Find the residues at the singularities of $frac{z^2 - z}{1-sin{z}}$.












1














I've been set a question where I'm asked to classify all of the singularities of



$$f(z) = frac{z^2 - z}{1-sin{z}}$$



and then calculate the residue of each of its singularities. I've found the singularities, all of which are of the form $ z_k =(2k + frac{1}{2})pi$, where $k$ is an integer, and shown each of these to be double poles.



I have the formula for the residue of $f$ at $z_k$ as



$$limlimits_{z to z_k} (frac{d}{dz}((z-z_k)^2f(z))$$



(since $z_k$ is a pole of order 2) but it ends up being really complicated and I can't seem to make it work. Is there a better way to do it?






complex-analysis limits analysis derivatives residue-calculus







share|cite|improve this question
























  • The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
    – Alonso Delfín
    Nov 19 '18 at 2:01






  • 1




    @Alonso I've given it a go but I couldn't do it - would you be able to walk me through (at least the start)?
    – pixuj
    Nov 19 '18 at 12:15












  • I just posted an answer with some computations for the pole $pi/2$. I'm sure you'll be able to generalize to all the other poles. Let me know if it helps.
    – Alonso Delfín
    Nov 20 '18 at 21:41


















1














I've been set a question where I'm asked to classify all of the singularities of



$$f(z) = frac{z^2 - z}{1-sin{z}}$$



and then calculate the residue of each of its singularities. I've found the singularities, all of which are of the form $ z_k =(2k + frac{1}{2})pi$, where $k$ is an integer, and shown each of these to be double poles.



I have the formula for the residue of $f$ at $z_k$ as



$$limlimits_{z to z_k} (frac{d}{dz}((z-z_k)^2f(z))$$



(since $z_k$ is a pole of order 2) but it ends up being really complicated and I can't seem to make it work. Is there a better way to do it?






complex-analysis limits analysis derivatives residue-calculus







share|cite|improve this question
























  • The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
    – Alonso Delfín
    Nov 19 '18 at 2:01






  • 1




    @Alonso I've given it a go but I couldn't do it - would you be able to walk me through (at least the start)?
    – pixuj
    Nov 19 '18 at 12:15












  • I just posted an answer with some computations for the pole $pi/2$. I'm sure you'll be able to generalize to all the other poles. Let me know if it helps.
    – Alonso Delfín
    Nov 20 '18 at 21:41
















1












1








1


0





I've been set a question where I'm asked to classify all of the singularities of



$$f(z) = frac{z^2 - z}{1-sin{z}}$$



and then calculate the residue of each of its singularities. I've found the singularities, all of which are of the form $ z_k =(2k + frac{1}{2})pi$, where $k$ is an integer, and shown each of these to be double poles.



I have the formula for the residue of $f$ at $z_k$ as



$$limlimits_{z to z_k} (frac{d}{dz}((z-z_k)^2f(z))$$



(since $z_k$ is a pole of order 2) but it ends up being really complicated and I can't seem to make it work. Is there a better way to do it?






complex-analysis limits analysis derivatives residue-calculus







share|cite|improve this question















I've been set a question where I'm asked to classify all of the singularities of



$$f(z) = frac{z^2 - z}{1-sin{z}}$$



and then calculate the residue of each of its singularities. I've found the singularities, all of which are of the form $ z_k =(2k + frac{1}{2})pi$, where $k$ is an integer, and shown each of these to be double poles.



I have the formula for the residue of $f$ at $z_k$ as



$$limlimits_{z to z_k} (frac{d}{dz}((z-z_k)^2f(z))$$



(since $z_k$ is a pole of order 2) but it ends up being really complicated and I can't seem to make it work. Is there a better way to do it?






complex-analysis limits analysis derivatives residue-calculus




complex-analysis limits analysis derivatives residue-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 '18 at 9:31

























asked Nov 19 '18 at 0:07









pixuj

62




62












  • The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
    – Alonso Delfín
    Nov 19 '18 at 2:01






  • 1




    @Alonso I've given it a go but I couldn't do it - would you be able to walk me through (at least the start)?
    – pixuj
    Nov 19 '18 at 12:15












  • I just posted an answer with some computations for the pole $pi/2$. I'm sure you'll be able to generalize to all the other poles. Let me know if it helps.
    – Alonso Delfín
    Nov 20 '18 at 21:41




















  • The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
    – Alonso Delfín
    Nov 19 '18 at 2:01






  • 1




    @Alonso I've given it a go but I couldn't do it - would you be able to walk me through (at least the start)?
    – pixuj
    Nov 19 '18 at 12:15












  • I just posted an answer with some computations for the pole $pi/2$. I'm sure you'll be able to generalize to all the other poles. Let me know if it helps.
    – Alonso Delfín
    Nov 20 '18 at 21:41


















The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
– Alonso Delfín
Nov 19 '18 at 2:01




The formula you have only works for poles of order 2. It might be a better idea to find the coefficient of $z^{-1}$ on the Laurent series of $f$ around $z_k$. You can do it by finding the Taylor series of $z^2 -z$ and $1-sin(z)$ around $z_k$ and then chasing the wanted coefficient by performing “long division”.
– Alonso Delfín
Nov 19 '18 at 2:01




