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For the straight line segment $[-3,1+i]$

I know to do this I need to convert to

$\int_{\gamma}\overline{z}dz=\int\overline{(\gamma(t))}\gamma'(t)dt$

which for this path I have got to be

$\int(-3+4t-it)(4+i)$

However I don't know what limits to integrate between, is it 0 and 1 or -3 and 1

Also am I correct so far?

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What parameterization of $\gamma$ are you using? The limits will depend on it. –  Sigur Feb 11 '13 at 22:27

2 Answers 2

Yes, it's right so far.

The path is being parametrized by $$ t\mapsto \gamma(t)=-3+t((1+i)-(-3))=-3+t(4+i). $$ The endpoints are at $\gamma(0)=-3$ and $\gamma(1)=1+i$, so the integral should be taken from $t=0$ to $t=1$.

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Since you are using $\gamma(t)=(-3+4t(1+i))$ you can see that it will be the required segment line only if $0\leq t\leq 1$.

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