Toronto Math Forum
MAT3342018F => MAT334Tests => End of Semester Bonussample problem for FE => Topic started by: Victor Ivrii on November 27, 2018, 03:56:33 AM

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.

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

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 $$h(z)=\sin^2 z, h(0)=0\\h'(z)=2\sin z\cos z=\sin 2z, h'(0)=0\\
h''(z)=2\cos 2z, h''z(0)\neq0, \therefore order=2.\\
20=2, \therefore \text{ it is a pole of order } 2.$$
$2)$
$\\$
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 $$h(z)=\sin^2 z, h(3\pi + 6k\pi)=0\\h'(z)=\sin 2z, h'(3\pi + 6k\pi)=0\\h''(z)=2\cos 2z, h''(3\pi + 6k\pi)\neq0\therefore order=2.\\21=1, \text{ order 1 simple pole}.$$
$3)$
$\\$
$$z=k\pi (k\neq 0), z\neq 3\pi + 6k\pi$$
Let $$g(z)=\cos(\frac{z}{6})+z\sin z,g(k\pi\neq 0),order=1$$
Let $$h(z)=\sin^2 z, h(k\pi)=0\\h'(z)=\sin 2z,h'(k\pi)=0\\h''(z)=2\cos 2z,h''(k\pi)\neq0, \therefore order=2$$
$$21=1, \text{ it is a simple pole}.$$

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

here are the residues. modified.

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

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.

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

So is there anything wrong with my calculation? Thanks.

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