Complex Analysis. Chapter IV. Complex Integration IV.8. Goursat s Theorem Proofs of Theorems. August 6, () Complex Analysis August 6, / 8

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1 Complex Analysis Chapter IV. Complex Integration IV.8. Proofs of heorems August 6, 017 () Complex Analysis August 6, / 8

2 able of contents 1 () Complex Analysis August 6, 017 / 8

3 . Let G be an open set and let f : G C be a differentiable function; then f is analytic in G. Proof. We need to show that f is continuous on each open disk in G, so without loss of generality we assume G is itself an open disk. Let = [a, b, c, a] be a triangular path in G and let be the closed set formed by and its interior. Notice that =. Use the midpoints of the sides of to form four triangles 1,, 3, 4 inside as: () Complex Analysis August 6, / 8

4 . Let G be an open set and let f : G C be a differentiable function; then f is analytic in G. Proof. We need to show that f is continuous on each open disk in G, so without loss of generality we assume G is itself an open disk. Let = [a, b, c, a] be a triangular path in G and let be the closed set formed by and its interior. Notice that =. Use the midpoints of the sides of to form four triangles 1,, 3, 4 inside as: () Complex Analysis August 6, / 8

5 . Let G be an open set and let f : G C be a differentiable function; then f is analytic in G. Proof. We need to show that f is continuous on each open disk in G, so without loss of generality we assume G is itself an open disk. Let = [a, b, c, a] be a triangular path in G and let be the closed set formed by and its interior. Notice that =. Use the midpoints of the sides of to form four triangles 1,, 3, 4 inside as: () Complex Analysis August 6, / 8

6 (continued 1) Proof (continued). he f (z) dz = 4 j=1 j f (z) dz where j = j. Let (1) be the j such that f (z) dz (1) j f (z) dz for j = 1,, 3, 4. he length of each j is half the length of : l( j ) = 1 l( ). Also, the diameter of j is half the diameter of : diam( j ) = 1 diam( ). By our choice of (1) we have f (z) dz 4 f (z) dz. We now iterate this process and produce a (1) sequence of triangles { (n) } n=1 such that (n) along with its interior, (n), we have: (1) () ; f (z) dz 4 f (z) dz, (n) (n+1) l( (n+1) ) = 1 l( (n) ); diam (n+1) = 1 diam (n). () Complex Analysis August 6, / 8

7 (continued 1) Proof (continued). he f (z) dz = 4 j=1 j f (z) dz where j = j. Let (1) be the j such that f (z) dz (1) j f (z) dz for j = 1,, 3, 4. he length of each j is half the length of : l( j ) = 1 l( ). Also, the diameter of j is half the diameter of : diam( j ) = 1 diam( ). By our choice of (1) we have f (z) dz 4 f (z) dz. We now iterate this process and produce a (1) sequence of triangles { (n) } n=1 such that (n) along with its interior, (n), we have: (1) () ; f (z) dz 4 f (z) dz, (n) (n+1) l( (n+1) ) = 1 l( (n) ); diam (n+1) = 1 diam (n). () Complex Analysis August 6, / 8

8 (continued ) Proof (continued). So inductively: f (z) dz 4n f (z) dz (8.6) (n) l( (n) ) = ( ) 1 n l( ); diam (n) = ( ) 1 n diam. Since each (n) is closed and diam (n) 0 as n 0, then the nestedness of the (n) s implies by Cantor s heorem (heorem II.3.6) that n=1 (n) = {z 0 }. () Complex Analysis August 6, / 8

9 (continued ) Proof (continued). So inductively: f (z) dz 4n f (z) dz (8.6) (n) l( (n) ) = ( ) 1 n l( ); diam (n) = ( ) 1 n diam. Since each (n) is closed and diam (n) 0 as n 0, then the nestedness of the (n) s implies by Cantor s heorem (heorem II.3.6) that n=1 (n) = {z 0 }. Let ε > 0. Since f is differentiable at z 0, we can find δ > 0 such that B(z 0 ; δ) G and f (z) f (z 0 ) f (z 0 ) z z 0 < ε whenever 0 < z z 0 < δ. () Complex Analysis August 6, / 8

10 (continued ) Proof (continued). So inductively: f (z) dz 4n f (z) dz (8.6) (n) l( (n) ) = ( ) 1 n l( ); diam (n) = ( ) 1 n diam. Since each (n) is closed and diam (n) 0 as n 0, then the nestedness of the (n) s implies by Cantor s heorem (heorem II.3.6) that n=1 (n) = {z 0 }. Let ε > 0. Since f is differentiable at z 0, we can find δ > 0 such that B(z 0 ; δ) G and f (z) f (z 0 ) f (z 0 ) z z 0 < ε whenever 0 < z z 0 < δ. () Complex Analysis August 6, / 8

11 (continued 3) Proof (continued). Alternatively, f (z) f (z 0 ) f (z 0 )(z z 0 ) ε z z 0 (8.9) whenever z z 0 < δ. Choose n such that diam (n) < (1/) n diam < δ. Since x 0 (n), this implies (n) B(z 0 ; δ). hen Cauchy s heorem (all versions!) implies that z dz dz = 0. Hence (n) (n) f (z) dz = f (z) dz f (z 0 ) dz (n) (n) } (n) {{} f (z 0 )z dz (n) } {{ } f (z 0 )z 0 dz } (n) {{} () Complex Analysis August 6, / 8 0

12 (continued 3) Proof (continued). Alternatively, f (z) f (z 0 ) f (z 0 )(z z 0 ) ε z z 0 (8.9) whenever z z 0 < δ. Choose n such that diam (n) < (1/) n diam < δ. Since x 0 (n), this implies (n) B(z 0 ; δ). hen Cauchy s heorem (all versions!) implies that z dz dz = 0. Hence (n) (n) f (z) dz = f (z) dz f (z 0 ) dz (n) (n) } (n) {{} f (z 0 )z dz (n) } {{ } f (z 0 )z 0 dz } (n) {{} () Complex Analysis August 6, / 8 0

13 (continued 4) Proof (continued). f (z) dz = (f (z) f (z 0 ) f (z 0 )(z z 0 )) dz (n) (n) f (z) f (z 0 ) f (z 0 )(z z 0 ) dz (n) ε z z 0 dz by (8.9) (n) ε diam (n) l( (n) ) ( ) 1 n ( ) 1 n = ε diam l( ) ( ) 1 n = ε diam l( ). 4 () Complex Analysis August 6, / 8

14 (continued 5). Let G be an open set and let f : G C be a differentiable function; then f is analytic in G. Proof (continued). Next, f (z) dz 4 n f (z) dz by (8.6) ( (n) ) 1 < 4 n ε diam l( ) 4 n = ε diam l( ). Since ε > 0 is arbitrary and diam, l( ) are fixed, it follows that f (z) dz = 0. So by Morera s heorem, f is analytic on G. () Complex Analysis August 6, / 8

15 (continued 5). Let G be an open set and let f : G C be a differentiable function; then f is analytic in G. Proof (continued). Next, f (z) dz 4 n f (z) dz by (8.6) ( (n) ) 1 < 4 n ε diam l( ) 4 n = ε diam l( ). Since ε > 0 is arbitrary and diam, l( ) are fixed, it follows that f (z) dz = 0. So by Morera s heorem, f is analytic on G. () Complex Analysis August 6, / 8