Laurent serie of $ frac { cos z}{ sin z + sinh z - 2z}$












1












$begingroup$


I'm working on an example given in my book of complex analysis:



$$ frac { cos z}{ sin z + sinh z - 2z}$$



but I can't figure out how he finded the residue in 0.
The few steps he is showing make me think he has computed the first few terms of its Laurent serie (like here : Laurent series $frac{1}{sin(z)}$ around $z=0$).



Nevertheless, I can't find the same result. Am I doing something wrong ?
So can you help find the laurent serie, and do you have another method to find the residue ?




I found that



$$ frac 1 { sin z + sinh z - 2z} = frac{ 5! } {2} z^{-5} - frac{(5!)^2}{2 cdot 9!} z^{-1} + o(z^2) $$



The book gives :



$$ frac {cos z} { sin z + sinh z - 2z} =
frac{ 5! } {2} z^{-5} (1 - z^2 /2 + 125/3024 z^4 +...) $$



The residue is equal to $ frac {625 i pi} {126} $






residue-calculus complex-integration laurent-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Isn't the pole of order $3?$
    $endgroup$
    – saulspatz
    Jan 4 at 23:36










  • $begingroup$
    Nop the z^3 term beneath is canceled between the sinus and the hyperbolic sinus
    $endgroup$
    – Marine Galantin
    Jan 4 at 23:37










  • $begingroup$
    Yes, you're right. My bad.
    $endgroup$
    – saulspatz
    Jan 4 at 23:38










  • $begingroup$
    I got the same thing you did. What does the book say the residue is?
    $endgroup$
    – saulspatz
    Jan 4 at 23:45










  • $begingroup$
    I ll edit the post. At least it s a good sign that you got the same thing as I did, thank you for this good news :)
    $endgroup$
    – Marine Galantin
    Jan 4 at 23:47


















1












$begingroup$


I'm working on an example given in my book of complex analysis:



$$ frac { cos z}{ sin z + sinh z - 2z}$$



but I can't figure out how he finded the residue in 0.
The few steps he is showing make me think he has computed the first few terms of its Laurent serie (like here : Laurent series $frac{1}{sin(z)}$ around $z=0$).



Nevertheless, I can't find the same result. Am I doing something wrong ?
So can you help find the laurent serie, and do you have another method to find the residue ?




I found that



$$ frac 1 { sin z + sinh z - 2z} = frac{ 5! } {2} z^{-5} - frac{(5!)^2}{2 cdot 9!} z^{-1} + o(z^2) $$



The book gives :



$$ frac {cos z} { sin z + sinh z - 2z} =
frac{ 5! } {2} z^{-5} (1 - z^2 /2 + 125/3024 z^4 +...) $$



The residue is equal to $ frac {625 i pi} {126} $






residue-calculus complex-integration laurent-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Isn't the pole of order $3?$
    $endgroup$
    – saulspatz
    Jan 4 at 23:36










  • $begingroup$
    Nop the z^3 term beneath is canceled between the sinus and the hyperbolic sinus
    $endgroup$
    – Marine Galantin
    Jan 4 at 23:37










  • $begingroup$
    Yes, you're right. My bad.
    $endgroup$
    – saulspatz
    Jan 4 at 23:38










  • $begingroup$
    I got the same thing you did. What does the book say the residue is?
    $endgroup$
    – saulspatz
    Jan 4 at 23:45










  • $begingroup$
    I ll edit the post. At least it s a good sign that you got the same thing as I did, thank you for this good news :)
    $endgroup$
    – Marine Galantin
    Jan 4 at 23:47
















1












1








1





$begingroup$


I'm working on an example given in my book of complex analysis:



$$ frac { cos z}{ sin z + sinh z - 2z}$$



but I can't figure out how he finded the residue in 0.
The few steps he is showing make me think he has computed the first few terms of its Laurent serie (like here : Laurent series $frac{1}{sin(z)}$ around $z=0$).



Nevertheless, I can't find the same result. Am I doing something wrong ?
So can you help find the laurent serie, and do you have another method to find the residue ?




I found that



$$ frac 1 { sin z + sinh z - 2z} = frac{ 5! } {2} z^{-5} - frac{(5!)^2}{2 cdot 9!} z^{-1} + o(z^2) $$



The book gives :



$$ frac {cos z} { sin z + sinh z - 2z} =
frac{ 5! } {2} z^{-5} (1 - z^2 /2 + 125/3024 z^4 +...) $$



The residue is equal to $ frac {625 i pi} {126} $






residue-calculus complex-integration laurent-series







share|cite|improve this question











$endgroup$




I'm working on an example given in my book of complex analysis:



$$ frac { cos z}{ sin z + sinh z - 2z}$$



but I can't figure out how he finded the residue in 0.
The few steps he is showing make me think he has computed the first few terms of its Laurent serie (like here : Laurent series $frac{1}{sin(z)}$ around $z=0$).



Nevertheless, I can't find the same result. Am I doing something wrong ?
So can you help find the laurent serie, and do you have another method to find the residue ?




