 |
 |  |
| FAQ
•
Buscar
•
Wiki
•
Apuntes
•
Planet
•
Mapa
•
Eyeon
•
Chat
Preferencias • Grupos de Usuarios Registrarse • Perfil • Entrá para ver tus mensajes privados • Login |
 | | |
Todas las horas son ART, ARST (GMT - 3, GMT - 2 Horas) Protected by CBACK CrackerTracker 365 Attacks blocked. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
![[tex]I = \int_\gamma \frac{\log(z-1)}{\sin(\pi z)} dz[/tex]](images/latex/286973960f8c72e1a65bdf33715864f5288ca970_0.png "[tex]I = \int_\gamma \frac{\log(z-1)}{\sin(\pi z)} dz[/tex]")
, siendo ![[tex]\gamma[/tex]](images/latex/80916f1ea9a71a2fbc9b3e960163b504c931febf_0.png "[tex]\gamma[/tex]")
la circunferencia de centro -2 y radio ![[tex]\frac{3}{2}[/tex]](images/latex/f18e00544688dee42f99fb3802586d7d27890e7b_0.png "[tex]\frac{3}{2}[/tex]")
recorrida en sentido antihorario. Se considera la determinación del ![[tex]\log(z-1)[/tex]](images/latex/4863bdafd1c0d17625a5f460e79b7c4f6a57152e_0.png "[tex]\log(z-1)[/tex]")
definida en ![[tex]\mathcal{C} - \{ z = 1 + iy, y \leq 0 \}[/tex]](images/latex/22160525381389451c2d661bd8a7f42151effc5d_0.png "[tex]\mathcal{C} - \{ z = 1 + iy, y \leq 0 \}[/tex]")
cuyo valor en ![[tex]z=2[/tex]](images/latex/663d70051670ffa9764b92892d36fea14212cfad_0.png "[tex]z=2[/tex]")
es 0.
![[tex]I = 2\pi + i 2 \ln(3/8)[/tex]](images/latex/85fb22d9578b39cf9710fcdc85cf740ad4959f8d_0.png "[tex]I = 2\pi + i 2 \ln(3/8)[/tex]")
![[tex]2 \pi i \left( \frac{1}{\pi} (log(-3)-log(-2)-log(-4)) \right)[/tex]](images/latex/561db70cb024b5c945cfac0db7f8cd6a92b88635_0.png "[tex]2 \pi i \left( \frac{1}{\pi} (log(-3)-log(-2)-log(-4)) \right)[/tex]")
que seguramente es lo mismo pero me da paja agrupar los logaritmos.![[tex] \nabla ^u \nabla_u \phi = g^{ij} \Big( \frac{\partial ^2 \phi}{\partial x^i \partial x^j} - \Gamma^{k}_{ij} \frac{\partial \phi}{\partial x^k} \Big)\\\\\frac{\partial \sigma^{ij}}{\partial x^i} + \sigma^{kj} \Gamma^i _{ki} + \sigma^{ik} \Gamma^j _{ki} = 0[/tex]](images/latex/99f2e51931163431f7feb4825fc711ebccd99981_0.png "[tex] \nabla ^u \nabla_u \phi = g^{ij} \Big( \frac{\partial ^2 \phi}{\partial x^i \partial x^j} - \Gamma^{k}_{ij} \frac{\partial \phi}{\partial x^k} \Big)\\\\\frac{\partial \sigma^{ij}}{\partial x^i} + \sigma^{kj} \Gamma^i _{ki} + \sigma^{ik} \Gamma^j _{ki} = 0[/tex]")
![[tex]z=-1, z=-2, z=-3[/tex]](images/latex/7c3bc65fd34e32527c6c7072eff9861b69331432_0.png "[tex]z=-1, z=-2, z=-3[/tex]")
.
![[tex]Res(f, z=-1) = \frac{Log(-2)}{\pi \cos(-\pi)} = \frac{\ln(2) + i\pi}{-\pi}[/tex]](images/latex/08ba3be14318c1d61c99cc30eb82142fdf75e90e_0.png "[tex]Res(f, z=-1) = \frac{Log(-2)}{\pi \cos(-\pi)} = \frac{\ln(2) + i\pi}{-\pi}[/tex]")
![[tex]Res(f, z=-2) = \frac{Log(-3)}{\pi \cos(-2\pi)} = \frac{\ln(3) + i\pi}{\pi}[/tex]](images/latex/e8f144caa396dc76da99640046bb2071334f7da5_0.png "[tex]Res(f, z=-2) = \frac{Log(-3)}{\pi \cos(-2\pi)} = \frac{\ln(3) + i\pi}{\pi}[/tex]")
![[tex]Res(f, z=-2) = \frac{Log(-4)}{\pi \cos(-3\pi)} = \frac{\ln(4) + i\pi}{-\pi}[/tex]](images/latex/17c5cd1fc866de68323d58ba39d42506dbb586c0_0.png "[tex]Res(f, z=-2) = \frac{Log(-4)}{\pi \cos(-3\pi)} = \frac{\ln(4) + i\pi}{-\pi}[/tex]")
![[tex]I = 2 \pi i \sum Res[/tex]](images/latex/2919db56456b9f3047aaf6a6bbff21ff219109f4_0.png "[tex]I = 2 \pi i \sum Res[/tex]")
![[tex]I = 2 \pi i \left ( \frac{\ln(2) + i\pi}{-\pi} + \frac{\ln(3) + i\pi}{\pi} + \frac{\ln(4) + i\pi}{-\pi} \right)[/tex]](images/latex/8591da0ef64025218437a4af9c9417be5fc91acc_0.png "[tex]I = 2 \pi i \left ( \frac{\ln(2) + i\pi}{-\pi} + \frac{\ln(3) + i\pi}{\pi} + \frac{\ln(4) + i\pi}{-\pi} \right)[/tex]")
![[tex]I = 2 \pi i \left ( \frac{\ln(2)}{-\pi} - i + \frac{\ln(3)}{\pi} + i + \frac{\ln(4)}{-\pi} - i \right)[/tex]](images/latex/9fde6352ab99c8f32b56a8198c82ceebf8c64c96_0.png "[tex]I = 2 \pi i \left ( \frac{\ln(2)}{-\pi} - i + \frac{\ln(3)}{\pi} + i + \frac{\ln(4)}{-\pi} - i \right)[/tex]")
![[tex]I = 2 i \left ( \ln(\frac{1}{2}) + \ln(3) + \ln(\frac{1}{4}) - i\pi \right)[/tex]](images/latex/aa6f050d5c2728d45838fa2c3b64a1b5b500af22_0.png "[tex]I = 2 i \left ( \ln(\frac{1}{2}) + \ln(3) + \ln(\frac{1}{4}) - i\pi \right)[/tex]")
![[tex]I = 2\pi + i 2 \ln(3/8)[/tex]](images/latex/02d35024275e12807ced63414deee411b5662ab9_0.png "[tex]I = 2\pi + i 2 \ln(3/8)[/tex]")
| Powered by phpBB2 Plus, phpBB Styles and Kostenloses Forum based on phpBB © 2001/6 phpBB Group :: FI Theme :: Mods y Créditos |
| Foros-FIUBA está hosteado en Neolo.com Cloud Hosting |