How calculate $ \int_C \bar {z} ^2dz$ if $C$ is a circle such that $ |z| = 1 $

Question

How calculate $ \int_C \bar {z} ^2dz$ if $C$ is a circle such that $ |z| = 1 $

for $z=x+iy \in C$

$\int\limits_{C}{\bar{z}^2 \ dz}=\int\limits_{C}({(x-iy) \ (dx+i\:dy))^2}= \int\limits_{C} \left(xa+yb+ixb-iya\right)^2= \\ \int\limits_{C}{\left(a^2x^2-b^2x^2+4abxy+b^2y^2-a^2y^2\right)}+i\int\limits_{C}{(\left(2abx^2-2a^2xy+2b^2xy-2aby^2\right)} $

I don't know if I'm on the right track and how to use the fact that the circumference has radius 1

Answer 1

Since $|z| = 1$, $z\bar z = 1$. Therefore, you may write $$\int_C \bar z^2 \, dz = \int_C \frac{dz}{z^2}$$ It is well known that $$\int_C \frac{dz}{z^2} = 0$$ since $\frac{1}{z^2}$ has a primitive for $z\ne 0$, namely $-\frac{1}{z}$.

Answer 2

Parameterize by $z=e^{i\theta}$, for $0\leq \theta \leq 2 \pi$. Then $z=\cos\theta+i\sin\theta$ so $\bar{z}=\cos\theta-i\sin\theta=\cos(-\theta)+i\sin(-\theta)=e^{-i\theta}.$ It is easy to see then that $z'(\theta)=ie^{i\theta}$ so \begin{align} \int_C\bar{z}^2dz=i\int_0^{2\pi}e^{-i\theta}d\theta=0. \end{align} Of course, as @peek-a-boo suggested, we also know that $\bar{z}^2=1/z^2,$ which has an antiderivative function around the unit circle, so integrates to zero.