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

FE Sample--Problem 3

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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:

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


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

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

$$z=k\pi (k\neq 0), z\neq 3\pi + 6k\pi$$
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.


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