256 Chapter5
We recall (Exercises 1.2, problem 24) that an equation of the form Lz +
Mz + N = 0 has no solution if ILi = IMI and LN =/=MN, one solution,
namely, the point
Zo =MN-LN
ILl^2 - IMl^2
if ILi -:/:-IMI, and infinitely many solution lying on a straight line if ILi =
IMI -:/:- 0 and LN = MN. Hence it follows that:
- The general bilinear function is defined at every point of <C if Lz +
Mz + N = 0 has no· solution in <C. It will not be defined at oo, unless
either B = M = 0 or A = L = 0, that is, unless it is either of the
form (5.11-1) or (5.12-1). - If ILi -:/:-IMI so that Lz + Mz + N = 0 has a solution z = zo and
Az 0 + Bz 0 + C -:/:-0, then w(zo) = oo. Conversely, solving (5.13-1)
for z, we have
z= -'---~-'--~___:.~----'-~----'-~----'-~----'-~~ (MN - LN)ww +(AN - CM)w +(CL-BN)w +(BC -AC)
(LL - MM)ww +(BM -AL)w +(BM - AL)w +(AA - BB)
(5.13-2)
which shows that under the assumption ILi -:/:-IMI the inverse image
of w = oo is zo = (MN - LN)/(ILl^2 - IMl^2 ).
The conditions ILi-:/:-IMI and Az 0 + Bzo + C-:/:- 0 imply thatA B C
E1 = L M N -:j:.OM L N
If ILi -:/:-IMI but E1 = 0, then w(zo) is undefined.
- If ILi = IMI -:/:- 0 and LN = MN, then Lz + Mz + N = 0 defines
a straight line, so the denominator of (5.13-1) vanishes at every point
of that line. If it happens that IAI = IBI while AC -:/:- BC, then
Az + Bz + C -:/:- 0 for all z, and every point on the line Lz + Mz +
N = 0 will be mapped into oo. If IAI -:/:- IBI, then the point z 1 =
(BC - AC)/(IAl^2 - IBl^2 ) satisfies Az1 + Bz1 + C = 0, and if we have
Lz1 + Mz1 + N = 0 also or
A B CE2= f3 A C =0
L M Nthen w(zi) is undefined. However, all remaining points of the line
Lz + Mz + N = 0 are mapped into oo.