198 Vector and complex-number methods Ch. 11
andsohave
Z 4 =
st
st+( 1 −s)( 1 −t)
Z 2 +
( 1 −s)( 1 −t)
st+( 1 −s)( 1 −t)
Z 3.
With the second pair of solutions above, we obtain that
Z 0 =
1 −vw
( 1 −v)( 1 −w)
Z 1 −
v( 1 −w)
( 1 −v)( 1 −w)
Z 2
w( 1 −v)
( 1 −v)( 1 −w)
Z 3 ,
andalsothat
Z 4 =
v( 1 −w)
v+w− 2 vw
Z 2 +
w( 1 −v)
v+w− 2 vw
Z 3 ,
so that
Z 0 =
1 −vw
( 1 −v)( 1 −w)
Z 1 −
v+w− 2 vw
( 1 −v)( 1 −w)
Z 4.
As the sum of the coefficients ofZ 1 andZ 4 is equal to 1,Z 1 Z 4 passes throughZ 0 .This
proves the result.
To obtain a formula forZ 0 we note that on solving the second pair of solutions
above forvandw, we obtain the pair of solutions
v= 1 ,w=1; v=−
st
1 −t
,w=−
( 1 −s)( 1 −t)
s
.
To see this, note that
1 −w=−
1 −v
v
t,w= 1 +
1 −v
v
t,
so that
1 −v
(
1 +^1 −vvt
)
1 −v
v t
=s, i.e.
1 −v−( 1 −v)t
−^1 −vvt
=s.
Thus eitherv=1 and consequentlyw=1, or
1 −t
−tv
=s,i.e.v=
st
1 −t
,
and hence
w=−
( 1 −s)( 1 −t)
s
.
The first pair lead toZ 5 =Z 2 ,Z 6 =Z 3 , another degenerate case, while the second
pair lead to
Z 0 =
s( 1 −t)
1 −t+st
Z 1 +
st
1 −t+st
Z 2 +
( 1 −s)( 1 −t)
1 −t+st
Z 3. (11.5.2)
Because of the condition (11.5.1) the coefficients in (11.5.2) could be given in
several different forms.