Question

3. Suppose f is an entire function. Show that any of the two criteria below imply that f is a constant function. (a) Imf(z) #0 for all z E C and Ref (2) is an entire function. Imf (2) [10] (b) -1 < Ref(x) < 1 for all 2 € C. [8]

Answer 1

3. Suppose, fin am entire function f: & ¢, analytic everywhere on & (a) Im fiz) = 0 , &Z€¢ and Le tez) Im Hz) is an entire function. Let g (2) = Refer) Im f1z) Since, Ref(): & IR. = Re fez) e JR , & 8€¢. - Abo, Im fez) : & IR & Im (43)) AIR, AZE¢. Hence, Refez) im tez) EIR. AZE¢. Ronge of g , R (9) CIR. 91¢) cannot be open Now, bo open mapping theorem implies.gce) i either a single loire set open connected. Honce, by , g2¢) = {e} (where e e¢) ce is constant). Im fee) Re fee) ire 2. Hence, f(3) = Re fez) i Im f3. (1 + %) Re fez) (by ® ) Re fez? = .f(2) 1+% Be h (2) = Re f(2) is an entire function. =) where, hlze= u(a,m) ti recaring) is entire function cu (a,m) = Re f2) (wherzantiy and v Englao. Now, by C. R equation we hace. au au ay are ord du ay In=o. 0

Hena, lí lzzan au + = 0, & Zek. are an hlz)=0 & 264 > h is constant. → Re fez) is constant , &Z€¢. Now, by a equation ③ fois constant. Hence, f in a constant function Proceed

(b) - 1 Retez) & 1 , t ze & Let g1zze e fez) & ze¢. son Now, of in entire function. = go in an entire function Ref(2) + i Imfezo 19 (8)1 = le Ref (2) ilm f(2) | le Retezll. 10 i Im fez? le Retez) 1.1. = e Now, е Re fez) le Now, -1 Re fez) 1 nelle Refez) Le' Now, by leovili's y =C - e' { 19 (2)ize , &Z€¢ theorem, we get g is constant function. g (8) = e tez) are for some constant (2) Rek. e Do888 cousteniz de tez) =0 de fez) f(2)=0 f' (2)=0,vzet. (as é Vzek. (as e tras) = f is constant functia Henc, f is on a constant function a