Determine the order of the pole of
- \({1\over z\sin(z)}\) at \(z_0 = 0\).
- \({e^{z^2}-1\over z^4}\) at \(z_0=0\)
Solution
Order 2: \[ \lim_{z\to 0}z^0 f(z) &= \infty\\ \lim_{z\to 0}z^1 f(z) &= \infty\\ \lim_{z\to 0}z^2 f(z) &= 1 \neq 0 ,\] using that \(z/\sin(z) \convergesto{z\to 0} 1\).
Order 2: apply L’Hopital as necessary \[ \lim_{z\to 0}z^0 f(z) &= \infty\\ \lim_{z\to 0}z^1 f(z) &= \infty\\ \lim_{z\to 0}z^2 f(z) &= 1 \neq 0 .\] Alternatively, take the Laurent expansion: \[ f(z) &= z^{-4}(1 + z^2 + {1\over 2!} (z^2)^2 + {1\over 3!} (z^2)^3 + \bigo(z^4) - 1) \\ &= z^{-2} + {1\over 2!} + {1\over 3!} z^2 + \bigo(z^4) .\]