Lemma 41 if f ab c is continuous then vextendsingle

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Lemma 4.1. If F : [ a,b ] C is continuous then vextendsingle vextendsingle vextendsingle vextendsingle integraldisplay b a F ( t ) dt vextendsingle vextendsingle vextendsingle vextendsingle ( b a ) sup t [ a,b ] | F ( t ) | . We summarise this result in a slogan 1 In the midst of the word he was trying to say, In the midst of his laughter and glee, He had softly and silently vanished away – For the Snark was a Boojum, you see. 9
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modulus integral length × supremum . Next we wish to define the contour integral. integraldisplay C f ( z ) dz where C is a path in C . Roughly speaking integraldisplay C f ( z ) dz N summationdisplay j =1 f ( z j )( z j z j 1 ) where the polygonal path joining z 0 , z 1 , . . . z N is a ‘good approximation to C ’. We formalise this idea as follows. (A function g : R C is said to be differentiable if g and g are. We write g ( t ) = ( g ) ( t ) + i ( g ) ( t ).) Definition 4.2. If γ : [ a,b ] C is a sufficiently smooth 2 function describing the path C and f : C C is continuous, we define integraldisplay C f ( z ) dz = integraldisplay b a f ( γ ( t )) γ ( t ) dt. If a path C is made up of a path C 1 followed by a path C 2 followed by a path C 3 . . . followed by a path C n with each path satisfying the conditions of our definition, then we take integraldisplay C f ( z ) dz = n summationdisplay r =1 integraldisplay C r f ( z ) dz. It is, more or less, clear that our definitions are unambiguous but a rigorous development would need to prove this. If a contour begins and ends at the same point we call it at closed contour. In this case, some older texts use the pleasant notation contintegraldisplay C f ( z ) dz = integraldisplay C f ( z ) dz. Lemma 4.1 now takes the following form. Lemma 4.3. Under the conditions above, vextendsingle vextendsingle vextendsingle vextendsingle integraldisplay C f ( z ) dz vextendsingle vextendsingle vextendsingle vextendsingle length C × sup z C | f ( z ) | . 2 Continuously differentiable will certainly do. 10
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The next result is very important. Lemma 4.4. Suppose a, w C and r R with r > 0 . Let C be the path w + r exp described as θ runs from 0 to 2 π (less formally, the circle radius r and centre w described once anticlockwise). Then integraldisplay C 1 z a dz = 2 πi if | a w | <r, integraldisplay C 1 z a dz = 0 if | a w | >r. Note that this illustrates an important point Change of contour is not change of variable. The next result has very little to do with the course but I could not resist including it. Lemma 4.5. If C is a closed contour which does not cross over itself and is described once anticlockwise then integraldisplay C z dz = 2 i × Area enclosed by C. We now come to our master theorem. Theorem 4.6 (Cauchy’s theorem) . Let Ω be an open, simply connected (that is all in a single piece and with no holes) set in C and f : Ω C be an analytic function. Then integraldisplay C f ( z ) dz = 0 Note Observe that Lemma 4.4 shows that the ‘no holes’ condition can not be dropped. Given a particular Ω it is usually trivial to check that it has
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  • Fall '08
  • Groah
  • Math, Analytic function, Q7, Cauchy, Lemma

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