Graduate Understanding the Order of Poles in Complex Functions

[Baibhab Bose]

When The denominator is checked, the poles seem to be at Sin(πz²)=0, Which means πz²=nπ ⇒z=√n for (n=0,±1,±2...)
but in the solution of this problem, it says that, for n=0 it would be simple pole since in the Laurent expansion of (z∕Sin(πz²)) about z=0 contains the highest negative power to be 1/z. But, in case of the other values of z=±√n,±i√n, it says the Laurent expansion of z∕Sin(πz²) about z=±√n,±i√n, contains the highest negative to be 1/(z±√n)² and 1/(z±i√n)² from which we can infer that for n≠0 it has second order poles at z=±√n,±i√n.
I couldn’t figure out how to check this, since I can't manage to expand this weird function in Laurent series about z=±√n,±i√n these singular points.
So, how do I proceed to do so?

 

[stevendaryl]

That answer doesn't seem correct to me. Let's look at a pole at $\sqrt{n}$. Let $\xi = z-\sqrt{n}$ and look at $sin(\pi z^2) = sin(\pi n + 2\pi \sqrt{n} \xi + \pi \xi^2)$. We can write that (using a trig identity): $sin(\pi n + 2\pi \sqrt{n} \xi + \pi \xi^2) = sin(\pi n) cos(2\pi \sqrt{n} \xi + \pi \xi^2) + cos(\pi n) sin(2\pi \sqrt{n} \xi + \pi \xi^2) = 0 + (-1)^n sin(2\pi \sqrt{n} \xi + \pi \xi^2)$ When $\xi$ is small, we can use $sin(x) \approx x$, so this is approximately $(-1)^n (2\pi \sqrt{n} \xi +$ higher order terms $)$. So $\frac{z}{sin(\pi z^2)} \approx \frac{\sqrt{n}}{(-1)^n 2 \pi \sqrt{n} \xi} = \frac{(-1)^n}{2\pi \xi}$. So I think it has a first order pole at $z = \sqrt{n}$

 

[Baibhab Bose]

Thank you so much @stevendaryl !
But again my confusion is, here we see that this function, at the singular point, which you have evaluated, doesn't blow up at all, rather it boils down to a constant value. So in that case at z=√n the function remains analytic, and z=√n is not a pole at all, isn't it?
Again if I'm not wrong, that last constant term (−1)^n/2πξ is not a constant, since the ξ is present there in the denominator, which actually approaches 0, thereby blowing up the function. Thus we can conclude that its a simple pole at z=rootn. Right?

 

[stevendaryl]

Thank you so much @stevendaryl !
But again my confusion is, here we see that this function, at the singular point, which you have evaluated, doesn't blow up at all, rather it boils down to a constant value.

No, assuming my approximation is accurate, near $z=\sqrt{n}$, the function $\frac{z}{sin(\pi z^2)} \approx \frac{1}{2\pi \xi} = \frac{1}{2\pi (z-\sqrt{n})}$. So it blows up when $z \rightarrow \sqrt{n}$

Again if I'm not wrong, that last constant term (−1)^n/2πξ is not a constant, since the ξ is present there in the denominator, which actually approaches 0, thereby blowing up the function. Thus we can conclude that its a simple pole at z=rootn. Right?

I think so. Saying that it has an order 2 pole seems wrong to me.

 

[Baibhab Bose]

Yes, now its clear! Thank you, it was really helpful sir! :D