MAT334-2018F > End of Semester Bonus--sample problem for FE

FE Sample--Problem 3

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Victor Ivrii:
Calculations of residue should be with explanation. Especially, at double poles simply wrong

Heng Kan:
For the residues, please see the attatched scanned pictures.  Also, as for infinity, since infinity is not an isolated singularity, the residue at infinity is not defined.

Victor Ivrii:

--- Quote from: Heng Kan on December 02, 2018, 04:10:22 PM --- Also, as for infinity, since infinity is not an isolated singularity, the residue at infinity is zero not defined!

--- End quote ---

Heng Kan:
So is there anything wrong with my calculation? Thanks.

Victor Ivrii:

--- Quote from: Heng Kan on December 02, 2018, 05:24:08 PM ---So is there anything wrong with my calculation? Thanks.

--- End quote ---
Difficult to check. You leave too little vertical space and overcomplicated. Everything is much more straightforward:
$\newcommand{\Res}{\operatorname{Res}}$

Consider term $\frac{\cos(z/6)}{\sin^2(z)}$. Since $\sin (z)= (-1)^n (z-n\pi) + O((z-n\pi)^3)$ near $z=n\pi$,
$$\Res (\frac{\cos(z/6)}{\sin^2(z)}, n\pi) = \Res (\frac{\cos(z/6)}{(z-n\pi)^2}, n\pi) = (\cos (z/6))'|_{z=n\pi}= -\frac{1}{6}\sin (n\pi/6).$$
On the other hand
$$\Res (\frac{z}{\sin (z)}, n\pi) = \frac{1}{\cos (n\pi)}= (-1)^n.$$
So
$$\Res (f(z), n\pi) = -\frac{1}{6}\sin (n\pi/6)+ (-1)^n.$$