Vipul: 2 revisions

2008-05-18T19:13:16Z

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Vipul at 18:50, 26 April 2008

2008-04-26T18:50:48Z

← Older revision		Revision as of 18:50, 26 April 2008
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			{{basic fact}}
	==Statement==		==Statement==

Vipul: New page: ==Statement== Suppose $U \subset \mathbb{C}$ is an open subset, and $f:U \to \mathbb{C}$ is complex-differentiable at ...

2008-04-19T21:36:20Z

New page: ==Statement== Suppose <math>U \subset \mathbb{C}</math> is an open subset, and <math>f:U \to \mathbb{C}</math> is complex-differentiable at ...

New page

==Statement==

Suppose <math>U \subset \mathbb{C}</math> is an open subset, and <math>f:U \to \mathbb{C}</math> is [[function complex-differentiable at a point|complex-differentiable]] at every point. Suppose <math>\triangle</math> is a triangle lying completely inside <math>U</math>. Then, we have:

<math>\int_{\partial \triangle} f(z) \, dz = 0</math>

==Proof==

Define:

<math>\alpha(\triangle) = \int_{\partial \triangle} f(z) \, dz</math>

Note that <math>\triangle</math> can be split into four congruent triangles by joining the midpoints of the sides. Since <math>\alpha(\triangle)</math> is the sum of <math>\alpha</math> evaluated for each of the pieces, at least one of the pieces <math>\triangle^1</math> must satisfy:

<math>|\alpha(\triangle^1)| \le |\alpha(\triangle)|</math>

Pick this triangle, and again divide it into four triangles. Repeating this process we get a descending sequence of triangles, each having side-lengths half of its predecessor:

<math>\triangle \supset \triangle^1 \supset \triangle^2 \supset \ldots</math>

and with the further property that:

<math>|\alpha(\triangle)| \le 4^n|\alpha(\triangle^n)|</math>

On the other hand, we have that:

<math>L (\partial \triangle^n) = \frac{1}{2^n} L(\partial \triangle)</math>

Now let <math>c \in \triangle</math> be the unique point in the intersection of all the <math>\triangle^n</math>s. Since <math>f</math> is differentiable at the point <math>c</math>, we have:

<math>\lim_{z \to c} \frac{f(z) - f(c)}{z - c} = 0</math>

Define now:

<math>g(z) := \frac{f(z) - f(c)}{z - c} - f'(c)</math>

Then <math>g(c) = 0</math>. Observe that:

<math>\int_{\partial \triangle^n} f(c) \, dz = \int_{\partial \triangle^n} f'(c) (z - c) = 0</math>

Thus, we get:

<math>\alpha(\triangle^n) = \int_{\partial \triangle^n} g(z)(z - c) \, dz</math>

Applying the length estimates and size estimates now gives the result. {{fillin}}

Goursat's integral lemma for complex-differentiable functions - Revision history

Vipul: 2 revisions

Vipul at 18:50, 26 April 2008

Vipul: New page: ==Statement== Suppose U \subset \mathbb{C} is an open subset, and f:U \to \mathbb{C} is complex-differentiable at ...

Vipul: New page: ==Statement== Suppose $U \subset \mathbb{C}$ is an open subset, and $f:U \to \mathbb{C}$ is complex-differentiable at ...