# How do I find the Laurent series expansion?

I want to find the order of the pole of $$\sqrt{1+\frac{1}{z}}$$ at 0. But how can I expand this function at 0? Can I take the Taylor expansion of $$\sqrt{1+z}$$ and plug in $$\frac{1}{z}$$ in the place of z? And how can one find the Laurent series expansion of a function in general? I mean since the function diverges toward infinity at poles, how can the Laurent series be "defined"?

• The function $\sqrt{1+1/z}$ has branch points at $z=-1$ and $z=0$. So, you can develop the Laurent series outside the unit disk. Jan 17 at 16:28

You're quite right that to find the Laurent series expansion for $$f(z)=\sqrt{1+\frac1z}$$ about $$0,$$ we need only start with the Taylor series expansion of $$g(z)=\sqrt{1+z}$$ about $$0,$$ then note that $$f(z)=g(1/z).$$ In fact, this will show you that it isn't a pole, but rather an essential singularity.
As for how the Laurent series can be thought of as being defined, recall that a power series is defined wherever it converges. In this case, it converges at all values of $$z$$ such that $$|z|>1.$$
• The Taylor series of $g$ around $z = 0$ converges for $|z| < 1$, so the corresponding Laurent series of $f$ converges for $|z| > 1$, not for all non-zero $z$. $f$ does not have a Laurent series expansion on $0 < |z| < r$, $z = 0$ is not an essential singularity of $f$. Jan 17 at 20:38
• To be clear, the new center of expansion $z = \infty$ is not an essential singularity of $f$ either. Jan 17 at 22:52