Abel's theorem in problems and solutions - School of Mathematics

Solutions 181
are continuous functions and
FIGURE 70
Let be the positive real value of Thus is a continuous
function (cf., 243.) and the continuous curves with parametric equations
are the continuous images of the curve under the mapping
(cf., 229). Since on these curves takes all values of one of
these curves will begin at the point
If the curve C passes through the point then the points at
which divide the curve C into segments. In this case we have, as
before, a continuous image for every segment of the curve, and we take
thus as the initial segment the image which starts from the point If
then also. Hence the images obtained can be joined
in one unique continuous curve, which is the required curve.
294. See solution 280 and 289. Answer.
295. See solution 290. Answer.
296. for every is one of the values of All the values
of for a given are (cf.,