<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

Show that the element of arc-length $\frac{c}{{\rm Im}(z)} |{\rm
dz}|$ on ${\bf H}$ is invariant under $B(z) =-\overline{z}$.
\medskip

{\bf Answer}

Let $f:[a,b] \longrightarrow {\bf{H}},\ f(t)=x(t)+iy(t)$ be a
piecewise differentiable path.  Then,

$$\rm{length}(f)=\ds\int_f \ds\frac{c}{\rm{Im}(z)}\,|dz|$$

$B \circ f(t)=- \overline{f(t)}=-x(t)+iy(t)$. Thus, Im$(B \circ
f(t))=\rm{Im}(f(t))$ and $|(B \circ f)'(t)|=|f'(t)|$, and so

\begin{eqnarray*} \rm{length}(B \circ f) &amp; = &amp; \ds\int_{B \circ f}
\ds\frac{c}{\rm{Im}(z)} |\,dz|\\ &amp; = &amp; \ds\int_a^b
\ds\frac{c}{\rm{Im}(B \circ f(t))} |(B \circ f)'(t)| \,dt\\ &amp; = &amp;
\ds\int_a^b \ds\frac{c}{\rm{Im}(f(t))} |f'(t)| \,dt\\ &amp; = &amp;
\rm{length}(f)\ \rm{as \ desired}. \end{eqnarray*}
\end{document}
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