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I'm having trouble trying to find this integral, where $C$ is the semicircle, centre $z = 1$, of radius $1$, lying in the upper half-plane $$ \int_C \bar{z}\ {dz} $$

Currently I have that, $c(t)=1+e^{it}$, where $t$ is in $(0,\pi)$, and where $c$ is a parametrisation of the curve $C$.

So then I have that

$$ \int_C \bar{z}\ {dz}=\int_C (1+e^{-it})ie^{it}\ {dz}=\int_C (ie^{it}+i)\ {dz}=i\pi $$

Is this correct? Probably not, but I can't see where im going wrong, any help would be greatly appreciated. Furthermore apologies for the bad notation.

A.P.
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sean
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1 Answers1

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$\int\limits_C\bar{z}dz=\int\limits_0^{\pi}(1+e^{-it})ie^{it}dt=\int\limits_0^{\pi}ie^{it}dt+\int\limits_0^{\pi}idt=e^{i\pi}-1+i\pi=-2+i\pi.$

math
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