Author Topic: FE-P3  (Read 644 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
$$
f(z)= \tan (z) + z\cot^2(z);
$$
infinity included.

Hanyu Zhou

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Re: FE-P3
« Reply #1 on: December 18, 2018, 11:13:44 AM »
f(z) = $\frac{\mathrm{sin}\mathrm{}(z)}{\mathrm{cos}\mathrm{}(z)}+z\frac{\mathrm{cos}\mathrm{}\wedge 2(z)}{\mathrm{sin}\mathrm{}\wedge 2(z)}$
      =$ \frac{{{\mathrm{sin}}^{\mathrm{3}}\left(z\right) +\ }{\mathrm{zcos}}^{\mathrm{3}}\left(z\right)\ \ \ \ =g}{{\mathrm{cos} \left(z\right)\ }{sin}^2\left(z\right)\ \ \ \ =h}$
     

Cos(z)sin$\mathrm{\wedge}$2(z) = 0

cos(z) = 0 or sin$\mathrm{\wedge}$2(z) = 0

so z =k$\pi $ or $z=\frac{\pi }{2}+k\pi $
 
1, when z = k$\pi $

g = ${sin}^3(z)+z{cos}^3\left(z\right)\neq 0$                            h= ${\mathrm{cos} \left(z\right)\ }{sin}^2\left(z\right)=0$
                                                                     $h^`=-{sin}^3\left(z\right)+2{cos}^2\left(z\right){\mathrm{sin} \left(z\right)\ }=0$                    ${\ h}^{''}\neq 0$


So pole of order = 2

 2, when z =$\frac{\pi }{2}+k\pi $               

g = ${sin}^3(z)+z{cos}^3\left(z\right)\neq 0$                            h= ${\mathrm{cos} \left(z\right)\ }{sin}^2\left(z\right)=0$
                                                                    $h^`=-{sin}^3\left(z\right)+2{cos}^2\left(z\right){\mathrm{sin} \left(z\right)\ }\ \neq 0$
                               
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)

${\rm f}\left({\rm z}\right){\rm =}{\tan  \left({\rm z}\right)\ }{\rm +z}{{\cot }^{{\rm 2}} \left(z\right)=\frac{{\sin  \left(z\right)\ }}{{\cos  \left(z\right)\ }}+\frac{z{{\cos }^2 \left(z\right)\ }}{{{\sin }^2 \left(z\right)\ }}\ }=\frac{{{\sin }^3 (z)\ }+z{{\cos }^3 (z)\ }}{{\cos  \left(z\right)\ }{{\sin }^2 \left(z\right)\ }}$
If ${\cos  \left(z\right)\ }=0$, then ${\rm z=}\frac{\pi}{2} + k\pi$

For numerator, ${{\sin }^3 (\frac{\pi}{2})\ }+\frac{\pi}{2}{{\cos }^3 \left(\frac{\pi}{2}\right)\ne 0\ }$, the numerator has zero order of zeros

For denominator, ${\cos  \left(\frac{\pi}{2}\right)\ }{{\sin }^2 \left(\frac{\pi}{2}\right)=0\ ,-{{\sin }^3 \left(\frac{\pi}{2}\right)\ }+\frac{\pi}{2}{{\cos }^2 \left(\frac{\pi}{2}\right)\ }{\sin  \left(\frac{\pi}{2}\right)\ }\ne 0\ }$, the denominator has zeros of order 1

Then ${\rm f}\left({\rm z}\right)$ has simple pole ${\rm z=}\frac{\pi}{2}+k\pi$

If ${{\sin }^{{\rm 2}} \left(z\right)\ }=0$, then ${\rm z=0}$ or ${\rm z=}k\pi $

When ${\rm z=0}$,

For numerator, ${{\sin }^3 (0)\ }+0{{\cos }^3 \left(0\right)=0\ },3{{\sin }^2 \left(0\right)+{{\cos }^3 \left(0\right)\ }+3{{\cos }^2 (0)\ }{\rm sin}?(0)\ }\ \ne 0\ $, the numerator has zero order of 1

