Complex line integral

**monster** · April 11th 2012, 01:24 AM

I'm doing a question where i have to calculate a the line integral; ∮z .dz over a simple closed contour, which as expected ( By Cauchy) came out as 0.
It's the supplementary questions that are confusing, i'm asked to calculate the integral with z replaced with the conjugate of z, and then to explain the relationship between these two integrals and the integrals over the same contour of just the real(z) and imaginary(z).
can someone shed some light on how they relate.
cheers

**jens** · April 11th 2012, 06:08 AM

Try this: if you decompose $z$ into its real and imaginary part, then what are the sum
$\oint z\;dz + \oint \bar{z}\;dz$
and the difference
$\oint z\;dz - \oint \bar{z}\;dz$
equal to?

**monster** · April 11th 2012, 05:42 PM

I think i see, is i that the countour integral of the real part plus the contour integral of the imag part = contour integral of z, and the difference of them = contour integral of conjugate of z ?

**jens** · April 13th 2012, 06:52 AM

Up to a factor, yes. Just write the integrand $z$ as $\mathfrak{R}\{z\} + i \mathfrak{I}\{z\}$ and use the fact that the integral is a linear operator, in the sense that
$\int \alpha f(x) dx = \alpha \int f(x) dx$
and
$\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx$

Thread: Complex line integral

LinkBack

Thread Tools

Display