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For the following functions, find its domain of analyticity:

a) $(z^2 - 1)^{\frac{1}{2}}$
b) $(1 - z^2)^{\frac{1}{2}}$
c) $(4 + z^2)^{\frac{1}{2}}$

d) $(z^2 - 4)^{\frac{1}{2}}$

For the first one, I factored out $z^2 - 1$ using z = x + iy and got $x^2 + 2ixy - y^2 - 1$, then tried to solve for ($x^2 - y^2 - 1) < 0$ but doesn't this just factor out to $(x - y) (x + y) < 1$, I'm not sure how to go from here. Also 2xy = 0, implies x, y = 0 or either or. If so does this mean both imaginary and real axis would then be branch cuts?

For the second one if I just considered just $1 - z^2 > 0$ and $1 - z^2 \neq 0$ then the function is not analytic outside $|x| \geq 1$. However I'm not sure if this is correct at all. Then I tried breaking it into Re and Im parts, with Re = $1 - x^2 + y^2 < 0$ and Im = -2ixy = 0. I got here but I'm having trouble after that finding the set of points where the function is not analytic, especially where the Re part of function < 0. Is it the points that do not satisfy the inequality or the ones that do?

For the third one I think it's not analytic when $|y| \geq 2$, but I think that's wrong.

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Do you mean 'not equal' where you type =/=? –  gnometorule Feb 18 '13 at 20:44
yes sorry about the formatting –  DJ_ Feb 18 '13 at 20:54
Why the branch cut? The branch cut depends on which choice of branch for $z^{1/2}$ you choose. (Judging from your work, it looks like you're assuming the principal branch, but it would be better if you say so explicitly.) –  mrf Feb 19 '13 at 10:12

2 Answers 2

General answer: each of the functions listed has two single-valued holomorphic branches in any simply-connected domain that does not contain the zeros of the expression in parentheses. Also, a domain that becomes simply-connected after adding $\infty$ is okay.

More specifically: you can cut the plane along any curve that either (i) joins both zeros to $\infty$, or (ii) simply joins them.

Even more specifically: if your goal is to never, ever, put negative numbers under square root (i.e., you stick to the principal branch of square root), then cut between the zeroes in (b) and (c), and from each zero to $\infty$ in (a) and (d). This removes the part of the plane where the expression in parentheses takes values in $[-\infty,0]$.

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Consider the following definitions taken from the literature of Theodore Gamelin and Joseph Bak:

Definition 1

$f$ is analytic at $z$ if $f$ is differentiable in a neighborhood of $z$. Similarly, $f$ is analytic on a set $S$ if $f$ is differentiable at all points of some open set containing $S$.

Definition 2

A function $f(z)$ is analytic on the open set $U$ if $f(z)$ is (complex) differentiable at each point of $U$ and the complex derivative $f'(z)$ is continuous on $U$.

The function

$\hspace{2in} a(z) = (z^2-1)^{1/2}$,

for example, has derivative

$\hspace{2in} a'(z) = \frac{z}{(z^2-1)^{1/2}}$

thus, by Definition 1 and Definition 2, $a(z)$ is only differentiable for all $z$ such that $z^2-1 \ne 0$, so the "domain of analyticity" for $a(z)$ is $\mathbb{C}$ \ {$z \ne -1$ and $z \ne 1$}.

Image 1 $\hspace{0.5in}$ $z \mapsto a'(z)$

$\hspace{2in}$ im1

Image 2 $\hspace{0.5in}$ $z \mapsto a'(z)$

$\hspace{2in}$ im2

This approach can be applied to the latter problems. Furthermore, note that $a'(z)$ is discontinuous at these values:

Image 3

$\hspace{2in}$ im3

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The complex log I thought you had to consider negative numbers as well –  DJ_ Feb 21 '13 at 8:03
This is not correct or at least not complete, despite all the pretty pictures. For example, in (1), you need to specify what you mean by $(z^2-1)^{1/2}$. (Which square root do you choose?) You're ignoring branch cuts completely. –  mrf Feb 21 '13 at 10:05

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