4.3. ARGUMENT PRINCIPLE 167 Theorem 4.3.4 (Hurwitz theorem) Let D C be a domain and fn be a sequence of holo- ⊂ morphic functions defined in D that converge uniformly on compact subsets of D to f. If every fn does not have zero in D, then either f has no zero in D, or f 0. ≡ 1 Remark For example, fn = nz is holomorphic and has no zero in ∆(1; 1), but f = lim fn = 0. Also such phenomenon does not hold for real-valued functions. Proof: Assume f 0. To show: f has no zero in D, 6≡ i.e., suppose f(a) = 0 for some a D, we want to prove that fn(a)=0 n n for ∈ ∀ ≥ 0 some n0 > 0, which is a contradiction. Take δ > 0 such that ∆(a; δ) ⋐ D and that f never vanish. |∂∆(a;δ) By uniformly convergence, we have fn(z) f(z) < f(z) z ∂∆(a; δ), | − | | | ∀ ∈ as n n for some n > 0. Applying Rouch´eTheorem, it implies that the number of zeros ≥ 0 0 of f in ∆(a; δ) is equal to the number of zeros of fn =(fn f)+ f inside ∆(a; δ). − Theorem 4.3.5 Let hn be a sequence of holomorphic functions in a domain D C such ⊂ that hn h uniformly on compact subsets. If all hn are one-to-one, then either h is one- to-one, or→ h constant. ≡ Remark This phenomenon does not hold for real-valued functions, for example, we can take x 1, if x [1, + ); − ∈ ∞ h(x)= 0, if x [ 1, 1); ∈ − x +1, if x ( , 1), − ∈ −∞ − and x 1+ 1 , if x [1, + ); − n ∈ ∞ h (x)= , 1 , 1 , if x , n the line segment connecting the points ( 1 n ) and (1 + n ) [ 1 1); 1 − − ∈ − x +1 n , if x ( , 1), − − ∈ −∞ − Proof: Assume h constant. Suppose that h is not one-to-one, i.e., h(z ) = h(z ) = w 6≡ 1 2 0 for some two distinct points z , z D. Then we can take two disjoint neighborhoods Uj for 1 2 ∈ 168 CHAPTER 4. RESIDUE THEOREM zj, j =1, 2, such that the function h(z) w has at least two zeros in U U and h(z) w − 0 1 ∪ 2 − 0 never vanish on ∂U ∂U . 1 ∪ 2 Let f(z)= h(z) w and g(z)= hn(z) h(z). We see − 0 − f(z) = h(z) w0 > hn(z) h(z) = g(z) , z ∂U1 ∂U2, | | | − | ∂U1∪∂U2 | − | | | ∀ ∈ ∪ for any n n for some sufficiently large n . By Rouch´eTheorem, the number of zeros ≥ 0 0 of f(z)+ g(z) = hn(z) w in U U is equal to two, which is the number of zeros of − 0 1 ∪ 2 f(z)= h(z) w . This implies hn are not one-to-one for n n , a contradiction. − 0 ≥ 0 Let f(z)= z2. We have f(0) = 0, i.e., f has order 2 at the zero z = 0, i.e., the equation f(z) = 0 has 2 zeros (counting multiplicity). Now consider the equation f(z) = z2 = w where w = 0 and w is near 0. The equation f(z)= w has 2 roots: z = √w. 6 ± Theorem 4.3.6 Let f be a meromorphic function defined in a domain D C. Let z D ⊂ 0 ∈ and f(z ) = w Cˆ. If w = , let m be the order of pole of f at z ; if w = , let m 0 0 ∈ 0 ∞ 0 0 6 ∞ be the order of the zero of f(z) w0. Then there exists a neighborhood V of z0 in D and neighborhood W of w in Cˆ such− that for any w W w , there exist exactly m distinct 0 ∈ −{ 0} points zj V such that f(zj) w =0 of order 1, 1 j m. ∈ − ≤ ≤ Remark We may say that z0 is a w0-point of f of order m. Proof: Assume that w = . Otherwise we can use the coordinate w. 0 6 ∞ By the definition of m, we write e m f(z)= w + am(z z ) + ..., z ∆(z ; r) 0 − 0 ∀ ∈ 0 where am = 0, i.e., 6 m f(z) w =(z z ) [am + am (z z )+ ...], z ∆(z ; r). − 0 − 0 +1 − 0 ∀ ∈ 0 ′ Since z0 cannot be accumulation point of f or f , we can find a constant δ > 0 such that ∆(z0, δ) f = w0 = z0 , ∩{ } ′ { } (∆(z0, δ) z0 ) f =0 = , −{ ′ } ∩{ } ∅ 1 f (z) dz m. 2πi ∂∆(z0;δ) f(z)−w0 = R The last equality holds because of Argument Principle. 