16 P. DEIFT, L. C. LI, AND C. TOMEI

Suppose A = ^2j Ajzi £ £* • We compute -K+A. For I g j ,

(Tt*+A,X) = (A,Tr+X)

= J tr A(z)(v+X)(z)^

= tt(ir+A)(0)(ir+X)(0) + (ir-A,X)- tr (v.A)(oo)(ir-X)(oo) ,(2.23)

where we have used Cauchy's theorem repeatedly. But for any constant matrix C,

i tr CX{z),_~~^_ = - tr C(n+X)(0) (2.24)

'(z-l)z

and

J tr CX{z)^±^ = tr C(TT_A')(CO) . (2.25)

Applying these formulae to (2.23), we find

A =

n_A.(jI±^B+^-A)^)

z - \ z - \

A0

(2.26)

Now

z - l

A,

suppose A = —^— G £* ,. Then a similar calculation shows that

2 - 1 4V

*+A = r - ^ T + 7(-4p)diag^i , (2.27)

where (Ap)diag is the diagonal part of Ap and ! i is the delta function at z = 1,

/ B(z)6l(z)az = B(l).

J\z\=l

Note that as X(l) is diagonal, only the diagonal part of Ap plays a role at 2 = 1 in (2.27).

We have proved the following result:

L e m m a 2.28. For A = Ap/(z - 1 ) 4 - ^reg € £in ,

=

^ - ( ^ r e g ) o

+

_

+

h^

Sl

(2.29)

2: — 1 4