Consider \[ f(z) \da {1\over \sin(z)} - {1\over z} + {2z\over z^2-\pi^2} .\] Show that on \(\abs{z} < 2\pi\), all singularities are removable, and find a Laurent expansion about \(z=0\).
Solution: Using L'Hopital and boundedness
Note that the singularities are \[ z = 0, \pi, -\pi .\]
That \(z=0\) is removable: \[ \lim_{z\to 0} f(z) &= \lim_{z\to 0} {z-\sin(z) \over z\sin(z)} \\ &\equalsbecause{\text{LH}} \lim_{z\to 0} {1 - \cos(z) \over \sin(z) + z\cos(z)} \\ &\equalsbecause{\text{LH}} \lim_{z\to 0} {\sin(z) \over \cos(z) + \cos(z) -z\sin(z) } \\ &= 0 < \infty ,\] so in particular \(f\) is bounded in a neighborhood of \(z=0\), making it removable.
That \(z=\pi\) is removable: \[ \lim_{z\to \pi} f(z) &= \lim_{z\to \pi} {1\over \sin(z)} - {1\over z} + {1\over z-\pi} + {1\over z+\pi}\\ &= c_1 + \lim_{z\to \pi} {1\over \sin(z)} + {1\over z-\pi} \\ &= c_1 + \lim_{z\to \pi} { (z-\pi) -\sin(z) \over (z-\pi) \sin(z) }\\ &= c_1 + \lim_{w\to 0} { w -\sin(w + \pi) \over w \sin(w+\pi) } \qquad w\da z-\pi \\ &= c_1 - \lim_{w\to 0} { w + \sin(w) \over w \sin(w) } \\ &\equalsbecause{\text{LH}} c_1 + 0 < \infty ,\] using the same L’Hopital argument as above. So this limit is bounded.
That \(z=-\pi\) is removable: \[ \lim_{z\to \pi} f(z) &= \lim_{z\to -\pi} {1\over \sin(z)} - {1\over z} + {1\over z-\pi} + {1\over z+\pi}\\ &= c_2 + \lim_{z\to -\pi} {1\over \sin(z)} + {1\over z+\pi}\\ &= c_2 + \lim_{z\to -\pi} {(z+\pi) - \sin(z) \over (z+\pi) \sin(z) } \\ &= c_2 - \lim_{z\to -\pi} {w + \sin(w) \over w \sin(w) } \qquad w \da z+\pi \\ &= c_2 + 0 < \infty ,\] again by the same argument.
For a Laurent expansion about \(z=0\), note \[ {1\over \sin(z) } &= {1\over z + c_3 z^3 + c_5 z^5 + \bigo(z^7)} \\ &= \inverseof{z}( 1 + c_3z^2 + (c_3^2-c_5)z^4 + \bigo(z^6)) \\ &= \inverseof{z} + {1\over 3!} z + \qty{ \qty{1\over 3!}^2 - {1\over 5!} }z^3 + \bigo(z^5) \\ &= \inverseof{z} + {1\over 6}z + {7\over 360}z^3 + \bigo(z^5) .\] and \[ {2z\over z^2-\pi^2} &= - {2z\over \pi^2} {1\over 1 - \qty{z\over \pi}^2 } \\ &= -{2z\over \pi^2}\sum_{k\geq 0}\qty{z\over \pi}^{2k} \\ &= -{2z\over \pi^2}\qty{1 + {1\over \pi^2} z^2 + {1\over \pi^4}z^4 + \bigo(z^6) } \\ &= -{2\over \pi^2}z -{2\over \pi^4}z^3 - {2\over \pi^6} z^5 - \bigo(z^7) ,\] so \[ f(z) &= \qty{\inverseof{z} + {1\over 6}z + {7\over 360}z^3 + \bigo(z^5)} \qty{-{2\over \pi^2}z -{2\over \pi^4}z^3 - {2\over \pi^6} z^5 - \bigo(z^7)} - \inverseof{z} \\ &= \qty{ {1\over 6} - {2\over \pi^2}}z + \qty{{7\over 360} - {2\over \pi^4} }z^3 + \bigo(z^5) .\]
Solution: Comparing orders of vanishing
Write \[ f(z) = {z-\sin(z) \over z\sin(z)} - {2z\over z^2-\pi^2} ,\]
For \(z=0\), the 2nd term doesn’t contribute to zero/pole order. For the first, take an expansion: \[ f_1(z) &= {z - \qty{ z + c_3z^3 + \bigo(z^5)} \over z \qty{z + c_3z^3 + \bigo(z^5)} } \\ \\ &= { -c_3z^3 + \bigo(z^5)\over z^2 + \bigo(z^4) } ,\] so there is a zero of order 3 in the numerator and of order 2 in the denominator, making the singularity removable. A similar argument works at \(z=\pm \pi\).