Laurent series for $\sqrt{z^{2}+a}$ I'm trying to figure out how to find the Laurent series for the complex function $z\mapsto\sqrt{z^{2}+a}$
  where a is just some constant. What I really need is to understand its behavior as $\left|z\right|$
  goes to infinity.
Specifically, wolphram-alpha gives me the expansion $z+(2t)/z-(2t^{2})/z^{3}+O(z{}^{-4})$, but how does one compute this?
 A: $\begin{array}\\
\sqrt{z^2+a}
&=z\sqrt{1+a/z^2}\\
&=z\sum_{n=0}^{\infty} \binom{1/2}{n}\left(\dfrac{a}{z^2}\right)^n\\
&=\sum_{n=0}^{\infty} \binom{1/2}{n}\dfrac{a^n}{z^{2n-1}}\\
&=z+\dfrac{a}{2z}-\dfrac{a^2}{8z^3}+\sum_{n=3}^{\infty} \binom{1/2}{n}\dfrac{a^n}{z^{2n-1}}\\
\end{array}
$
This disagrees with Wolfies answer,
which may reflect a
fundamental misunderstanding on my part.
A: To expand a function about $z=\infty$, we rewrite it as a function of $w=1/z$ and expand about $w=0$:
$$ \sqrt{z^2+a} = \sqrt{\frac{1}{w^2}+a}=\frac{1}{w} \sqrt{1+aw^2} $$
N.B.: I have chosen a branch by saying that $\sqrt{1/w^2}=1/w$; I could equally have chosen $-1/w$. I can now use the binomial theorem, if $aw^2<1$:
$$ \frac{1}{w} \sqrt{1+aw^2} = \frac{1}{w} \sum_{k=0}^{\infty} \binom{1/2}{n}(aw^2)^k = \frac{1}{w} + \frac{aw}{2} - \frac{a^2w^3}{8} + O(w^5). $$
Converting back to $z$ gives the series you want,
$$ \sqrt{z^2+a} = z + \frac{a}{2z} - \frac{a^2}{8z^3} + O(z^{-5}). $$
By the way, I'm not sure you gave W|A the right input: it gives me the answer I got.
