# Laurent expansion of square root

I have the following two part problem:

(a) Prove that $$(z^2 - 1)^{-1}$$ has an analytic square root in $$\mathbb{C} - [-1,1]$$

(b) Find the Laurent expansion of an analytic square root from part (a) on a domain $$\{a: |z| > 1 \}$$, centered at $$z = 0$$.

For part (a), I note that the mobius transformation $$F(z) = \frac{z-i}{z+i}$$ maps the $$\mathbb{C} - [-1,1]$$ onto $$\mathbb{C}-(-\infty,0]$$. Since $$\mathbb{C} - (-\infty,0]$$ is simply connected and $$F$$ is nonzero on $$\mathbb{C} - [-1,1]$$, we can define a single-valued analytic branch of $$\sqrt{F(z)}$$ on $$\mathbb{C} - [-1,1]$$. Then, by a quick computation

$$G(z) = \frac{1}{(z+i)^2\sqrt{F(z)}}$$

is an analytic square root of $$(z^2 - 1)^{-1}$$ in $$\mathbb{C} - [-1,1]$$.

However, I do not know how to go about part (b). Any help would be appreciated.

• The Laurent expansion of $s(z)$ in an unbounded annulus is the same as the usual power series expansion of $s(1/w)$ around $w=0$, with $w$ subsequently replaced by $1/z$. Can you find the power series expansion of $\sqrt{(1/w)^2-1)^{-1}}$? – Greg Martin Aug 22 '20 at 19:58
• @GregMartin Strictly speaking, $\sqrt {w^2/(1 - w^2)}$ doesn't have a power series expansion at zero ($w/\sqrt {1 - w^2}$ with a suitable choice of the square root does). – Maxim Aug 22 '20 at 23:03

By part $$(a)$$ because $$|z|>1$$, if $$z=re^{i\theta}: -\pi<\theta< \pi,$$ we can use the principal branch of the logarithm, and choose $$\sqrt {w^2}=w.$$ Then, with $$Z=1/z^2$$ and noting that the binomial theorem is valid for $$|z|>1,$$ we compute

$$\sqrt {(z^2 - 1)^{-1}}=\sqrt {(z^2 - 1)^{-1}}=\frac{1}{z}\sqrt{\frac{1}{1-Z}}=\frac{1}{z}(1-Z)^{-1/2}=$$

$$\frac{1}{z}( 1 + Z/2 + 3 Z^2/8 + 5 Z^3/16 + 35 Z^4/128 + 63 Z^5/256 + 231 Z^6/1024 + 429 Z^7/2048 + 6435 Z^8/32768 + 12155 Z^9/65536 + 46189 Z^{10}/262144 + O(Z^{11}))$$

If $$\theta$$ lies on the negative real axis, then choose the branch cut accordingly and repeat the above calculation for $$0<\theta<2\pi$$.

I also think we can get $$(a)$$ by elementary means. We have by definition,

$$\sqrt{(z^2 - 1)^{-1}}=e^{-\frac{1}{2}\log (z^2-1)}$$. This function has branch points at $$1$$ and $$-1$$ but not $$\infty$$ so we may implement the diagram

setting $$z + 1 = r_1e^{i\theta_1}$$ and $$z -1 = r_2e^{i\theta_2}$$ and $$\pi<\theta_1,\theta_2<\pi$$

and prove analyticity by direct calculation.It comes down to considering cases.

• It's true that $(z \sqrt {1 - z^{-2}})^{-1}$ with the principal value of the square root is one of the required analytic branches on $|z| > 1$, but $\operatorname {Re}((z^2 - 1)^{-1})$ is not necessarily greater than $0$ for $|z| > 1$. – Maxim Aug 22 '20 at 15:01