \sqrt{r_1r_2}e^{\frac{\theta_1+\theta_2}{2} + (k_1+k_2)\pi } $$ And if we loop once around them both we get $k_1 = 1, k_2 = 1$, therefore yielding $e^{i2\pi} = 1$, which does not cause the real part of the square root to change sign. There is something wrong in all of this though, the branch cuts do not allow me to go around all combinations of the two branch points, although this would be allowed if my previous reasoning makes sense. Now to the second part: the computation of the contour integral might seem easy to you but it wasn't for me. The reason for that particular choice of branch cuts was also dictated by the second part of the exercise: the curve doesn't loop around any branch cuts and things become easier. I applied the residue theorem: $\int_\gamma f(z)dz = 2\pi i Res(f,0)$. Now the problem is that I don't know how to compute the Laurent series of $f(z)$ and then find the $c_{-1}$ coefficient
$\begingroup$
$\endgroup$
5
-
1$\begingroup$ Note that the integrand is of the form $g(z)/z$, where $g$ is analytic inside the contour. $\endgroup$– eyeballfrogCommented Jun 21, 2023 at 13:51
-
$\begingroup$ If $g(z)$ is analytic then I could expand it via Taylor series, but our function $g$ seems to be excessively complicated due to the all the square roots. $\endgroup$– ClaudioCommented Jun 21, 2023 at 14:15
-
1$\begingroup$ True, but how many terms of the power series of $g$ do you really need? $\endgroup$– eyeballfrogCommented Jun 21, 2023 at 14:17
-
$\begingroup$ I guess only the costant term coefficient since it would give the $c_{-1}$ coefficient. So I need to compute the taylor series of $\sqrt(...)(1-z^2/6 +z^4/120-...)$ up to the first power of $z$. No, obvioulsy there's something Im missing. $\endgroup$– ClaudioCommented Jun 21, 2023 at 14:27
-
$\begingroup$ Ok I need to find the value of $g(z)$ at $z_0=0$, namely the first term of a taylor expansion at $z_0$. So $g(0) = \sqrt{3}$ is the residue, therefore the integral is equal to $2\sqrt{3} \pi i $ $\endgroup$– ClaudioCommented Jun 21, 2023 at 14:59
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
There are three branch points, $z=3, z=2, z=-2$, and they are all outside the contour $\gamma$, hence you can choose any branch cuts, starting from these three branch points, as long as those cuts do NOT pass through the unit circle.
As shown in the figure, hence, $z=0$ is the only singularity inside the contour, and it is a simple pole. The residue is
$$\text{Res}(z=0)=\lim_{z\to0}z\cdot f(z)=\sqrt3\cdot\lim_{z\to0}\frac{\sin z}{z}=\sqrt3$$
The integral equals
$$\int_\gamma f(z)~dz=2\pi\sqrt3~i$$
-
$\begingroup$ Ok.I suppose my choice to be equally valid then? Again, the doubt remains: would it be possible to loop once around, lets say, $2,3$? Also why didnt you consider the behavior of the function at infinity? I thought It was a branch point too, since the function looses analyticity there due to the essential singularity $\endgroup$– ClaudioCommented Jun 21, 2023 at 21:58
-
$\begingroup$ For a branch cut, you need to completely cut the plane, start from branch point and shoot a line to infinity, this is a cut. So your $[2, 3]$ cut is ambiguous. You may see some dogbone shaped cut, but that is two cuts, $[2, \infty)$ and $[3,\infty)$. Under some conditions, two cuts might be sealed (canceled) at some overlaped region. Infinity is outside the unit circle, so no need to consider the infintiy point. $\endgroup$– MathFailCommented Jun 21, 2023 at 22:06