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With $f(z)$ I denote the branch of $(z^2-1)^{1/2}$ defined by branch cuts in the $z$-plane along the real axis from $-1$ to $-\infty$ and from $1$ to $\infty$ with $f(z)$ real and positive above the latter cut.

$g(z)$ denotes the branch of $(z^2-1)^{1/2}$ defined by a cut along the real axis from $-1$ to $+1$ with $g(z)$ real and positive for $(x-1)$ real and positive.

Now I do not understand why $f(z)=f(-z)$ and $g(z)=-g(-z)$? Any ideas how this can be derived?

Ulrich Otto
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1 Answers1

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The function $s(z) = z^2 - 1$ is an even function. Thus, if $D\subset \mathbb{C}$ is a symmetric domain ($D = -D$), and $h$ a branch of $\sqrt{s(z)}$ defined on $D$, $h$ must be either an even or an odd function, for

$$h(-z)^2 = s(-z) = s(z) = h(z)^2,$$

so there is a continuous, hence constant, $\sigma \colon D \to \{-1,1\}$ such that $h(-z) = \sigma(z) \cdot h(z)$.

Since $s(0) = -1 \neq 0$, if $0 \in D$, any branch of $\sqrt{s(z)}$ on $D$ must be even. That shows that $f(-z) = f(z)$.

For $g$, we can consider the complement of the closed unit disk, $A = \{ z \in \mathbb{C} : \lvert z\rvert > 1\}$. On $A$, we have

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

with the principal branch of the square root used in $\sqrt{1-z^{-2}}$. Since $\sqrt{1-z^{-2}}$ is even and $z$ is odd, that shows $g$ is odd on $A$, and by continuity on all of $\mathbb{C}\setminus [-1,1]$.

Daniel Fischer
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  • Thank you, very clear answer, the only thing I do not fully understand is where to do you get the expression for $g(z)$ in the second part, I mean $g(z)$ on $A$ – Ulrich Otto Jan 27 '14 at 20:06
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    Write $$z^2 - 1 = z^2 \left(1 - \frac{1}{z^2}\right),$$ and pull the first factor out from under the root. Since $\lvert 1/z^2\rvert < 1$ on $A$, we have $\sqrt{1-1/z^2}$ on $A$ with the principal branch of the square-root. Then $z\sqrt{1-1/z^2}$ and $-z\sqrt{1-1/z^2}$ are two branches of $\sqrt{s(z)}$ on $A$, and since $A$ is connected, $g(z)$ must be one of the two. Looking at the values for real $z > 1$, we see that it must be $z\sqrt{1-1/z^2}$. – Daniel Fischer Jan 27 '14 at 20:17