- #1
ibensous
- 10
- 0
Ok, so I'm suppose to be able to remove the singularity to find the residue of the function
[tex](z)cos{\frac{1}{z}[/tex]
I tried to see how "bad" the singularity was by taking the limit, but I can't figure out if
[tex]
\lim_{ z \to 0 } (z)cos{\frac{1}{z}
[/tex]
goes to 0 or if it is not bounded. If it goes to zero I should be able to remove the singularity. Since I couldn't figure that out I used the Taylor Series of cos(z) to expand the function and I got a Laurent Series.
[tex](z)cos{\frac{1}{z} = \sum_{n=0}^\infty \frac{(-1)^{(n)}}{(2n)!}z^{(-2n+1)} [/tex]
This is where I'm stuck. I don't see how I can possibly remove the singularity or if its even removable. Any ideas?
[tex](z)cos{\frac{1}{z}[/tex]
I tried to see how "bad" the singularity was by taking the limit, but I can't figure out if
[tex]
\lim_{ z \to 0 } (z)cos{\frac{1}{z}
[/tex]
goes to 0 or if it is not bounded. If it goes to zero I should be able to remove the singularity. Since I couldn't figure that out I used the Taylor Series of cos(z) to expand the function and I got a Laurent Series.
[tex](z)cos{\frac{1}{z} = \sum_{n=0}^\infty \frac{(-1)^{(n)}}{(2n)!}z^{(-2n+1)} [/tex]
This is where I'm stuck. I don't see how I can possibly remove the singularity or if its even removable. Any ideas?