254with lal f: 0, lbl f: 0, lei f: 0. Then
be= lbl ieieih'-,8)and this is real if'Y-/3=mr
for some integer n. Also, we must have, whether b = 0 or not
ae + cd = lailel [ei(i-a) + ei(o-1)] = 0which implies that'Y = ~ (a+ l5) + (2m + 1 )7r
2 2
for some integer m.Chapter 5(5.12-5)(5.12-6)Taking up the case b = 0, we have ad = 1, and it follows that lai =
ldl = 1 and ei(aH) = 1, so that a+ 15 = 2p7r, for some integer p, and
'Y = p7r + (2m + 1 )7r /2 = (2p + 2m + 1 )7r /2. Hence for this case
a=eia, b=O, e=±iiei, d=e-ia=a
and the transformation has the formaz
w = ±ilelz + e.-ia
which has the circle of fixed points±ilelz +a
·. lz ± i 1:11 = l!I
the sign corresponding to that of ilcl in (5.12-7).
Now for the case b f: 0 we have
ab+ bd = iallblei(a-,8) + lallblei(,8-o) = 0
implying that a - /3 = 7r + /3 - 15 + 2q7r, or
/3 = ~ ( (Y + 15) - ( 2q + 1 )7r
2 2for some integer q. Adding (5.12-6) and (5.12-8), we get
Hence/3 + 'Y = a + /5 + ( m - q )7r'
ad - be= lai^2 ei(aH) - lblleiei(,B+i)
= ei(aH) [ial2 ± lbllelJ = 1
(5.12-7)(5.12-8)