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1. Inverse Function Theorem for Holomorphic Functions the Field Of 1. Inverse Function Theorem for Holomorphic Functions 2 The field of complex numbers C can be identified with R as a two dimensional real vector space via x + iy 7! (x; y). On C; we define an inner product hz; wi = Re(zw): With respect to the the norm induced from the inner product, C becomes a two dimensional real Hilbert space. Let C1(U) be the space of all complex valued smooth functions on an open subset U of ∼ 2 C = R : Since x = (z + z)=2 and y = (z − z)=2i; a smooth complex valued function f(x; y) on U can be considered as a function F (z; z) z + z z − z F (z; z) = f ; : 2 2i For convince, we denote f(x; y) by f(z; z): We define two partial differential operators @ @ ; : C1(U) ! C1(U) @z @z by @f 1 @f @f @f 1 @f @f = − i ; = + i : @z 2 @x @y @z 2 @x @y A smooth function f 2 C1(U) is said to be holomorphic on U if @f = 0 on U: @z In this case, we denote f(z; z) by f(z) and @f=@z by f 0(z): A function f is said to be holomorphic at a point p 2 C if f is holomorphic defined in an open neighborhood of p: For open subsets U and V in C; a function f : U ! V is biholomorphic if f is a bijection from U onto V and both f and f −1 are holomorphic. A holomorphic function f on an open subset U of C can be identified with a smooth 2 2 mapping f : U ⊂ R ! R via f(x; y) = (u(x; y); v(x; y)) where u; v are real valued smooth functions on U obeying the Cauchy-Riemann equation ux = vy and uy = −vx on U: 2 2 For each p 2 U; the matrix representation of the derivative dfp : R ! R with respect to 2 the standard basis of R is given by ux(p) uy(p) dfp = : vx(p) vy(p) In this case, the Jacobian of f at p is given by 2 2 0 2 J(f)(p) = det dfp = ux(p)vy(p) − uy(p)vx(p) = ux(p) + vx(p) = jf (p)j : Theorem 1.1. (Inverse Function Theorem for holomorphic Functions) Let f be a holomor- phic function on U and p 2 U so that f 0(p) 6= 0: Then there exists an open neighborhood V of p so that f : V ! f(V ) is biholomorphic. Proof. Since f is holomorphic on U; we can represent f by f = f(z) on U: Since f 0(p) 6= 0; the Jacobian of f at p is J(f)(p) = jf 0(p)j2 > 0: By inverse function theorem, there exists an open neighborhood V of p so that f : V ! W = f(V ) is a C1-diffeomorphism and J(f)(q) 6= 0 for all q 2 U: Let g : W ! V be the inverse map of f: Then g is smooth on V and hence we may represent g by g(w; w). Then g(f(z); f(z)) = z for all z 2 V and 1 2 f(g(w; w)) = w for all w 2 W: To show that g is holomorphic on W; we only need to show @g=@w = 0: In fact, by chain rule, @z @ @g @f @g @f 0 = = g(f(z); f(z)) = + : @z @z @w @z @w @z Since f is holomorphic on U; @f=@z = 0 on U: This implies @f @g 0 = : @z @w Since f : V ! W is a diffeomorphism, J(f)(z) = jf 0(z)j2 > 0 for all z 2 V: Then @f=@z = f 0(z) is nonzero on V; and hence @f @f = 6= 0; on V: @z @z We obtain @g=@w = 0 on W: This shows that g is holomorphic. Let Ω be a region in C: A holomorphic function f :Ω ⊂→ C is univalent if it is an injection. A locally univalent function on Ω is a holomorphic function f :Ω ! C so that every point a of Ω has an open neighborhood U so that f : U ! C is univalent. Corollary 1.1. Let f :Ω ! C be a holomorphic function. Let U be the subset of Ω 0 consisting of points a so that f (a) 6= 0: Then U is open and f : U ! C is locally univalent. 0 Theorem 1.2. Let U be an open set of C and f be a univalent function on U: Then f 6= 0 on U and f : U ! f(U) is biholomorphic. Since f is holomorphic on U, f 0 is also holomorphic on U: Since f is a nonconstant function, f(U) is open and f 0(z) is not the zero function on U: The zeros of f 0 forms a 0 discrete subset of U: Hence if f (z0) = 0; there exists an open neighborhood V = D(z0; δ) 0 of z0 so that f (z) 6= 0 for all z 6= z0 on V: Define a new function F (z) = f(z) − f(z0) on U: 0 0 0 Then F (z0) = F (z0) = 0 and F (z) = f (z) 6= 0 for z 6= z0 on V: Since F is holomorphic and nonconstant, we can choose δ small enough so that F (z) 6= 0 for 0 < jz − z0j < δ: Let C be the circle jz − z0j = δ=2: Then F is nonzero on C: Since C is compact, F is continuous on C; jF (z)j with z 2 C has minimum. Let 1 m = minfjF (z)j : z 2 Cg: 2 Then m > 0: For any 0 < jwj < m; jwj < jF (z)j on C: Rouche's theorem implies that F (z) − w and F (z) have the same number of zeros (counting multiplicities) in D(z0; δ=2): Hence #ZC (F (z) − w) = #ZC (F (z)) ≥ 2: This shows that F (z) = w has at least two solutions in D(z0; δ=2): In other words, F is not univalent on D(z0; δ=2) which leads to a contradiction to F being univalent. Now, let us prove that f −1 : f(U) ! U is holomorphic. To do this, we need the following lemma. Lemma 1.1. Let f be a holomorphic function on jzj < R: Suppose f is nonzero on 0 < 0 jzj < ρ for some ρ < R: Assume that f(0) = 0 and f (0) 6= 0: Let m0 = minfjf(z)j : jzj = ρg and j!j < m0: Then the solution of f(z) = ! is given by 1 I f 0(ζ) z = ζ dζ: 2πi C f(ζ) − ! Here C is the circle jzj = ρ 3 Proof. Let F (z) = f(z) − ! on jzj < R: Then F is holomorphic on jzj < R: Since j!j < m0; F (z) and f(z) has the same number of zero in jzj < ρ by Rouche's theorem. We know f(z) = 0 has a unique zero 0 in jzj < ρ. Hence F (z) = 0 has a unique solution. Denote this solution by z∗: Then 1 I F 0(ζ) 1 I f 0(ζ) z∗ = ζ dζ = ζ dζ 2πi C F (ζ) 2πi C f(ζ) − ! which proves our assertion. Now, assume that f : U ! C is univalent. To show that the inverse map g : f(U) ! U is holomorphic, we only need to show that given any point !0 2 f(U); the function z = g(!) has a Taylor expansion in a neighborhood of !0: Let f(z0) = !0: Choose ρ > 0 so that f(z) 6= !0 for 0 < jz − z0j < ρ. Denote m0 = minfjf(z) − !0j : jz − z0j = ρg as above. For all j! − !0j < m0(1 − δ); the infinite series converges uniformly 1 n 1 1 X (! − !0) = = : f(ζ) − ! (f(ζ) − ! ) − (! − ! ) (f(ζ) − ! )n+1 0 0 n=0 0 By the uniform convergence, we obtain 1 1 I f 0(ζ) X 1 I f 0(ζ) g(!) = ζ dζ = ζ dζ (! − ! )n: 2πi f(ζ) − ! 2πi (f(ζ) − ! )n+1 0 C n=0 C 0 This shows that g is holomorphic at !0: Since !0 is arbitrary in U; g is holomorphic on U:.
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