Stein 0h3 Ex 1 23.7

If z n is an integer it's easy to check thatsin na _0

Conversely if sincz O then etiZ eTIZ

e2TiZ z

which is only possible if zaiz ziti n fornek

To check the orderof the Zero at 2 R sufficeto

check that sinctz z n to

since't t tcosCTZ lz n ITC l o2an

Resz.ms azT sinTazTTzn C D T

2_ Evaluate If 4 DX

The pole of 11 24 is at the zeros of 2412 1 0 C 24 1 2 e J j o 1 2,3i3 TEe e e

Resz.eiq.lttziIheiE ei

4 I z ei Fei

Resz e 4 I 2I z ei3 ta e

I ziti I Reszf ziti f Ly e EI ei

Residueintheupperhalf za j EI tiJ plane

FEIsameargumentas in the book

3 Show that ftp.EEadx T eaI for a o

Let I denote the integral then

Itiamf.FI dx Re1kigIiIaTdxComplete the contour by adding au upper semi circle

and denote the total contour as C C ER R CR

FadZ Res iaE az zaia R O R

e zitiziaa e

a

Ifc.FI zdZ Ept7az 2TR 0 as R D

Hence Lim I E aadx I a.dz Ling Ea.dz

Rao IR e CR

tea Note that it is already a real numberHence I q a

Let e0 Z then dieO dze it i do DZdo iz DZ

2 YzAud cos ei0z

ztzf.ca EzI fzI die1211

Next we find root for the denominator2212oz 11 0 2 a 2 a 1 0

Eta5 AHEta Hatz a at

Let 2 at Tat then 121 1 IZ I 1Then we have an order 2 pole at Z 2 within 1714

2234 iz a

Reszez Fizz.E II

4it 4I.IE lz z

C 4i taa iceIyz

i I 2Ti Resz integrand zti.fi k C2aIfysI

ok this is a messy problem

51 Let Pot an znt doAn Z Z Z Zz Z Zn

for Z Zn in ID by assumption

w Cwt dwThen f dz

p dnf fait141HH HH cow

CCW CW11 clockwise

counterclockwise

claim i Fcw l Z w C 1 Zaw

is a polynomial of degree En with no roots in Iwkl

PI if Zi are all distinct and none Zero thenFcw Zi Zn FD w Yz Cw Yzna degree n polynomials with roots all outside

sayof 14 1 If some 2i are zero Emei Zu are Zero then

ofFcw l Z w C1 2m w hence is a degree m

polynomial And the rest is the same as before

Wh2 dw 0AnMa

Wn 2since is a holomorphic function inside

Intel

Extra version deg QE degP 2 say deg P n degQ m

E 895 da FEET do aw

EH

Then