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.

  • $\begingroup$ 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 $\sqrt{(1/w)^2-1)^{-1}}$? $\endgroup$ – Greg Martin Aug 22 '20 at 19:58
  • $\begingroup$ @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). $\endgroup$ – 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

enter image description here

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.

  • $\begingroup$ 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 $|z| > 1$. $\endgroup$ – Maxim Aug 22 '20 at 15:01

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