Find complex residue and Laurent series expansion












0












$begingroup$


$$ w = sin(z) * sin(frac{1}{:z}) $$



special point is $$ z_0 = 0 $$
$$ lim _{zto 0}left(sinleft(zright)cdot :sin:left(frac{1}{z}right)right) $$ isn't exist, next I have to decompose both functions in Laurent series.
$$sin z=z-frac{z^3}{3!}+frac{z^5text{}}{5!}-... $$
$$ sinfrac{1}{z}=frac{1}{z}-frac{1}{z^33!}+frac{1}{z^55!} $$
after, multiplication series I have to find $$ c^{-1} $$ addend in Laurent series, and coefficients with it will be residue?
but
$$ 1+frac{1}{3!cdot 3!}+frac{1}{5!cdot 5!} $$






complex-analysis taylor-expansion residue-calculus laurent-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    $LaTeX$ advice: use sin for sine function.
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:22










  • $begingroup$
    thanks for hint
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 15:41
















0












$begingroup$


$$ w = sin(z) * sin(frac{1}{:z}) $$



special point is $$ z_0 = 0 $$
$$ lim _{zto 0}left(sinleft(zright)cdot :sin:left(frac{1}{z}right)right) $$ isn't exist, next I have to decompose both functions in Laurent series.
$$sin z=z-frac{z^3}{3!}+frac{z^5text{}}{5!}-... $$
$$ sinfrac{1}{z}=frac{1}{z}-frac{1}{z^33!}+frac{1}{z^55!} $$
after, multiplication series I have to find $$ c^{-1} $$ addend in Laurent series, and coefficients with it will be residue?
but
$$ 1+frac{1}{3!cdot 3!}+frac{1}{5!cdot 5!} $$






complex-analysis taylor-expansion residue-calculus laurent-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    $LaTeX$ advice: use sin for sine function.
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:22










  • $begingroup$
    thanks for hint
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 15:41














0












0








0





$begingroup$


$$ w = sin(z) * sin(frac{1}{:z}) $$



special point is $$ z_0 = 0 $$
$$ lim _{zto 0}left(sinleft(zright)cdot :sin:left(frac{1}{z}right)right) $$ isn't exist, next I have to decompose both functions in Laurent series.
$$sin z=z-frac{z^3}{3!}+frac{z^5text{}}{5!}-... $$
$$ sinfrac{1}{z}=frac{1}{z}-frac{1}{z^33!}+frac{1}{z^55!} $$
after, multiplication series I have to find $$ c^{-1} $$ addend in Laurent series, and coefficients with it will be residue?
but
$$ 1+frac{1}{3!cdot 3!}+frac{1}{5!cdot 5!} $$






complex-analysis taylor-expansion residue-calculus laurent-series







share|cite|improve this question











$endgroup$




$$ w = sin(z) * sin(frac{1}{:z}) $$



special point is $$ z_0 = 0 $$
$$ lim _{zto 0}left(sinleft(zright)cdot :sin:left(frac{1}{z}right)right) $$ isn't exist, next I have to decompose both functions in Laurent series.
$$sin z=z-frac{z^3}{3!}+frac{z^5text{}}{5!}-... $$
$$ sinfrac{1}{z}=frac{1}{z}-frac{1}{z^33!}+frac{1}{z^55!} $$
after, multiplication series I have to find $$ c^{-1} $$ addend in Laurent series, and coefficients with it will be residue?
but
$$ 1+frac{1}{3!cdot 3!}+frac{1}{5!cdot 5!} $$






complex-analysis taylor-expansion residue-calculus laurent-series




complex-analysis taylor-expansion residue-calculus laurent-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 15:21









Chinnapparaj R

5,5072928




5,5072928










asked Dec 17 '18 at 14:48









Dmitry SokolovDmitry Sokolov

526




526












  • $begingroup$
    $LaTeX$ advice: use sin for sine function.
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:22










