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I am trying to find residues for all singularities of the function: $$f(z)= \frac{\tanh z}{z^2}$$ Here is what I did: $$f(z)= \frac{\sinh z}{z^2\cosh z}$$ when $$\cosh z=0$$ then $z_k =i( \frac{π}{2}+πk)$, for $k \in\mathbb{Z}.$

Let $p(z) = \dfrac{\sinh z}{z^2}$ and $q(z)=\cosh z$

Since $z_k$ is a simple pole, then:

$$\operatorname{Res}(f(z),z_k)= \frac{p(z_k)}{q'(z_k)}= \frac{\sinh z}{z_k^2\sinh z}=\frac{1}{z_k^2}= \frac{1}{(i( \frac{π}{2}+πk))^2}= \frac{-1}{( \frac{π}{2}+πk)^2} $$ Is my solution correct? What about the residue when $z=0$?

Michael Hardy
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Fabian
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  • I rejected an edit that I thought was trying to change $\cosh z$ to $\cos hz$, but now I'm not sure that's what it was doing. I presume that $\cosh z$ was intended, not $\cos hz$. The problem is now I can't do any of the obvious copy-editing that is needed unless someone else acts on this proposed edit. ${}\qquad{}$ – Michael Hardy Aug 10 '15 at 21:21
  • @MichaelHardy I also reject the same edit. Also before I did myself an edit since the OP wants to use the hyperbolic functions – Alonso Delfín Aug 10 '15 at 21:24

3 Answers3

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The residues at $z=i(2n+1)\pi/2$ are given by $-\frac{4}{(2n+1)^2\pi^2}$ as already evaluated in the posted question.

There is a pole of order $1$ at $z=0$ since

$$\frac{\sinh z}{z^2} =\frac1z+\sum_{n=1}^{\infty}\frac{z^{2n-1}}{(2n+1)!}$$

Therefore, the residue at $z=0$ of $f(z)=\frac{\tanh z}{z^2}$ is simply $1$ since $\cosh (0)=1$. To be explicit

$$\lim_{z\to 0} \left(z\dfrac{\tanh z}{z^2}\right)=\lim_{z\to 0} \left(\dfrac{\tanh z}{z}\right)=\lim_{z\to 0}\frac{\text{sech}^2 z}{1}=1$$

Mark Viola
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  • Thank you Dr. MV. How did we know that the order of the pole z=0 of f(z) is one just by expanding sinhz and dividing it by z^2? Why didn't we divide them by the expansion of coshz? And why did we compute cosh(0), is it just to ensure that cosh(0) doesn't equal zero? – Fabian Aug 11 '15 at 00:48
  • @ Dr. MV thank you for being kind. If cosh(0) were equal to say "c", would the residue be 1/c ? If the problem were $$cotanhz/z^2$$ would this way lead me to calculate the residue at z = 0? – Fabian Aug 11 '15 at 01:47
  • @Fabian You're quite welcome. My pleasure. To answer your first question, "Yes." For the second question, to evaluate the residue of $g(z) = \frac{coth z}{z^2}$ at $z=0$, we note that $\cosh (z)=1+\frac12z^2+O(z^4)$ and $z^2\sinh z=z^3(1+\frac16 z^2+O(z^4))$. Therefore, the residue of $g$ at $z=0$ is $\frac12-\frac16=\frac13$, originating from the coefficent of the quadratic term of $z^3g(z)$, divided by $z^3$. – Mark Viola Aug 11 '15 at 02:02
  • thank you so much, that is really helpful. – Fabian Aug 11 '15 at 02:22
  • @ Dr. MV Your answer is perfect, but I am not familiar with residues, that's why it wasn't clear for me at first. It is me who need to be improved :). Thanks a lot. – Fabian Aug 11 '15 at 02:32
  • ٍSure, you deserve 1000 up votes, I tried but it did not count. They say "I need to earn a total of 15 reputation" . I am sorry :( – Fabian Aug 11 '15 at 02:45
  • @ Dr. MV I did ^_^. Thanks. – Fabian Aug 11 '15 at 03:30
  • You're welcome and thank you!! Let me know if I can help with other problems if you like. – Mark Viola Aug 11 '15 at 03:31
  • @ Dr. MV Yes please, would like you to check my question http://math.stackexchange.com/questions/1392667/a-question-about-poles , This problem also is important for me if you could please look at it and I am going to vote for your answers http://math.stackexchange.com/questions/1391435/a-tough-problem-about-residue – Fabian Aug 11 '15 at 03:34
  • @Fabian For the other problem, what else is given about $f$? – Mark Viola Aug 11 '15 at 03:40
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You can get the residue $$\mathrm{Res}\left[f(z),0\right]$$ by Laurent expansion which is equal to the coefficient of $$\frac{1}{z}.$$ And this is also same as the coefficient of $z$ in the expansion of $$\tanh(z)=z-\frac{1}{3}z^3+\frac{2}{15}z^5+\cdots$$.

so, $$\mathrm{Res}_{z=0} f(z)=1$$

Jungmin Kim
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You've already found out that the singularities of $f$ are: $$ 0, z_k=i\left(\frac\pi2+k\pi\right) \quad k\in \mathbb{Z}, $$ therefore: \begin{eqnarray} \mathrm{Res}(f,0)&=&\lim_{z\to0}(z-0)f(z)]=\lim_{z\to0}\frac{\tanh z}{z}=\lim_{z\to0}\frac{(\tanh z)'}{(z)'}=\lim_{z\to0}\frac{1-\tanh^2z}{1}=1\\ \mathrm{Res}(f,z_k)&=&\lim_{z\to z_k}\frac{\sinh(z)/z^2}{(\cosh z)'}=\lim_{z\to z_k}\frac{\sinh z}{z^2\sinh z}=\lim_{z\to z_k}\frac{1}{z^2}=\frac{1}{z_k^2}=-\frac{2}{(2k+1)\pi}. \end{eqnarray}

HorizonsMaths
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  • The reside at $z=0$ is a simple pole (i.e., order $1$). You are applying a formula as if it were a pole of order $2$. The answer is correct, but the means is not. – Mark Viola Aug 10 '15 at 22:34