GOVT BILASA GIRLS P G COLLEGE BILASPUR C
GOVT. BILASA GIRLS P. G. COLLEGE BILASPUR (C. G. ) Topic – The Range of an Analytic functions and univalent functions SESSION- 2019 -20 Presented By: Dr. Aradhna Sharma
Index S. Noi. Topic 1. Defination- Branch of the logarithm 2. Lemma(1)- let G be a simply connected region and let f be an analytic function defined in G that does not assume the values 0 or 1. then there is an analytic function defined in G such that f(z) = -exp{i π cosh[2 g(z)]} for z in G. 3. Lemma (2) – Let G, f and g be as in Lemma 1. Then g(G 0 contains no disc of radius 1. 4. Little Picard Theorem- Let f be an entire function that omits two values then f is a constant. 5. Definationn- Univalent Function 6. Definiation – Biberbach Conjecture.
Branch of the Logarithm Defination – If G is an open connected set in. C and f: G- C is a continuous function such that z= expf(z) z ϵ G. Then f is called a branch of the logarithm. Lemma(1) – Let G be a simpluy connected region and let f be an analytic functin defined in G that does not assume the values 0, 1. Then there is an analytic function g defined in G. such that f(z) = -exp{i π cosh[2 g(z)]} for z in G. Proof – Since f(z)≠ 0 for any z in G, there is a branch h of log f(z) defined in G: that is, exp h(z) = f(z)
Let F(z) = h(z)/2 л i z. Thus, if F(Z 0) = n for some integr n then f(Z 0) = exph(Z 0) = exp{2 π i F(Z 0)} = exp(2 π in)=1 Which can not happen because f does not assume the value 1 Hence F does not assume any integer values. Again, since F cannot assume the value 0 and 1, it is possible to define H(z) = √ F(z) - √ F(z)-1 The defination of H shows that H(z) ≠ 0 for any z So that it is possible to deine a branch g of log Hh in G. Hence, Cash (2 g)+1 = ½ (e 2 g+e-2 g)+1 = ½(e 2 g+e-2 g)
Cogh(2 g)+1=1/2 (eg+e-2 g)2 Cash (2 g)+1 = ½(H+1/H) 2, by the definationn of branch of the logarithm. Cosh(2 g)+1= 2 F Cosh (2 g)+1=1/πi h Thus, h = πicosh(2 g)+1 This gives f = eh = exp[π icosh (2 g)+ πi] = exp[πicosh(2 g)]. exp[πi] = -exp[i π cosh(2 g)]
That is f(z) = -exp[i πcosh[2 g(z)]} for z in G. Hence Proved.
Lemma(2) – Let G, f and g be as in lemma 1. Then g(G) contains no disc of radius 1. Proof- Suppose f and g are as in Lemma 1, n is a positive integer, and m is any integer, now suppose there is a point z 0ϵG. With g(Z 0) = ± log (√n + √n-1 + лim/2) then 2 cosh [2 g (Z 0)] = e 2 gzo+e-2 g(zo) 1)2 = eimn (√n + √n-1) ± 2 + (e-imл√n + √n)± 2 = (-1) [(√n + √n-1) ± 2 + (√n + √n)-2] =(-1)m{(√n +√n-1)2+[(√n-√n-1)2]/(√n+√n-1)2)(√n-√n= (-1)m {(√n + √n-1)2+[(√n-√n-1)2]/[(√n)2 -√n-1)2] = (-1)m {[(√n + √n-1)2 ]+[(√n - √n-1)2/(n- n+1)2]
Univalent Function Defination- A functionn f(z) is said to be univalent in a domain D if it is analytic in d and assume no value more than once in D. Thus for a univalent functionn the condition z 1≠ z 2 implies f(z 1) ≠ f(z 2). A univalent function provides a one-to-one map of D onto f(D) and hence has a single-valued inverse on f(D)
Bieberbach Conjecture Defination- If f ϵ D, and f(z) = z+∑ 2 ∞ anzn Then (a) | an | ≤ n for all n ≥ 2, and (b) f(D) ᴐ{w: |w| < 1/4} The second assertion is that f(D) contains all W in an open disc with centre at 0 and of radius ¼. Moreover, if (an) = n for just one n ≥ 2, then f is one of the koebe functions.
- Slides: 9