1




1




@Alonso I've given it a go but I couldn't do it - would you be able to walk me through (at least the start)?
– pixuj
Nov 19 '18 at 12:15






@Alonso I've given it a go but I couldn't do it - would you be able to walk me through (at least the start)?
– pixuj
Nov 19 '18 at 12:15














I just posted an answer with some computations for the pole $pi/2$. I'm sure you'll be able to generalize to all the other poles. Let me know if it helps.
– Alonso Delfín
Nov 20 '18 at 21:41






I just posted an answer with some computations for the pole $pi/2$. I'm sure you'll be able to generalize to all the other poles. Let me know if it helps.
– Alonso Delfín
Nov 20 '18 at 21:41












1 Answer
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I'm going to write the computation for the pole $z_0:=frac{pi}{2}$. You should be able to generalize it for all the other poles. First of all we need to check that indeed $f$ has a pole of order $2$ around $z_0$. To do this let $g(z):=z^2-z$ and $h(z):=1-sin(z)$. Then, notice that near $z_0$ one has
$$
f(z)=frac{g(z)}{frac{h''(z_0)}{2!}(z-z_0)^2+frac{h^{(4)}(z_0)}{4!}(z-z_0)^4 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^6 + cdots}
$$
Thus,
$$
(z-z_0)^2 f(z)= frac{g(z)}{frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots},
$$
and therefore
$$
lim_{z to z_0} (z-z_0)^2 f(z) = 2frac{g(z_0)}{h''(z_0)}=frac{pi^2-2pi}{2}
$$
This shows that $f$ has indeed a pole of order $2$ around $z_0.$ Now if we let
$$
k(z):=frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots,
$$
we have that $(z-z_0)^2 f(z)=frac{g(z)}{k(z)}$, and therefore
$$
frac{d}{dz}(z-z_0)^2 f(z) = frac{g'(z)k(z)-g(z)k'(z)}{(k(z))^2}
$$
Note that $k(z_0)=frac{h''(z_0)}{2!}=frac{1}{2}$, $k'(z_0)=0$ and that $g'(z_0)=pi-1$. Hence, the residue at $z_0$ is
$$
lim_{z to z_0} left( frac{d}{dz}(z-z_0)^2 f(z) right) = frac{g'(z_0)frac{h''(z_0)}{2!}}{left( frac{h''(z_0)}{2!} right)^2}=frac{frac{pi-1}{2}}{left( frac{1}{2} right)^2} = 2(pi-1)
$$






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    I'm going to write the computation for the pole $z_0:=frac{pi}{2}$. You should be able to generalize it for all the other poles. First of all we need to check that indeed $f$ has a pole of order $2$ around $z_0$. To do this let $g(z):=z^2-z$ and $h(z):=1-sin(z)$. Then, notice that near $z_0$ one has
    $$
    f(z)=frac{g(z)}{frac{h''(z_0)}{2!}(z-z_0)^2+frac{h^{(4)}(z_0)}{4!}(z-z_0)^4 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^6 + cdots}
    $$
    Thus,
    $$
    (z-z_0)^2 f(z)= frac{g(z)}{frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots},
    $$
    and therefore
    $$
    lim_{z to z_0} (z-z_0)^2 f(z) = 2frac{g(z_0)}{h''(z_0)}=frac{pi^2-2pi}{2}
    $$
    This shows that $f$ has indeed a pole of order $2$ around $z_0.$ Now if we let
    $$
    k(z):=frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots,
    $$
    we have that $(z-z_0)^2 f(z)=frac{g(z)}{k(z)}$, and therefore
    $$
    frac{d}{dz}(z-z_0)^2 f(z) = frac{g'(z)k(z)-g(z)k'(z)}{(k(z))^2}
    $$
    Note that $k(z_0)=frac{h''(z_0)}{2!}=frac{1}{2}$, $k'(z_0)=0$ and that $g'(z_0)=pi-1$. Hence, the residue at $z_0$ is
    $$
    lim_{z to z_0} left( frac{d}{dz}(z-z_0)^2 f(z) right) = frac{g'(z_0)frac{h''(z_0)}{2!}}{left( frac{h''(z_0)}{2!} right)^2}=frac{frac{pi-1}{2}}{left( frac{1}{2} right)^2} = 2(pi-1)
    $$






