1550251515-Classical_Complex_Analysis__Gonzalez_

Integration 505

(

( z( )

= ( - z + R^2 - z( d1/J

(

( z )

= (-z + (-z d1f-i

R2 -r2

= I( - zl2 d1/J

so that (7.28-12) becomes

u(rei^9 ) = R2 - r212... · u(Rei.P) ---d1/J

27r a IC -zl2

From Cauchy's formula for the upper half-plane (Exercises 7.3, problem

12) we can derive, in a similar fashion, Poisson's formulas for that region.

Corollary 7.19 If f = u +iv is analytic on Imz 2 0 and such that

lf(z)j < Alzl-m(A > 0, m > 0), then

for y = Im z > 0.

Proof We have

u ( x,y ) = -^1 1+= yu(t, I 0) 12 d t

7r -oo t - z

v ( x,y ) = -^1 1+= (x -I t)u(t, 12 0) d t

7r -oo t - z

f(z) = ~ 1+= f(t)dt

27ri -oo t - z

for Imz > 0. Also,

0 = ~ 1+= f(t) ~t

27ri -oo t - z

(7.28-14)

(7.28-15)

(7.28-16)

Hence, by subtracting, then adding, (7.28-15) and (7.28-16), we get

.f(z) = ~ 1+= (-

1

• 
  
  
  
  • ~) f(t)dt
    27ri -oo t - z t - z
  
  
  
  

= .!. j+oo yj(t) dt

7r -oo it -zl2 (7.28-17)

f(z) = -.^1 1+= ( -^1 + -_ 1 ) f(t) dt

27ri -oo t - z t - z

= 2-1+= (t - x )f(t) dt

7ri _ 00 jt - zl^2

(7.28-18)