Calculating residue of $z\sin{ \frac {z+1}{z-1}}$

Question

Let $f=z\sin{ \frac {z+1}{z-1} }$. Calculate the residue of $f$ in $z=1$. I think $f$ has an essential singularity at $z=1$ so the only way I can proceed is with Laurent series. I've defined $g(z)=f(z+1)=(z+1)\sin{z+2\over z}$ And tried to use the laurent series of $g$ around $z=0$ but got nowhere. I would like a hint.

Answer 1

Some ideas:

$$z\sin\frac{z+1}{z-1}=(z-1)\sin\left(1+\frac2{z-1}\right)+\sin\left(1+\frac2{z-1}\right)=$$

$$=(z-1)\left(1+\frac2{z-1}-\frac16\left(1+\frac2{z-1}\right)^3+\frac1{120}\left(1+\frac2{z-1}\right)^5-\ldots\right)+$$

$$+\left(1+\frac2{z-1}-\frac16\left(1+\frac2{z-1}\right)^3+\frac1{120}\left(1+\frac2{z-1}\right)^5-\ldots\right)=$$

Taking the terms in the inner parentheses of the first big parentheses above with quadratic exponent (in order to get

$$(z-1)\left(k_1\frac4{(z-1)^2}+k_2\frac4{(z-1)^2}+\ldots\right)$$

which will give the coefficient of $\;\frac1{z-1}\;$) and the linear coefficients in the second big parentheses, we get

$$2^2\left(-\frac16\cdot3+\frac1{120}\cdot 10-\frac1{7!}\cdot 21+\ldots\right)+2\left(1-\frac16\cdot 3+\frac1{120}\cdot 5-\frac1{7!}\cdot 7+\ldots\right)=$$

$$2^2\left(-\frac16\cdot3+\frac1{120}\cdot 10-\frac1{7!}\cdot 21+\ldots\right)+2\left(1-\frac12+\frac14-\frac16+\ldots\right)$$

In both parentheses we get a convergent series (in the second one a rather nice and reasonably well-known series...)

Answer 2

As for a hint:

$$(z+1)\sin \frac{z+2}{z} = (z+1) \sin \left(1 + \frac{2}{z}\right) = (z+1)\sin 1 \cos \frac{2}{z} + (z+1) \cos 1 \sin \frac{2}{z}.$$

Perhaps not the most efficient, but simple and easy to do right.