Geometry with Trigonometry

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.