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I would like to find the residue of $$f(z)=\frac{e^{iz}}{z\,(z^2+1)^2}$$ at $z=i$. One way to do it is simply to take the derivative of $\frac{e^{iz}}{z\,(z^2+1)^2}$. Another is to find the Laurent expansion of the function.

I managed to do it using the first way, and the answer is $-3/(4e)$. However, I'm out of ideas as to how to find the expansion.

Any help is greatly appreciated.

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3 Answers

$$e^{iz} = e^{i(z-i) + i^2} = \dfrac{e^{i(z-i)}}{e} = \dfrac1e \sum_{k=0}^{\infty} \dfrac{i^k(z-i)^k}{k!}$$ $$\dfrac1z = \dfrac1{z-i+i} = \dfrac1i \dfrac1{1 + \dfrac{z-i}i} = \dfrac1i \sum_{k=0}^{\infty} (-1)^k\left(\dfrac{z-i}i \right)^k = \sum_{k=0}^{\infty} i^{k-1} (z-i)^k$$ $$\dfrac1{(z+i)^2} = \dfrac1{(z-i+2i)^2} = -\dfrac14 \dfrac1{\left(1 + \dfrac{z-i}{2i} \right)^2} = -\dfrac14 \sum_{k=0}^{\infty} (-1)^k (k+1) \left(\dfrac{z-i}{2i}\right)^k\\ = -\sum_{k=0}^{\infty} \dfrac{i^k (k+1)}{2^{k+2}} \left(z-i\right)^k$$ Hence, we have $$\dfrac1{(z-i)^2} \left(\dfrac1e \sum_{k=0}^{\infty} \dfrac{i^k(z-i)^k}{k!}\right) \left(\sum_{k=0}^{\infty} i^{k-1} (z-i)^k \right) \left(-\sum_{k=0}^{\infty} \dfrac{i^k (k+1)}{2^{k+2}} \left(z-i\right)^k\right)$$ Hence, the coefficient of $\dfrac1{z-i}$ is nothing but the coefficient of $(z-i)$ in $$\left(\dfrac1e \sum_{k=0}^{\infty} \dfrac{i^k(z-i)^k}{k!}\right) \left(\sum_{k=0}^{\infty} i^{k-1} (z-i)^k \right) \left(-\sum_{k=0}^{\infty} \dfrac{i^k (k+1)}{2^{k+2}} \left(z-i\right)^k\right)$$ Hence, the answer is $$- \dfrac1e \times \left(\dfrac{i}{1!} \times \dfrac1i \times \dfrac14 + 1 \times 1 \times \dfrac14 + 1 \times \dfrac1i \times \dfrac{2i}{8} \right) = -\dfrac3{4e}$$

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Use the formula for the first term of the Laurent series:$$a_{-1}=\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}\left((z-z_0)^mf(z)\right)$$ Where m is the order of the pole. You can get to this formula by taking the Taylor expansion of the function $f(z)(z-z_0)^m$ since it is holomorphic, and making the Laurent expansion without finding the coefficients.

In your problem the pole has order two so the formula is:$$a_{-1}=\frac{d}{dz}\left(\frac{e^{iz}}{z(z+i)^2}\right)=-\frac{3}{4e}$$

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Substitute $z=w+i$ to get $$ \begin{align} \frac{e^{iz}}{z\,(z^2+1)^2} &=\frac{e^{iw-1}}{(w+i)\,(w^2+2iw)^2}\\ &=\frac{i}{4ew^2}\frac{e^{iw}}{(1-iw)(1-\frac{i}{2}w)^2}\\ &=\frac{i}{4ew^2}\frac{1+iw+\dots}{(1-iw)(1-iw+\dots)}\\ &=\frac{i}{4ew^2}(1+3iw+\dots) \end{align} $$ Thus, the residue is the coefficient of $\dfrac1w$ which is $\ -\dfrac3{4e}$.

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