FE Sample--Problem 3

[Victor Ivrii]

Find all singular points, classify them, and find residues at these points of
$$
f(z)= \frac{\cos(z/6)}{\sin^2(z)} + \frac{z}{\sin(z)}.
$$
infinity included.

[Ziqi Zhang]

f(z)=(cos(z/6)+zsinz)/(sinz)^2

case 1: when z=3pi+6k*pi (where k is an integer), they have poles of order 1 because numerator has order 1 and denominator has order 2.

case 2: z=k*pi (where k is an integer and does not equal to 3+6n where n is an integer) have poles of order 2

let w=1/z

f(w)=(wcos(6/w)+sin(1/w))/w(sin(1/w))^2

when w approach 0,

f(w)=1/(wsin(1/w)) so f(w) approach infinity

so at infinity it is a pole

[Xiaoning Han]

Let $$f(z)=\frac{\cos(\frac{z}{6})+z\cdot\sin z}{\sin^2z}$$
$1)$
$\\$
Let $$g(z)=\cos(\frac{z}{6})+z\sin z,g(z)\neq 0,\therefore order =0.$$
Let $$h(z)=\sin^2 z, h(0)=0\\h'(z)=2\sin z\cos z=\sin 2z, h'(0)=0\\
h''(z)=2\cos 2z, h''z(0)\neq0, \therefore order=2.\\
2-0=2, \therefore \text{ it is a pole of order } 2.$$

$2)$
$\\$
Let $$z=3\pi + 6k\pi\\g(3\pi + 6k\pi)=0 \\ g'(z)=-\frac{1}{6}\sin(\frac{z}{6}) +z\cos z\\
g'(3\pi + 6k\pi)\neq 0,\therefore order =1.$$
Let $$h(z)=\sin^2 z, h(3\pi + 6k\pi)=0\\h'(z)=\sin 2z, h'(3\pi + 6k\pi)=0\\h''(z)=2\cos 2z, h''(3\pi + 6k\pi)\neq0\therefore order=2.\\2-1=1, \text{ order 1 simple pole}.$$

$3)$
$\\$
$$z=k\pi (k\neq 0), z\neq 3\pi + 6k\pi$$
Let $$g(z)=\cos(\frac{z}{6})+z\sin z,g(k\pi\neq 0),order=1$$
Let $$h(z)=\sin^2 z, h(k\pi)=0\\h'(z)=\sin 2z,h'(k\pi)=0\\h''(z)=2\cos 2z,h''(k\pi)\neq0, \therefore order=2$$
$$2-1=1, \text{ it is a simple pole}.$$

[Victor Ivrii]

Xiaoning, too many words. Ziqi , I read what you wrote, it is correct but formatting is terrible.

In fact, it is very simple:
1) If the first term has a pole of order $2$, the second term, which has poles of order $1$ at most, cannot cancel it. And as mentioned, poles of order $2$ are as $\sin(z)=0$ (that means $z=\pi n$) but $\cos (z)\ne 0$ which means $n$ is not divisible by $3$ and odd: $n\ne 6m+3$.

2) If $z=(6m+3)\pi$ we need to check that two simple poles do not cancel one another, namely, that indeed $[\cos(z/6)+z\sin(z)]'\ne0$ which was done:
$$
\bigl[\cos(\frac{z}{6})+z\sin(z)\bigr]'=-\frac{1}{6}\sin (\frac{z}{6}) + \sin(z)-z\cos(z)= \frac{1}{6}(-1)^{m+1}  + (6m+3)\pi \ne 0.
$$
Finally, $\infty$ is not isolated.

However, residues so far are not found

[Zechen Wang]

here are the residues. modified.

[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]

Also, as for infinity, since infinity is not an isolated singularity, the residue at infinity is zero not defined!

[Heng Kan]

So is there anything wrong with my calculation? Thanks.

[Victor Ivrii]

So is there anything wrong with my calculation? Thanks.

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
$$
\Res (f(z), n\pi) =  -\frac{1}{6}\sin (n\pi/6)+ (-1)^n.
$$