I'm quite new to complex analysis and have this question:
Evaluate $\ \int_C \bar z dz$, where C is the unit circle in a counterclockwise direction
What I have done is: $$\ \int_C \bar z dz = \int_0^{2\pi} \overline {e^{it}}. ie^{it}dt = \int_0^{2\pi} {e^{-it}}. ie^{it}dt = \int_0^{2\pi}idt $$ $$\ =2\pi i$$
Is this right or have I gone wrong somewhere, I think this seems too straightforward?
On
$$ \oint_{C}\bar{z}\,\mathrm{d}z = \oint_{C}{\bar{z}\,z \over z}\,\mathrm{d}z = \oint_{C}{\mathrm{d}z \over z} = \ \bbox[10px,border:1px dotted navy]{\displaystyle{2\pi\,\mathrm{i}}} $$
Note that $\displaystyle{\bar{z}\,z = \left\vert\, z\,\right\vert^{2} = 1}$ because $\displaystyle{C}$ is the $\underline{\mbox{unit circle}}$.
On
on the unit circle $\bar{z}=\frac{1}{z}$, so $\int_{C}\bar{z} \, dz$ is $\int_{C}z^{-1} \, dz$, and it is a standard theorem in complex analysis that $\int_{C}z^{-1} \, dz$ is $2\pi i$. And, to prove $2\pi i=\int_{C}z^{-1} \, dz$, the idea would be what you just wrote above (let $z=e^{it}$) and so on.
Yes, your computation is correct.
Note that if $\gamma$ is a simple bounded regular closed curve counterclockwise oriented then $$\int_\gamma \bar z dz=\int_\gamma (x-iy) d(x+iy)=\int_\gamma (xdx+ydy)+i\int_\gamma(xdy-ydx)\\\stackrel{GT}{=}\iint_D (0-0)dxdy+i\iint_D (1-(-1))dxdy=2i\cdot \mbox{Area}(D)$$ where we used the Green's theorem (GT) and $D$ is the region bounded by $\gamma$.