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Let $C$ be the unit square with vertices $0, 1, 1+i, i$ with the counterclockwise orientation.

a) Parameterize the contour $C$. For this, I parameterized the 4 line's which make up this unit square. With $C_1: f_1(t)=t , C_2: f_2(t)=1+it , C_3: f_3(t)=1-t+i , C_4: f_4(t)=i-it$ , where $t$ is from $[0,1]$

b) Using your parameterization of $C$, compute the value of the contour integral: (sorry not sure how to insert math type)

$\displaystyle\oint_C \bar{z} dz$, (contour integral over $C$ of $\bar{z} dz$) For this part I integrated $C_1, C_2, C_3, C_4$ separately, all going from $[0,1]$ and added them together. I got a final answer of $2i$, not sure if I made a mistake anywhere during the integration, but can anyone confirm this answer?

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You can compute this integral ($=: J$) a second time without parametrization of the $C_k$: On $C_1$ one has $\bar z=z$, on $C_2$ one has $\bar z=2-z$, and similarly for $C_3$, $C_4$. Therefore $$\eqalign{J &=\int_{C_1}z\ dz +\int_{C_2}(2-z)\ dz+\int_{C_3}(z-2i)\ dz+\int_{C_4}(-z)\ dz\cr &={z^2\over2}\Biggr|_0^1 +\Bigl(2z-{z^2\over2}\Bigr)\Biggr|_1^{1+i}+\Bigl({z^2\over2}-2iz\Bigr)\Biggr|_{1+i}^i +\Bigl(-{z^2\over2}\Bigr)\Biggr|_i^0 \cr &=2i\ .\cr}$$

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+1 for the smiley (=: – draks ... Apr 19 '12 at 8:48

Yeah that's what I get as well, you can always use Wolfram|Alpha to check these sorts of things.

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