Direct substitution in the Möbius transformation gives
Neither substitution meets the excluded denominator point .
For real , the numerator and denominator have equal complex modulus:
The denominator is always nonzero. Hence
Thus the real axis maps into the unit circle in the complex plane.
Differentiate the polar form of a complex number using the product rule and chain rule:
A second differentiation gives
The two contributions to arise separately from differentiating and from differentiating the moving exponential factor.
Write . Squaring the nonnegative complex modulus on each side gives
The linear terms cancel, leaving . Therefore the locus is
Its real and imaginary axis intersections are and .
Figure 1.
The circular complex locus and an anticlockwise tangential velocity at a general point
. Details.
On this circle the radial coordinate is constant, , so . The velocity is a tangent vector to the circle in the anticlockwise direction, represented by when . Substituting into part (b),
The inward component is the centripetal acceleration; the transverse component allows the speed to change. Anticlockwise motion does not imply uniform speed or . The preceding figure includes the required tangent arrow.
The perpendicular unit vectors represented by and point radially outwards and transversely in the increasing-angle direction. Reading off their real coefficients in part (b) gives the signed components
These are velocity in polar coordinates and acceleration in polar coordinates. In particular is a transverse speed component; alone is an angular velocity. The moving basis, rather than any extra force assumption, accounts for the additional acceleration terms. At the origin the polar directions themselves are undefined, although the Cartesian velocity and acceleration can remain well defined.

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