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

158 Problems of Chapter 2
Answer. a) b) (here is the real positive
root); c) where d)
where (here is
the real positive root).
231. Hint. See formula (3.5) of the solution of Problem 229;
(cf., 227).
232. Hint. If and then
Answer. (cf., 231).
233. By virtue of the result of 221 where O is the origin of
the coordinates. Answer. a) the distance of the point from the origin
of the coordinates; b) the angle between the positive side of the axis
and the ray c) the distance between the points and (because
(cf., 221); d) the angle between rays and (cf.,
226).
234. See 233. Answer. a), b) The circle of radius 1 (and R) with
centre at the origin of the coordinates; c) the circle of radius R and centre
at the point d) the disc of radius R with centre together with its
bounding circle; e) the straight line perpendicular to the segment joining
points and and passing through its middle point; f) the negative
side of the real axis; g) the bisector of the first quadrant of the plane; h)
the ray defining the angle with the positive side of the axis
235. See 229 and 232. Answer. In the vertices of a regular
with centre at the origin of the coordinates.
236. Let a point and an arbitrary real number be given.
Choose (independently of and of Thus for every satisfying
the condition the inequality
is satisfied. Consequently the function is continuous for every
value of the argument (For one may take an arbitrary positive real
number).
237. Let a point and an arbitrary real number be given.
Choose (independently of Thus for every satisfying the
condition the inequality is
satisfied, i.e., we will have Consequently the function of
complex argument is continuous for every value of the argument.
Considering only the real values of the argument we obtain that the