For denominator, ${\cos  \left(0\right)\ }{{\sin }^2 \left(0\right)=0\ ,-{{\sin }^3 \left(0\right)\ }+2{{\cos }^2 \left(0\right)\ }{\sin  \left(0\right)\ }=0\ },\ -3{{\sin }^2 \left(0\right){\cos  \left(0\right)\ }-4{\cos  \left(0\right)\ }{{\sin }^{{\rm 2}} \left(0\right)\ }+2cos^3\left(0\right)\ne 0\ }$, the denominator has zeros of order 2

Then ${\rm f}\left({\rm z}\right)$ has simple pole ${\rm z=0}$

When ${\rm z=}k\pi $,

For numerator, ${{\sin }^3 (\pi )\ }+0{{\cos }^3 \left(\pi \right)\ne 0\ }$ the numerator has zero order of 0

For denominator, ${\cos  \left(\pi \right)\ }{{\sin }^2 \left(\pi \right)=0\ ,-{{\sin }^3 \left(\pi \right)\ }+2{{\cos }^2 \left(\pi \right)\ }{\sin  \left(\pi \right)\ }=0\ },\ -3{{\sin }^2 \left(\pi \right){\cos  \left(\pi \right)\ }-4{\cos  \left(\pi \right)\ }{{\sin }^{{\rm 2}} \left(\pi \right)\ }+2cos^3\left(\pi \right)\ne 0\ }$, the denominator has zeros of order 2

Then ${\rm f}\left({\rm z}\right)$ has pole ${\rm z=}k\pi $ with order 2


${\rm Res}\left({\rm f}\left({\rm z}\right),\frac{\pi}{2}+k\pi\right)=\frac{{{\sin }^3 (\frac{\pi }{{\rm 2}})\ }+\frac{\pi}{2}{{\cos }^3 (\frac{\pi}{2})\ }}{-{{\sin }^3 \left(\frac{\pi }{{\rm 2}}\right)\ }+2{{\cos }^2 \left(\frac{\pi }{{\rm 2}}\right)\ }{\sin  \left(\frac{\pi }{{\rm 2}}\right)\ }}=\frac{1}{-1}=-1$
${\rm Res}\left({\rm f}\left({\rm z}\right),k\pi\right)=\frac{3{{\sin }^2 \left(\pi\right)+{{\cos }^3 \left(\pi\right)\ }+3{{\cos }^2 (\pi)\ }{\rm sin}(\pi)\ }}{1!}=\frac{-1}{1}=-1$
Since ${\rm f}\left({\rm z}\right)=\frac{{{\sin }^3 (z)\ }+z{{\cos }^3 (z)\ }}{{\cos  \left(z\right)\ }{{\sin }^2 \left(z\right)\ }}=\frac{{\left(\int^{\infty }_0{{\left(-1\right)}^n\frac{z^{2n+1}}{\left(2n+1\right)!}}\right)}^3+\int^{\infty }_0{{\left(-1\right)}^n\frac{z^{2n+1}}{\left(2n\right)!}}}{\int^{\infty }_0{{\left(-1\right)}^n\frac{z^{2n}}{\left(2n\right)!}}\ {\left(\int^{\infty }_0{{\left(-1\right)}^n\frac{z^{2n+1}}{\left(2n+1\right)!}}\right)}^2}$ has no term of ${\left({\rm z-0}\right)}^{{\rm -}{\rm 1}}$, then
${\rm Res}\left({\rm f}\left({\rm z}\right),0\right)=0$

Yifei Wang

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Re: FE-P3
« Reply #6 on: December 18, 2018, 12:55:26 PM »
let $w = \frac{1}{z}$ then $ Z = \frac{1}{w}$
${Z\to\infty} $ ${w\to0}$

$f(z) = f(\frac{1}{w}) = \frac{sin{\frac{1}{w}}}{cos\frac{1}{w}}+\frac{1}{w}\frac{cos^2{\frac{1}{w}}}{sin^2{\frac{1}{w}}}$

at ${w\to0}$, 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.