1550251515-Classical_Complex_Analysis__Gonzalez_

714

+ 1R f(x)(Inxt dx + j f(z)(logzt dz

r+

m

= 21l"i L !~~ f(z)(logzt

k=l

Now for z = rei^0 , 0 < r < 1, 1l" 2:: () 2:: 0, we have

lim Jzf(z)(logztl::; lim Jf(z)Jr(JlnrJ + 7r)n = 0

JzJ-+oo r-+O

since limr-+O Jf(z)J = Jf(O)J, and

r 11 I r J ln rJ r -1/r o

/!;~/ nr = r~ 1/r = r~ -1/r2 =

lim rJ ln rJ^2 = l~m J ln/rJ

2

= 2 lim rJ ln rJ = 0

r-+0 r-+0 1 r r-+O

and so on. Hence by Lemma 9.4,

r-+0 lim J f(z)(log zr dz=^0

,-

Also, from condition ( 4) we have

K

Jf(z)J::; r.;p (a> 1)

Chapter9

(9.11-43)

(9.11-44)

for JzJ 2:: .M > 1, say. Thus for z =Reio, R 2:: M, 0 ::; ()::; 11", we get

since

lim lzf(z)(logztJ::; lim RK_ 1 (lnR+1rt=O

JzJ-+oo R-+oo a

lim _lnR = lim __ l/_R __ = -^1 - lim -^1 - = 0

R--+oo R^0 -^1 R-+oo (a - l)R^0 -^2 a - 1 R-+oo R^0 -^1

lim (lnR)

2

= lim 2lnR(l/R)

R-+oo R<>-l R-+oo (a - l)R<>-2

= 2 lim lnR =0

a - 1 R-+oo Ra-l

and so on, and it follows from Lemma 9.3 that

lim J f(z)(log zr dz= o

R-+oo

(9.11-45)

r+