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I suppose that you meant :

$$ y(z)\,=\,\frac{2 \, \sqrt{\operatorname{arsinh}\left(\frac{z}{2 \, a}\right)}}{\sqrt{z^{2} + 2 \, a}}$$

What's the point of working the Taylor series when the analytic answer is available ? Run :

var("z, a")
y(z)=arcsinh(z)*sqrt(z/2*a)/sqrt((z^2+2*a))

then :

sage: Poles=[u.rhs() for u in y(z).denominator().solve(z)] ; Poles
[-sqrt(2)*sqrt(-a), sqrt(2)*sqrt(-a)]
sage: [maxima_calculus.residue(y(z), z, u)._sage_() for u in Poles]
[0, 0]

BTW,

sage: (sinh(z).exponentialize()==x).solve(z)
[z == log(x - sqrt(x^2 + 1)), z == log(x + sqrt(x^2 + 1))]

No need for Taylor series here either, but a need for branch choice...

HTH,

I suppose that you meant :

$$ y(z)\,=\,\frac{2 \, \sqrt{\operatorname{arsinh}\left(\frac{z}{2 \, a}\right)}}{\sqrt{z^{2} + 2 \, a}}$$

What's the point of working the Taylor series when the analytic answer is available ? Run :

var("z, a")
y(z)=arcsinh(z)*sqrt(z/2*a)/sqrt((z^2+2*a))
y(z)=sqrt(arcsinh(z/(2*a))/sqrt((z^2+2*a))

then :

sage: Poles=[u.rhs() for u in y(z).denominator().solve(z)] ; Poles
[-sqrt(2)*sqrt(-a), sqrt(2)*sqrt(-a)]
sage: [maxima_calculus.residue(y(z), z, u)._sage_() for u in Poles]
[0, 0]

BTW,

sage: (sinh(z).exponentialize()==x).solve(z)
[z == log(x - sqrt(x^2 + 1)), z == log(x + sqrt(x^2 + 1))]

No need for Taylor series here either, but a need for branch choice...

HTH,