I found that



$$ frac 1 { sin z + sinh z - 2z} = frac{ 5! } {2} z^{-5} - frac{(5!)^2}{2 cdot 9!} z^{-1} + o(z^2) $$



The book gives :



$$ frac {cos z} { sin z + sinh z - 2z} =
frac{ 5! } {2} z^{-5} (1 - z^2 /2 + 125/3024 z^4 +...) $$



The residue is equal to $ frac {625 i pi} {126} $






residue-calculus complex-integration laurent-series




residue-calculus complex-integration laurent-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 23:52









José Carlos Santos

173k23133241




173k23133241










asked Jan 4 at 23:29









Marine GalantinMarine Galantin

955419




955419












  • $begingroup$
    Isn't the pole of order $3?$
    $endgroup$
    – saulspatz
    Jan 4 at 23:36










  • $begingroup$
    Nop the z^3 term beneath is canceled between the sinus and the hyperbolic sinus
    $endgroup$
    – Marine Galantin
    Jan 4 at 23:37










  • $begingroup$
    Yes, you're right. My bad.
    $endgroup$
    – saulspatz
    Jan 4 at 23:38










  • $begingroup$
    I got the same thing you did. What does the book say the residue is?
    $endgroup$
    – saulspatz
    Jan 4 at 23:45










  • $begingroup$
    I ll edit the post. At least it s a good sign that you got the same thing as I did, thank you for this good news :)
    $endgroup$
    – Marine Galantin
    Jan 4 at 23:47




















  • $begingroup$
    Isn't the pole of order $3?$
    $endgroup$
    – saulspatz
    Jan 4 at 23:36










  • $begingroup$
    Nop the z^3 term beneath is canceled between the sinus and the hyperbolic sinus
    $endgroup$
    – Marine Galantin
    Jan 4 at 23:37










  • $begingroup$
    Yes, you're right. My bad.
    $endgroup$
    – saulspatz
    Jan 4 at 23:38










  • $begingroup$
    I got the same thing you did. What does the book say the residue is?
    $endgroup$
    – saulspatz
    Jan 4 at 23:45










  • $begingroup$
    I ll edit the post. At least it s a good sign that you got the same thing as I did, thank you for this good news :)
    $endgroup$
    – Marine Galantin
    Jan 4 at 23:47


















$begingroup$
Isn't the pole of order $3?$
$endgroup$
– saulspatz
Jan 4 at 23:36




$begingroup$
Isn't the pole of order $3?$
$endgroup$
– saulspatz
Jan 4 at 23:36












$begingroup$
Nop the z^3 term beneath is canceled between the sinus and the hyperbolic sinus
$endgroup$
– Marine Galantin
Jan 4 at 23:37




$begingroup$
Nop the z^3 term beneath is canceled between the sinus and the hyperbolic sinus
$endgroup$
– Marine Galantin
Jan 4 at 23:37












$begingroup$
Yes, you're right. My bad.
$endgroup$
– saulspatz
Jan 4 at 23:38




$begingroup$
Yes, you're right. My bad.
$endgroup$
– saulspatz
Jan 4 at 23:38












$begingroup$
I got the same thing you did. What does the book say the residue is?
$endgroup$
– saulspatz
Jan 4 at 23:45




$begingroup$
I got the same thing you did. What does the book say the residue is?
$endgroup$
– saulspatz
Jan 4 at 23:45












$begingroup$
I ll edit the post. At least it s a good sign that you got the same thing as I did, thank you for this good news :)
$endgroup$
– Marine Galantin
Jan 4 at 23:47






$begingroup$
I ll edit the post. At least it s a good sign that you got the same thing as I did, thank you for this good news :)
$endgroup$
– Marine Galantin
Jan 4 at 23:47












1 Answer
1






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oldest

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3












$begingroup$

You have$$sin(z)+sinh(z)-2z=frac{z^5}{60}+frac{z^9}{181,440}+cdots=z^5left(frac1{60}+frac{z^4}{181,440}+cdotsright),$$where the $cdots$ only has terms whose exponent is a multiple of $4$. So, you should compute$$frac{cos z}{frac1{60}+frac{z^4}{181,440}+cdots}=a_0+a_2z^2+a_4z^4+cdots$$This means that you havebegin{align}1-frac{z^2}2+frac{z^4}{24}+cdots&=left(frac1{60}+frac{z^4}{181,440}+cdotsright)left(a_0+a_2z^2+a_4z^4+cdotsright)\&=frac{a_0}{60}+frac{a_2}{60}z^2+left(frac{a_4}{60}+frac{a_0}{181,440}right)z^4+cdotsend{align}So, you know that $a_0=60$ and now you get $a_4$ by solving the equation$$frac{a_4}{60}+frac{a_0}{181,440}=frac1{24}.$$So,$$frac{cos z}{sin(z)+sinh(z)-2z}=frac{a_0}{z^5}+frac{a_4}z+cdots$$and the residue that you're after is $a_4=dfrac{625}{252}$.