    share|cite|improve this answer


























      0














      I'm going to write the computation for the pole $z_0:=frac{pi}{2}$. You should be able to generalize it for all the other poles. First of all we need to check that indeed $f$ has a pole of order $2$ around $z_0$. To do this let $g(z):=z^2-z$ and $h(z):=1-sin(z)$. Then, notice that near $z_0$ one has
      $$
      f(z)=frac{g(z)}{frac{h''(z_0)}{2!}(z-z_0)^2+frac{h^{(4)}(z_0)}{4!}(z-z_0)^4 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^6 + cdots}
      $$
      Thus,
      $$
      (z-z_0)^2 f(z)= frac{g(z)}{frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots},
      $$
      and therefore
      $$
      lim_{z to z_0} (z-z_0)^2 f(z) = 2frac{g(z_0)}{h''(z_0)}=frac{pi^2-2pi}{2}
      $$
      This shows that $f$ has indeed a pole of order $2$ around $z_0.$ Now if we let
      $$
      k(z):=frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots,
      $$
      we have that $(z-z_0)^2 f(z)=frac{g(z)}{k(z)}$, and therefore
      $$
      frac{d}{dz}(z-z_0)^2 f(z) = frac{g'(z)k(z)-g(z)k'(z)}{(k(z))^2}
      $$
      Note that $k(z_0)=frac{h''(z_0)}{2!}=frac{1}{2}$, $k'(z_0)=0$ and that $g'(z_0)=pi-1$. Hence, the residue at $z_0$ is
      $$
      lim_{z to z_0} left( frac{d}{dz}(z-z_0)^2 f(z) right) = frac{g'(z_0)frac{h''(z_0)}{2!}}{left( frac{h''(z_0)}{2!} right)^2}=frac{frac{pi-1}{2}}{left( frac{1}{2} right)^2} = 2(pi-1)
      $$






      share|cite|improve this answer
























        0












        0








        0






        I'm going to write the computation for the pole $z_0:=frac{pi}{2}$. You should be able to generalize it for all the other poles. First of all we need to check that indeed $f$ has a pole of order $2$ around $z_0$. To do this let $g(z):=z^2-z$ and $h(z):=1-sin(z)$. Then, notice that near $z_0$ one has
        $$
        f(z)=frac{g(z)}{frac{h''(z_0)}{2!}(z-z_0)^2+frac{h^{(4)}(z_0)}{4!}(z-z_0)^4 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^6 + cdots}
        $$
        Thus,
        $$
        (z-z_0)^2 f(z)= frac{g(z)}{frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots},
        $$
        and therefore
        $$
        lim_{z to z_0} (z-z_0)^2 f(z) = 2frac{g(z_0)}{h''(z_0)}=frac{pi^2-2pi}{2}
        $$
        This shows that $f$ has indeed a pole of order $2$ around $z_0.$ Now if we let
        $$
        k(z):=frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots,
        $$
        we have that $(z-z_0)^2 f(z)=frac{g(z)}{k(z)}$, and therefore
        $$
        frac{d}{dz}(z-z_0)^2 f(z) = frac{g'(z)k(z)-g(z)k'(z)}{(k(z))^2}
        $$
        Note that $k(z_0)=frac{h''(z_0)}{2!}=frac{1}{2}$, $k'(z_0)=0$ and that $g'(z_0)=pi-1$. Hence, the residue at $z_0$ is
        $$
        lim_{z to z_0} left( frac{d}{dz}(z-z_0)^2 f(z) right) = frac{g'(z_0)frac{h''(z_0)}{2!}}{left( frac{h''(z_0)}{2!} right)^2}=frac{frac{pi-1}{2}}{left( frac{1}{2} right)^2} = 2(pi-1)
        $$






        share|cite|improve this answer












        I'm going to write the computation for the pole $z_0:=frac{pi}{2}$. You should be able to generalize it for all the other poles. First of all we need to check that indeed $f$ has a pole of order $2$ around $z_0$. To do this let $g(z):=z^2-z$ and $h(z):=1-sin(z)$. Then, notice that near $z_0$ one has
        $$
        f(z)=frac{g(z)}{frac{h''(z_0)}{2!}(z-z_0)^2+frac{h^{(4)}(z_0)}{4!}(z-z_0)^4 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^6 + cdots}
        $$
        Thus,
        $$
        (z-z_0)^2 f(z)= frac{g(z)}{frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots},
        $$
        and therefore
        $$
        lim_{z to z_0} (z-z_0)^2 f(z) = 2frac{g(z_0)}{h''(z_0)}=frac{pi^2-2pi}{2}
        $$
        This shows that $f$ has indeed a pole of order $2$ around $z_0.$ Now if we let
        $$
        k(z):=frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots,
        $$
        we have that $(z-z_0)^2 f(z)=frac{g(z)}{k(z)}$, and therefore
        $$
        frac{d}{dz}(z-z_0)^2 f(z) = frac{g'(z)k(z)-g(z)k'(z)}{(k(z))^2}
        $$
        Note that $k(z_0)=frac{h''(z_0)}{2!}=frac{1}{2}$, $k'(z_0)=0$ and that $g'(z_0)=pi-1$. Hence, the residue at $z_0$ is
        $$
        lim_{z to z_0} left( frac{d}{dz}(z-z_0)^2 f(z) right) = frac{g'(z_0)frac{h''(z_0)}{2!}}{left( frac{h''(z_0)}{2!} right)^2}=frac{frac{pi-1}{2}}{left( frac{1}{2} right)^2} = 2(pi-1)
        $$







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        answered Nov 20 '18 at 21:40









        Alonso Delfín

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