  • $begingroup$
    thanks for hint
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 15:41


















  • $begingroup$
    $LaTeX$ advice: use sin for sine function.
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:22










  • $begingroup$
    thanks for hint
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 15:41
















$begingroup$
$LaTeX$ advice: use sin for sine function.
$endgroup$
– Chinnapparaj R
Dec 17 '18 at 15:22




$begingroup$
$LaTeX$ advice: use sin for sine function.
$endgroup$
– Chinnapparaj R
Dec 17 '18 at 15:22












$begingroup$
thanks for hint
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 15:41




$begingroup$
thanks for hint
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 15:41










1 Answer
1






active

oldest

votes


















4












$begingroup$

The residue is $0$ since two odd terms $z^m$ and $z^{-k}$ will always make an even term when multiplied: $z^mz^{-k} = z^{m-k}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, Should I find residue at infinity?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:10










  • $begingroup$
    *Must I have to find?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:29










  • $begingroup$
    @DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 18:13












  • $begingroup$
    I mean:Is Infinity isolated singular point?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 18:22












  • $begingroup$
    @DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 21:04











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The residue is $0$ since two odd terms $z^m$ and $z^{-k}$ will always make an even term when multiplied: $z^mz^{-k} = z^{m-k}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, Should I find residue at infinity?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:10










  • $begingroup$
    *Must I have to find?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:29










  • $begingroup$
    @DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 18:13












  • $begingroup$
    I mean:Is Infinity isolated singular point?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 18:22












  • $begingroup$
    @DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 21:04
















4












$begingroup$

The residue is $0$ since two odd terms $z^m$ and $z^{-k}$ will always make an even term when multiplied: $z^mz^{-k} = z^{m-k}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, Should I find residue at infinity?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:10










  • $begingroup$
    *Must I have to find?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:29










  • $begingroup$
    @DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 18:13












  • $begingroup$
    I mean:Is Infinity isolated singular point?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 18:22












  • $begingroup$
    @DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 21:04














4












4








4





$begingroup$

The residue is $0$ since two odd terms $z^m$ and $z^{-k}$ will always make an even term when multiplied: $z^mz^{-k} = z^{m-k}$.






share|cite|improve this answer









$endgroup$



The residue is $0$ since two odd terms $z^m$ and $z^{-k}$ will always make an even term when multiplied: $z^mz^{-k} = z^{m-k}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 15:13









ploosu2ploosu2

4,6451024




4,6451024












  • $begingroup$
    Thanks, Should I find residue at infinity?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:10










  • $begingroup$
    *Must I have to find?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:29










  • $begingroup$
    @DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 18:13












  • $begingroup$
    I mean:Is Infinity isolated singular point?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 18:22












  • $begingroup$
    @DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 21:04


















  • $begingroup$
    Thanks, Should I find residue at infinity?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:10










  • $begingroup$
    *Must I have to find?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:29










  • $begingroup$
    @DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 18:13












  • $begingroup$
    I mean:Is Infinity isolated singular point?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 18:22












  • $begingroup$
    @DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 21:04
















$begingroup$
Thanks, Should I find residue at infinity?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 16:10




$begingroup$
Thanks, Should I find residue at infinity?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 16:10












$begingroup$
*Must I have to find?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 16:29




$begingroup$
*Must I have to find?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 16:29












$begingroup$
@DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
$endgroup$
– ploosu2
Dec 17 '18 at 18:13






$begingroup$
@DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
$endgroup$
– ploosu2
Dec 17 '18 at 18:13














$begingroup$
I mean:Is Infinity isolated singular point?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 18:22






$begingroup$
I mean:Is Infinity isolated singular point?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 18:22














$begingroup$
@DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
$endgroup$
– ploosu2
Dec 17 '18 at 21:04




$begingroup$
@DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
$endgroup$
– ploosu2
Dec 17 '18 at 21:04


















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