- #1
Baibhab Bose
- 34
- 3
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?
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?