Question

Problem 5. Suppose that f: +C is analytic on an open set 12 containing the closed half plane H = {2€ C: Im(x) > 0} and that there is a finite constant M with f() < M for all z H. 1. Show that da = f(i) x² +1 +00 2. Show that if o is a point in C with Im(a) > 0, then I (a) Im(a)' 22-2Re(a)x+ lajar (3) deduce sin (Bx) where 870

Answer 1

(O ſto) dh rfi) 12+ integration with contour e -- 0 2+1 R Consider the Complen c= [-R R] U TR (2) dz. с. fcz) f(2) F (Z) = trti as simple pole. ani Residue [F(2) fli 2 i the only z=i (2+i) (2-1) has 2-i] S F (2) 22 - с - ari atli) fez) dz = cz fezl dz s + . Now fez) Z24 ERRI 22 22t1 Ге on FR 1 f(2) 22t If (2) 12²+1 R >> It (2) <M and 12h+11 >/1212-11 = R²1 M 1 f12) 2) 1 < R²_1 z2A Ź (TUR) Mr R R21 Is f12 dal 22+1 га f( 2) da ' Lim mn Rya R R²1 lim z2t f(2) Znt ilim Rugo 5 fen X241 and limitom dz ERR 22t

Hence lim s fraz dz = 1 fem du Z24 nt fino da = t Hi) net 2. S du © dh fins n²2 Redon t 1912 er nos com as (n-o) (hma) Consider the complen integration, fee) dz (2-4) (2-2) - e=TRUERR] с R>> 141 R O R --- I fea) (2-2) (22) Imaginary Part (d) yo. frzy so Z=2 he only simple pole of F(2) 7 Residue of F (2) (at 2=4) = lim frag fcal) 2 (2-2) 2 i 9m (d) f (2) od fea .xani zi am le gm (4) 142 | f(2) (2-2) (2-3) HE) ISM 12-23 (2–2)| = 12–4112-31 121-11)? f (2) M on Fr. (22) (2-x) (R-1412 TR fez) (2-2) (2-2) R MR = 2 (R- Lim Rya Hre

fee) RO 20 f(n) ch na 2 Refagntlar с an fon nr 2 Red) n+1012 an r fa) gm(a), Sin pn dir (n-ti Sinßn Sin ßn n2 ant 2 2² - 2 Reo.nt 1212 e An. dati Consider f(x)= A(Z) = ė B(x+ iyi ipn-By 1 +231 = leipny lety s le pay <r. B, 470 result (2) Prom the eißn te eißliti) n2 2 Re (2) n + 1912 cosent i sin ßn Heiß eß du 2_2 Recain 7 1912 Cosßnt i sin ßn du = rep ( cesp + ê Simp) n? - 2n +2 sinßn du reß sinß. (7) 2+1 -