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For a regular curve, the curvature and torsion definition givesIt measures the rate at which the unit tangent turns per unit arclength.

HereThus $|{\gamma }^{\prime }|=3$, $|{\gamma }^{\prime \prime }|=4$, and ${\gamma }^{\prime }\cdot {\gamma }^{\prime \prime }=0$. HenceThe arc-length parametrization from $t=0$ is $s=3t$, so the curvature remains the constant $\kappa (s)=4/9$ for $0\le s\le 3\pi$. Therefore the total curvature is

The curve $\tilde{\gamma }$ is a unit circle traversed once. Its curvature is one and its length is $2\pi$, so without further calculationThe helical curve $\gamma$ devotes part of its unit tangent to the constant vertical direction. Its tangent therefore turns more slowly on the unit sphere than the tangent to the planar circle, which explains the smaller total curvature.

A Möbius transformation is generated by translations $z\mapsto z+b$, nonzero scalings $z\mapsto az$, and inversion $z\mapsto 1/z$. Translations and scalings plainly take circles and lines to circles and lines. More uniformly, every circle or line has an equationwhere $\alpha ,\gamma \in \mathbb{R}$; the case $\alpha =0$ is a line. Under inversion, put $z=1/w$ and multiply by $|w{|}^2$ to obtainwhich is again such an equation. The generators, and hence every Möbius transformation, therefore preserve Generalized circle under a Möbius transformation.

Now letIf $c\ne 0$, its finite pole $-d/c$ lies on some Euclidean circle, and that circle maps to a line because its image contains infinity. Thus $f$ takes every circle to a circle exactly when $c=0$, so the subgroup in question is the Affine subgroup of the Möbius group, consisting of $z\mapsto az+b$ with $a\ne 0$.

It is not normal. If $I(z)=1/z$ and $T_b(z)=z+b$ with $b\ne 0$, then $T_b$ is affine buthas a finite pole and is not affine. Hence conjugation by $I$ does not preserve the subgroup.