Complex Analysis R.G. Halbud R.Halbud@ucl.ac.uk Depamen of Mahemaics Univesiy College London 202 The shoes pah beween wo uhs in he eal domain passes hough he complex domain. J. Hadamad
Chape The fis fundamenal heoem. Inoducion A funcion ha is analyic a evey poin of C is said o be enie. A funcion ha is analyic a evey poin of C excep a poins whee i has poles is said o be meomophic. Many commonly used funcions ae meomophic such as he igonomeic and hypebolic funcions, polynomials and aional funcions. In his chape we begin he sudy of he value disibuion of meomophic funcions. This is known as Nevanlinna heoy..2 Jensen s fomula Suppose f is analyic and nowhee vanishing in he disc D = {z : z }. Then log f(z) is analyic in D. In paicula noe ha log f(z) is no banched alhough we do have o choose a paicula banch o have a well-defined funcion since is imaginay pa is defined up o he addiion of inege muliples of 2π. Taking he eal pa of Cauchy s inegal fomula fo log f, log f(0) = 2πi z = log f(z) dz, z gives log f(0) = 2π log f(e iθ ) dθ. (.) Now le f be any meomophic funcion which is no idenically zeo. Then f has finiely many zeos a,...,a m and poles b,...,b n in D \{0}. Fuhemoe, suppose ha he Lauen seies expansion of f a z = 0 has he fom f(z) =αz ν +, whee α = 0. Define ilc 0 f = α and od 0 f = ν. Wih hese definiions, he funcion g(z) :=z od B(aj,z) 0f B(bk,z) f(z), whee B(a, z) = 2 āz (z a), 2
CHAPTER. THE FIRST FUNDAMENTAL THEOREM 3 has no zeos o poles in D. Jensen s fomula esuls fom wiing (.) fo g. log ilc 0 f = 2π log f(e iθ ) dθ + log b k log a j (od 0f) log. The saing poin fo Nevanlinna heoy is he ealizaion ha he Jensen fomula can be wien in a vey symmeic way in which he zeos and he poles of f play an eual ole. Fo any x>0, define log + x := max(log x, 0). Then log x = log + x log + (x ). Jensen s fomula can now be wien as 2π f(e log + iθ ) dθ + log = 2π log + f(e iθ ) dθ + log + log ilc 0 f, + n(0,f) log b k + n(0, /f) log (.2) a j whee n(0,f) is he ode of he pole of f a z = 0, if i has one. I follows ha n(0, /f) ishe ode of he zeo of f a z = 0, if i has one..3 The Nevanlinna chaaceisic The poximiy funcion is m(, f)= 2π log + f(e iθ ) dθ, 2π 0 whee log + x := max(log x, 0). The enumeaive funcion is N(, f):= 0 n(, f) n(0,f) d + n(0,f) log = n log + n(0,f) log, b k whee n(, f) is he numbe of poles of f (couning mulipliciies) in z. The Nevanlinna chaaceisic funcion T (, f)=m(, f)+n(, f) measues he affiniy of f fo infiniy. Similaly, fo all a C, T, = m, + N, f a f a f a measues he affiniy of f fo he value a. Jensen s fomula becomes k= T (, f)=t (, /f) + log ilc 0 f. (.3) We will see ha if f is no consan hen T (, f) as. So euaion (.3) says ha asympoically, he affiniy of f fo is appoximaely he same as he affiniy of f fo 0. Nevanlinna s fis main heoem below exends his o all values a C.
CHAPTER. THE FIRST FUNDAMENTAL THEOREM 4 Lemma (Elemenay popeies of log + ) log + log + a j, log + a j a j log + max a j j log a = log + a log + (/a), log a = log + a + log + (/a), log + a log + b, a b. log + a j + log, Lemma 2 (Elemenay popeies of he Nevanlinna funcions) n, n(, ) n, n(, ) N, N(, ), N, N(, ), m, m(, ) + log, m, m(, ), T, T (, ) + log, T, T (, ). Poof: execise. Theoem 3 (Nevanlinna s Fis Main Theoem) Fo any meomophic funcion f and any a C, we have T, = T (, f)+o(),, f a whee f a. Poof: Similaly Hence T (, f a) T (, f)+t (, a) + log 2 = T (, f). T (, f) T (, f a)+t (, a) + log 2. T (, f a) T (, f) T (, a) + log 2 = log + a + log 2.
CHAPTER. THE FIRST FUNDAMENTAL THEOREM 5 So, T (, f) T T (, f) T (, f a) + f a log + a + log 2 + log + ilc a f., T (, f a) T f a Example: The funcion exp(z) is neve 0 o bu i says nea hese values on lage pas of he cicle z = fo >>. The ode of meomophic funcions The ode of a meomophic funcion f is σ(f) =limsup log T (, f), log Lemma 4 Fo all K>, n(, f) N(K,f) (.4) log K Poof: N(K,f) = = K 0 K K n(, f) n(, f) n(0,f) d + n(0,f) log K n(, f) n(0,f) d + n(0,f) log K n(, f) d + n(0,f) log K d = n(, f) log K. Theoem 5 Le f be a non-consan enie funcion. Le > 0 be sufficienly lage ha M(, f) := max z = f(z). Then fo all finie R>we have T (, f) log M(, f) R + T (R, f). R Poof: See execises..4 Raional funcions in Nevanlinna heoy Recall ha a aional funcion is a aio of polynomials. Raional funcions ae he simples kinds of meomophic funcions. Theoem 6 A meomophic funcion f is aional if and only if T (, f)=o(log ). Poof: Fis we assume ha f is aional. Then f has finiely many poles, say ν, and so hee is an 0 such ha n(, f)=ν fo all > 0, fom which i follows ha N(, f)=o(log ). Similaly m(, f)=o(log ) since any aional funcion can be appoximaed by αz µ on lage cicles. (Deails ae lef o he eade.)
CHAPTER. THE FIRST FUNDAMENTAL THEOREM 6 Now assume ha f is a meomophic funcion such ha T (, f)=o(log ). This means ha hee exiss C such ha N(, f) T (, f) C log. Theefoe, using he ineualiy (.4) wih K =, we have fo > n(, f) N( 2,f)/ log 2C. Hence f has only finiely many poles. So hee is a polynomial p(z) such ha F (z) :=p(z)f(z) is an enie funcion. Moeove, fom he poof of he fis pa of his heoem we know ha T (, p) =O(log ). Hence T (, F )=O(log ). I follows fom Theoem 5 ha log M(, F )=O(log ). Theefoe F (e iθ ) M(, F ) C fo some C. Hence by Liouville s heoem, F mus be a polynomial. Execises. Fill in he deails showing ha if f is aional hen T (, f)=o(log ). 2. Le a 0,...,a d be aional funcions and le f be a meomophic funcion. Show ha T (, a 0 + a f + + a d f d ) dt (, f)+o(log ). 3. Le f be meomophic on he closed disc D of adius R>0, f 0. Denoe he zeos and poles of f in D by a µ and b ν especively. Pove he Poisson-Jensen fomula: log f(z) = 2π Re log f(re iφ iφ + z ) Re iφ dφ z log R 2 ā µ z R(z a µ ) + log R 2 b µ z R(z b µ ). a µ <R b µ <R Hin: Fix z and le F (w) =f(z), whee w =(Re iφ z)/(r ze iφ ). Noe ha w =0 coesponds o z = Re iφ. Fo z <R, w =. 4. Pove heoem 5. Hin: use he Poisson-Jensen fomula.