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I am currently working with line integral in the complex plane. I want to compute $\int_\gamma zdz$ when $\gamma$ is a square with vertices at $1+i, 1-i, -1+i, -1-i$.

I know how to parametrize lines, circle and do line integral for those, but I am a little lost with this one. Please give me some insight?

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You can break up the integral into the sum of four integrals, one for each side of the square. Each side of the square is a line segment, which you know how to parameterize.

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This is a fact of life,so-called Cauchy's theorem, that $\int_ \gamma f(z)dz=0$ provided $f$ is analytic in $int(\gamma)$ with no singularity on $\gamma$. So in your our case $\int_\gamma zdz=0$.Well for sake of computational math here is how we can compute the given integral by parameterizing of square: $\int_\gamma zdz=i\int_{-1}^1(1+it)dt+\int_{1}^{-1}(t+i)dt+i\int_{1}^{-1}(-1+ti)dt+\int_{-1}^1(t-idt)=2i-2i+2i-2i$.

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  • $\begingroup$ P.S Suppose we have two points $a,b\in\mathbb{C}$ then the affine combination of these two points is a parameterization for the segment connecting $a$ and $b$ i.e $ta+(1-t)b$ where $t\in[0,1]$ $\endgroup$ – BigM Feb 28 '14 at 5:54

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