1550251515-Classical_Complex_Analysis__Gonzalez_

Elementary Functions 253

For e -:j:. 0, lal = ldl, the second inequality implies that be - be -:j:. 0.

However, the point

Zo =

ab+bd

be-be

is not fixed since ez 0 + d = (b(ae + ed)/(be - be) = 0 while az 0 + b -:j:. 0,

so that U(z 0 ) = oo.

If ae + ed = 0 and ldl^2 - Jal^2 +be -be= 0, again we must consider two

possibilities, e = 0 or e -:j:. 0. If e = 0, the second equation gives lal = ldl,

and from ad= 1 we get lalldl = 1, so that Jal = ldl = 1. In this case the

transformation is of the form

w = a^2 z +ab

with lal = 1, and the only fixed point is z = oo, unless b = i).. with ).. real.

In this case there is a line of fixed points

z - a^2 z - ia).. = 0 or - iaz + iaz - ).. = 0

which is a line through the point z 0 =^1 Ma>-. in the direction of -a. It

should be noted that in the present case ab + bd = 0.

If e -:j:. 0, the first equation gives lal = ldl, and the second reduces to

be - be = O, which yields b = 0 or e/e = b/b. In either case the constant
term in (5.12-4) vanishes. For b = 0 it is obvious, and for b -:j:. 0 it suffices to
replace e/e by b/b in the equation a( e/c) + d = 0 followed by a conjugation.
Hence if e -:j:. 0, ae + ed = 0 and be - be = 0, all three coefficients
in (5.12-4) vanish, so the equation is satisfied identically. In remains to
prove that in this case either there is no fixed point or that the locus of
fixed points is a circle or straight line.
As already noted, the first two conditions imply that lal = Id!- Suppose
that lal = ldl = 0. Then we get a= d = 0 and -be= 1, orb= -1/e, so
that the transformation reduces to

If we let w = z, we obtain

1 1

w=---e2 z

lzl2 = -1

e2

Thus there is no fixed point, except for e = ±i/k (k > 0 real), in which

case every point on the circle lzl = k is fixed.

Next, suppose that lal = Jdl -:j:. 0. Since be - be = 2i Im(be) = 0, it

follows that be is real. Of course, this may happen if b = O; otherwise, let