4.3. ARGUMENT PRINCIPLE 169 This implies f(z) w >ǫ> 0, z ∂∆(z , δ) | − 0| ∀ ∈ 0 for some constant ǫ> 0. Let V = ∆(z0, δ) and W = ∆(w0, ǫ). We notice the function f(z) w and w w satisfy − 0 0 − w w = ǫ< f(z) w , w ∂∆(w , ǫ). | 0 − | | − 0| ∀ ∈ 0 Then by Rouch´e’s theorem, we see that the number of zeros of f(z) w insider ∆(z , δ) − 0 0 is equal to the number of zeros of (f(z) w0)+(w0 w) = f(z) w insider ∆(z0, δ). So f(z) w has m zeros in ∆(z , δ). By our− construction,− since f ′ never− vanish in ∆(z , δ), − 0 0 each zero of f(z) w has order 1. − Theorem 4.3.7 (Open mapping theorem) Let f be a non constant meromorphic function defined in a domain in D Cˆ. Then f(D) Cˆ is also a domain. ⊂ ⊂ Proof: Since D is connected and f is continuous, f(D) is connected. It is enough to prove that f(D) is open, i.e., for any w f(D), there exists a disk 0 ∈ ∆(w , δ) f(D). 0 ⊂ In other words, to show: ∆(w , δ) such that w ∆(w, δ ), z such that f(z) = w. ∃ 0 ∀ ∈ 0 ∃ This was proved by the theorem above. Theorem 4.3.8 (Inverse Function Theorem) Let f be a holomorphic function defined in a neighborhood of z C. If f ′(z ) =0, there exist a neighborhood U of z and a neighborhood 0 ∈ 0 6 0 W of w0 = f(z0) in C such that f(U)= W and that the restriction f U : U W | → has a holomorphic inverse function g : W U of f U with g(w )= z . → | 0 0 Proof: Step 1. Define g Since f ′(z ) = 0, by Theorem above, z is a w -point of 0 6 0 0 order 1. Then there exists a neighborhood U of z0 and a neighborhood W of w0 such that for any w W , there exists a unique point z U with f(z) = w. This defines a function ∈ ∈ g(w)= z, which is an inverse function of f with g(w0) = z0. It remains to show that such g is holomorphic. 170 CHAPTER 4. RESIDUE THEOREM Step 2. To show: g is continuous Fix any w′ W with z′ = g(w′), we want to show: ∈ g(w) g(w′), asw w′, → → i.e., to show: for w near w′, we have g(w) g(w′) constant w w′ . | − | ≤ | − | In fact, f(z) f(z′)= a (z z′)+ a (z z′)2 + ... =(z z′)[a + a (z z′)+ ...]=(z z′)h(z) − 1 − 2 − − 1 2 − − ′ where a1 = 0 so that h is a holomorphic function at z that never vanish in a neighborhood ′ 6 of z . Then the desired inequality is proved so that g is continuous. Step 3. To show: g is holomorphic To show: for any w′ W , g′(w′) exists, i.e., the ∈ limit ′ g(w) g(w ) ′ − exists, as w w . w w′ → − ′ ′ In fact, by Step 2, we have f(z) f(z )=(z z )h(z), i.e., − − w w′ =(g(w) g(w′))h(g(w)). − − Since h never vanish near w′, it implies ′ g(w) g(w ) 1 1 ′ − = , asw w . w w′ h(g(w)) → h(gw′) → − In real analysis, if a smooth real-valued function f(x) is one-to-one, its inverse may not be smooth. For example, f(x)= x3 is smooth and one-to-one, but its inverse g(y)= √3 y is not differentiable at 0. Nevertheless, we have the following: Theorem 4.3.9 Let f Mer(D) where D Cˆ is a domain. Let f be one-to-one. Then there exists a function g∈ Mer(W ) where W⊂ = f(D) is a domain in Cˆ such that g is a ∈ meromorphic inverse function of f. Proof: Since f is one-to-one, f = constant. By the Open Mapping Theorem, f(D)= W 6 is a domain on which g is well defined. To show: for any w W with g(w )= z , g is meromorphic at w . 0 ∈ 0 0 0 Since f is one-to-one, z0 is a w0-point of order 1. Assume z0 = 0. If z = and w = , f ′(z ) = 0 so that by the Inverse Function6 Theorem, g = f −1 is 0 6 ∞ 0 6 ∞ 0 6 holomorphic at w0. If z = but w = , then h(z) = 1 f(z) = 1 is one-to-one and with finite value, 0 6 ∞ 0 ∞ z ◦ f(z) as above, h is holomorphic so that f is meromorphic at z0. .