4.3: Teorema Fundamental para Integrales Complejas de Línea
- Page ID
- 109812
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Esto es exactamente análogo al teorema fundamental del cálculo.
Si\(f(z)\) es una función analítica compleja en una región abierta\(A\) y\(\gamma\) es una curva en\(A\) de\(z_0\) a\(z_1\) entonces
\[\int_{\gamma} f'(z) \ dz = f(z_1) - f(z_0). \nonumber\]
- Prueba
-
Esta es una aplicación de la regla de la cadena. Tenemos
\[\dfrac{df(\gamma (t))}{dt} = f'(\gamma (t)) \gamma '(t). \nonumber\]
Entonces
\[\int_{\gamma} f'(z) \ dz = \int_{a}^{b} f'(\gamma (t)) \gamma '(t)\ dt = \int_{a}^{b} \dfrac{df(\gamma (t))}{dt} \ dt = f(\gamma (t)) \vert_{a}^{b} = f(z_1) - f(z_0) \nonumber\]
Otra forma equivalente de afirmar el teorema fundamental es: si\(f\) tiene un antiderivado\(F\), i.e\(F' = f\).
\[\int_{\gamma} f(z)\ dz = F(z_1) - F(z_0). \nonumber\]
Rehacer\(\int_{\gamma} z^2\ dz\), con\(\gamma\) la línea recta de 0 a\(1 + i\).
Solución
Podemos verificar por inspección que\(z^2\) tenga un antiderivado\(F(z) = z^3/3\). Por lo tanto, el teorema fundamental implica
\[\int_{\gamma} z^2\ dz = \left. \dfrac{z^3}{3} \right\vert_{0}^{1 + i} = \dfrac{(1 + i)^3}{3} = \dfrac{2i(1 + i)}{3}. \nonumber\]
Rehacer\(\int_{\gamma} z^2 \ dz\), con\(\gamma\) el círculo unitario.
Solución
Nuevamente, dado que\(z^2\) tenía antiderivado\(z^3/3\) podemos evaluar la integral tapando los puntos finales de\(\gamma\) en el\(z^3/3\). Como los puntos finales son los mismos la diferencia resultante será 0!