I need a little help with the following problem. I've tried many ways, but i didnt succeed. I think there needs to be a trick or something, some transformation. The task is to find the residue of the function at its singularity e.g. z=-3 \begin{equation} f(z)=\cos\left(\frac{z^2+4z-1}{z+3}\right) \end{equation}

I tried to write it as \begin{align} \cos\left(\frac{z^2+4z-1}{z+3}\right)=1-\frac{1}{2!}\left((z+1)-\frac{4}{z+3}\right)^2+\frac{1}{4!}\left((z+1)-\frac{4}{z+3}\right)^4-\frac{1}{6!}\left((z+1)-\frac{4}{z+3}\right)^6+\ldots \end{align} and collect the coefficients at $\frac{1}{z+3}$ using binomial expansion of the brackets, but it seems to be a dead end, because there is to much of them and well hidden. If somebody could give me a hint, that would be great. Thanks.


$$ \begin{align} &\cos\left(\frac{z^2+4z-1}{z+3}\right)\\ &=\cos\left((z+3)-2-\frac4{z+3}\right)\\ &=\cos\left((z+3)-\frac4{z+3}\right)\cos(2)+\sin\left((z+3)-\frac4{z+3}\right)\sin(2)\tag{1} \end{align} $$ Since $\cos\left((z+3)-\frac4{z+3}\right)$ is an even function of $z+3$, its residue at $z=-3$ is $0$.

The $(z+3)^{-1}$ term of $$ \frac{(-1)^n}{(2n+1)!}\left((z+3)-\frac4{z+3}\right)^{2n+1}\tag{2} $$ is $$ \begin{align} &\frac{(-1)^n}{(2n+1)!}\binom{2n+1}{n}(z+3)^n\left(-\frac4{z+3}\right)^{n+1}\\[6pt] &=\frac{-1}{(2n+1)!}\binom{2n+1}{n}\frac{4^{n+1}}{z+3}\tag{3} \end{align} $$ Summing and multiplying by $\sin(2)$, we get the residue to be $$ \begin{align} -\sin(2)\sum_{n=0}^\infty\frac{4^{n+1}}{n!(n+1)!} &=-2\sin(2)I_1(4)\\ &=-17.7485131\tag{4} \end{align} $$ where $I_n(z)$ is the modified Bessel function of the first kind.

| cite | improve this answer | |
  • $\begingroup$ Yeah, you were right. (+1) $\endgroup$ – Ron Gordon Apr 2 '15 at 11:59
  • $\begingroup$ Nice move at (1). Thank you very much for you help. $\endgroup$ – David Apr 2 '15 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.