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I am reading the lecture notes for my one of my physics courses on scattering theory and in it, I came across a statement that I really do not quite get. Basically we start with a function $f$ that is supposed to be real analytic so satisfies the condition $$f(\bar{s}) = \overline{f(s)}$$ in the complex $s$-plane. Now because the function $f$ is conjectured to be analytic, we perform a series of conformal mappings that basically allow us to study certain properties of $f$ more easily. The conformal maps that are performed have the form - $$x = \frac{s-a}{s+a},\quad y = x^2,\quad v= \frac{y-C}{y-B},\quad u = \sqrt{v},\quad z = - \frac{u-i\beta}{u+i\beta},$$ where $a$, $B$, $C$, and $\beta$ are positive real numbers. After performing these mappings, the text goes on to state "these mappings are such that they preserve the real analyticity of $f$". So it seems to claim that we satisfy the same condition above in the $z$-plane but I do not understand how this is case. I tried to show this but I was unable to do so. But they arrive at the right answer by doing this.

I am unfortunately unable to send a link to the lecture notes but I have added a link to a paper that makes same claim.(on page 3, last paragraph they seem to explicitly suggest that it does)

I apologise if this is a very basic question but I'm an experimental physicist so I have never studied these things in great depth as it doesn't have much to do with my day to day tasks. Any help/advice is appreciated.

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  • $\begingroup$ Dear Alex Francisco, if you look at the paper I linked, then on page 3, last paragraph they seem to explicitly suggest that it does. $\endgroup$ – Nameless Paladin Oct 22 '18 at 13:12
  • $\begingroup$ No $\beta$ is chosen to be real. I can send you the paper if you want $\endgroup$ – Nameless Paladin Oct 22 '18 at 13:16
  • $\begingroup$ No matter whether the final one acts on $f$ or on its domain, real analyticity will be lost unless $f=0$. $\endgroup$ – Saad Oct 22 '18 at 13:31
  • $\begingroup$ Dear Alex Francisco, is there any other condition that $f$ might satisfy that would make it real analytic in final mapping besides it being 0 everywhere. $\endgroup$ – Nameless Paladin Oct 22 '18 at 13:56
  • $\begingroup$ The other condition is that $β=0$, which is also degenerate. $\endgroup$ – Saad Oct 22 '18 at 14:02
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$\def\i{\mathrm{i}}\def\e{\mathrm{e}}$Note that if $g$ and $h$ are both real-analytic, then so is $g \circ h$ since$$ (g \circ h)(\overline{z}) = g(h(\overline{z})) = g(\overline{h(z)}) = \overline{g(h(z))} = \overline{(g \circ h)(z)}. \quad \forall z \in \mathbb{C} $$ Now it suffices to verify that if $a, b, c, d \in \mathbb{R}$ and $ad - bc ≠ 0$, then$$ g_1(z) = \frac{az + b}{cz + d},\ g_2(z) = z^2,\ g_3(z) = \sqrt{z} $$ are all real-analytic. For any $z = r\e^{\i θ}\in \mathbb{C}\ (r > 0,\ θ \in (-π, π])$,\begin{gather*} g_1(\overline{z}) = \frac{a\overline{z} + b}{c\overline{z} + d} = \frac{\overline{az + b}}{\overline{cz + d}} = \overline{\left( \frac{az + b}{cz + d} \right)} = \overline{g_1(z)},\\ g_2(\overline{z}) = \overline{z}^2 = \overline{z^2} = \overline{g_2(z)},\\ g_3(\overline{z}) = g_4(r\e^{-\i θ}) = \sqrt{r} \e^{-\i \frac{θ}{2}} = \overline{\sqrt{r} \e^{\i \frac{θ}{2}}} = \overline{g_4(z)}. \end{gather*} Therefore, $g_1, g_2, g_3$ are all real-analytic.

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  • $\begingroup$ You write "if $g$ and $h$ are both real-analytic, then so is $g \circ h$ " That is true, but it has nothing to do with the conjugation property. $\endgroup$ – zhw. Oct 25 '18 at 16:38

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