- #1
Gale
- 684
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I'm studying for a math exam for Complex Variable Analysis. We had a bunch of problems, and we keep having to let an integral just equal zero, and we're not sure why, but it always works.
What we do is this...
we're integrated over a closed contour. and we're supposed to chose a point [tex]z_0[/tex] which is inside the countour and the we use the equation:[tex] \int_C \frac {\frac {1}{f(z)}}{z-z_0} = 2\pi i f(z_0) [/tex] or something like that. there's also a formula using the derivatives of f. so actually, i don't quite understand any of this... cause I'm not sure why we use the above equation when we need to get the integral of f(z) over a countour C... but anyway, we use that, and that's always the answer. We've been using partial fractions in order to get it into that form i guess. oh yeah, we use partial fractions to change f(z) into something that looks like the above, and then we apply that formula to each fraction and sum them... right.
anyway, sometimes we get it so that [tex]z_0[/tex] isn't in the interior of C and then the integral equals zero. I've looked through the book, and i don't quite understand why, but its always the case. So i was hoping someone could explain it to me.
What we do is this...
we're integrated over a closed contour. and we're supposed to chose a point [tex]z_0[/tex] which is inside the countour and the we use the equation:[tex] \int_C \frac {\frac {1}{f(z)}}{z-z_0} = 2\pi i f(z_0) [/tex] or something like that. there's also a formula using the derivatives of f. so actually, i don't quite understand any of this... cause I'm not sure why we use the above equation when we need to get the integral of f(z) over a countour C... but anyway, we use that, and that's always the answer. We've been using partial fractions in order to get it into that form i guess. oh yeah, we use partial fractions to change f(z) into something that looks like the above, and then we apply that formula to each fraction and sum them... right.
anyway, sometimes we get it so that [tex]z_0[/tex] isn't in the interior of C and then the integral equals zero. I've looked through the book, and i don't quite understand why, but its always the case. So i was hoping someone could explain it to me.
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