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Author Topic: FE-P3  (Read 10964 times)

Victor Ivrii

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FE-P3
« on: December 18, 2018, 06:14:31 AM »
Find all singular points, classify them, and find residues at these points of
f(z)=tan(z)+zcot2(z);
infinity included.

hz12

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Re: FE-P3
« Reply #1 on: December 18, 2018, 11:13:44 AM »
f(z) = sin(z)cos(z)+zcos2(z)sin2(z)
      =sin3(z)+ zcos3(z)    =gcos(z) sin2(z)    =h
     

Cos(z)sin2(z) = 0

cos(z) = 0 or sin2(z) = 0

so z =kπ or z=π2+kπ
 
1, when z = kπ

g = sin3(z)+zcos3(z)0                            h= cos(z) sin2(z)=0
                                                                     h=sin3(z)+2cos2(z)sin(z) =0                     h0


So pole of order = 2

 2, when z =π2+kπ               

g = sin3(z)+zcos3(z)0                            h= cos(z) sin2(z)=0
                                                                    h=sin3(z)+2cos2(z)sin(z)  0
                               
So pole of order = 1
« Last Edit: December 18, 2018, 11:51:34 AM by Hanyu Zhou »

Ziqi Zhang

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Re: FE-P3
« Reply #2 on: December 18, 2018, 11:17:50 AM »
I think when z=0, it should have pole of order 1. Because at z=0, the numerator is not 0 when taking derivative once and denominator is not 0 when taking derivative twice.

Ziqi Zhang

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Re: FE-P3
« Reply #3 on: December 18, 2018, 11:28:38 AM »
Residue at z=0: 0+cos2(0)=1

Residue at z=kπ, but k≠0: 0-2(kπ)cos(kπ)sin(kπ)+cos2(kπ)=(-1)n

Residue at z=0.5π+kπ: -1+0=-1
« Last Edit: December 18, 2018, 11:36:43 AM by Ziqi Zhang »

Zixuan Miao

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Re: FE-P3
« Reply #4 on: December 18, 2018, 11:37:21 AM »
for infinity case, it is a non-isolated singularity

Siying Li

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Re: FE-P3
« Reply #5 on: December 18, 2018, 12:45:02 PM »
I think the residue at 0 should be 0..
Here's my answer: (in calculation I ignore k for convenience)

f(z)=tan(z) +zcot2(z)=sin(z) cos(z) +zcos2(z) sin2(z)  =sin3(z) +zcos3(z) cos(z) sin2(z) 
If cos(z) =0, then z=π2+kπ

For numerator, sin3(π2) +π2cos3(π2)0 , the numerator has zero order of zeros

For denominator, cos(π2) sin2(π2)=0 ,sin3(π2) +π2cos2(π2) sin(π2) 0 , the denominator has zeros of order 1

Then f(z) has simple pole z=π2+kπ

If sin2(z) =0, then z=0 or z=kπ

When z=0,

For numerator, sin3(0) +0cos3(0)=0 ,3sin2(0)+cos3(0) +3cos2(0) sin?(0)  0 , the numerator has zero order of 1

For denominator, cos(0) sin2(0)=0 ,sin3(0) +2cos2(0) sin(0) =0 , 3sin2(0)cos(0) 4cos(0) sin2(0) +2cos3(0)0 , the denominator has zeros of order 2

Then f(z) has simple pole z=0

When z=kπ,

For numerator, sin3(π) +0cos3(π)0  the numerator has zero order of 0

For denominator, cos(π) sin2(π)=0 ,sin3(π) +2cos2(π) sin(π) =0 , 3sin2(π)cos(π) 4cos(π) sin2(π) +2cos3(π)0 , the denominator has zeros of order 2

Then f(z) has pole z=kπ with order 2


Res(f(z),π2+kπ)=sin3(π2) +π2cos3(π2) sin3(π2) +2cos2(π2) sin(π2) =11=1
Res(f(z),kπ)=3sin2(π)+cos3(π) +3cos2(π) sin(π) 1!=11=1
Since f(z)=sin3(z) +zcos3(z) cos(z) sin2(z) =(0(1)nz2n+1(2n+1)!)3+0(1)nz2n+1(2n)!0(1)nz2n(2n)! (0(1)nz2n+1(2n+1)!)2 has no term of (z0)1, then
Res(f(z),0)=0

Yifei Wang

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Re: FE-P3
« Reply #6 on: December 18, 2018, 12:55:26 PM »
let w=1z then Z=1w
Z w0

f(z)=f(1w)=sin1wcos1w+1wcos21wsin21w

at w0, we have a non-isolated singularity

Victor Ivrii

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FE-P3 official
« Reply #7 on: December 20, 2018, 05:04:59 AM »
\newcommand{\Res}{\operatorname{Res}}
(a) Singular points are of g(z)=\tan(z) and h(z)=z\cot ^2(z), that is z_n= (n+\frac{1}{2})\pi and w_n = \pi n.

(b) z_n are simple poles and \Res (f,z_n)= \Res (\tan(z), z_n)= \frac{\sin(z)}{(\cos (z))'}\bigr|_{z=z_n} =-1.

w_0 is a simple pole
\Res (f,w_0)= \Res (z\cot^2(z), w_0)= \Res (z\cot (z) \times \cot(z), 0)= \Res (\cot(z),0)= 1
because \lim _{z\to 0} z\cot(z)=1.

w_n with n\ne 0 are double poles and
\begin{align*} \Res (f,w_n)=&\Res (z\cot^2 (z), w_n)= \Res ((\pi n +w)\cot^2 (w) , 0) =\\ &\pi n\Res (\cot^2 (w) , 0) +\Res (w\cot^2(w),0) =1 \end{align*}
because \Res (\cot^2 (w) , 0)=0 (since \cot^2(w) is an even function) and \Res (w\cot^2(w),0) =1 we already calculated.

\infty is a not isolated singularity and therefore residue here is not defined.