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    1 Answer
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    1 Answer
    1






    active

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    active

    oldest

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    $begingroup$

    You have$$sin(z)+sinh(z)-2z=frac{z^5}{60}+frac{z^9}{181,440}+cdots=z^5left(frac1{60}+frac{z^4}{181,440}+cdotsright),$$where the $cdots$ only has terms whose exponent is a multiple of $4$. So, you should compute$$frac{cos z}{frac1{60}+frac{z^4}{181,440}+cdots}=a_0+a_2z^2+a_4z^4+cdots$$This means that you havebegin{align}1-frac{z^2}2+frac{z^4}{24}+cdots&=left(frac1{60}+frac{z^4}{181,440}+cdotsright)left(a_0+a_2z^2+a_4z^4+cdotsright)\&=frac{a_0}{60}+frac{a_2}{60}z^2+left(frac{a_4}{60}+frac{a_0}{181,440}right)z^4+cdotsend{align}So, you know that $a_0=60$ and now you get $a_4$ by solving the equation$$frac{a_4}{60}+frac{a_0}{181,440}=frac1{24}.$$So,$$frac{cos z}{sin(z)+sinh(z)-2z}=frac{a_0}{z^5}+frac{a_4}z+cdots$$and the residue that you're after is $a_4=dfrac{625}{252}$.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      You have$$sin(z)+sinh(z)-2z=frac{z^5}{60}+frac{z^9}{181,440}+cdots=z^5left(frac1{60}+frac{z^4}{181,440}+cdotsright),$$where the $cdots$ only has terms whose exponent is a multiple of $4$. So, you should compute$$frac{cos z}{frac1{60}+frac{z^4}{181,440}+cdots}=a_0+a_2z^2+a_4z^4+cdots$$This means that you havebegin{align}1-frac{z^2}2+frac{z^4}{24}+cdots&=left(frac1{60}+frac{z^4}{181,440}+cdotsright)left(a_0+a_2z^2+a_4z^4+cdotsright)\&=frac{a_0}{60}+frac{a_2}{60}z^2+left(frac{a_4}{60}+frac{a_0}{181,440}right)z^4+cdotsend{align}So, you know that $a_0=60$ and now you get $a_4$ by solving the equation$$frac{a_4}{60}+frac{a_0}{181,440}=frac1{24}.$$So,$$frac{cos z}{sin(z)+sinh(z)-2z}=frac{a_0}{z^5}+frac{a_4}z+cdots$$and the residue that you're after is $a_4=dfrac{625}{252}$.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        You have$$sin(z)+sinh(z)-2z=frac{z^5}{60}+frac{z^9}{181,440}+cdots=z^5left(frac1{60}+frac{z^4}{181,440}+cdotsright),$$where the $cdots$ only has terms whose exponent is a multiple of $4$. So, you should compute$$frac{cos z}{frac1{60}+frac{z^4}{181,440}+cdots}=a_0+a_2z^2+a_4z^4+cdots$$This means that you havebegin{align}1-frac{z^2}2+frac{z^4}{24}+cdots&=left(frac1{60}+frac{z^4}{181,440}+cdotsright)left(a_0+a_2z^2+a_4z^4+cdotsright)\&=frac{a_0}{60}+frac{a_2}{60}z^2+left(frac{a_4}{60}+frac{a_0}{181,440}right)z^4+cdotsend{align}So, you know that $a_0=60$ and now you get $a_4$ by solving the equation$$frac{a_4}{60}+frac{a_0}{181,440}=frac1{24}.$$So,$$frac{cos z}{sin(z)+sinh(z)-2z}=frac{a_0}{z^5}+frac{a_4}z+cdots$$and the residue that you're after is $a_4=dfrac{625}{252}$.






        share|cite|improve this answer











        $endgroup$



        You have$$sin(z)+sinh(z)-2z=frac{z^5}{60}+frac{z^9}{181,440}+cdots=z^5left(frac1{60}+frac{z^4}{181,440}+cdotsright),$$where the $cdots$ only has terms whose exponent is a multiple of $4$. So, you should compute$$frac{cos z}{frac1{60}+frac{z^4}{181,440}+cdots}=a_0+a_2z^2+a_4z^4+cdots$$This means that you havebegin{align}1-frac{z^2}2+frac{z^4}{24}+cdots&=left(frac1{60}+frac{z^4}{181,440}+cdotsright)left(a_0+a_2z^2+a_4z^4+cdotsright)\&=frac{a_0}{60}+frac{a_2}{60}z^2+left(frac{a_4}{60}+frac{a_0}{181,440}right)z^4+cdotsend{align}So, you know that $a_0=60$ and now you get $a_4$ by solving the equation$$frac{a_4}{60}+frac{a_0}{181,440}=frac1{24}.$$So,$$frac{cos z}{sin(z)+sinh(z)-2z}=frac{a_0}{z^5}+frac{a_4}z+cdots$$and the residue that you're after is $a_4=dfrac{625}{252}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 23:59

























        answered Jan 4 at 23:51









        José Carlos SantosJosé Carlos Santos

        173k23133241




        173k23133241






























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