Contour integral for $\bar{z}^3$

Question

Consider the curve $C: [0,\frac{\pi}{2}] \to \mathbb{C},\,C(t) = 2\exp(-it)$. Is it true that $$ \begin{align*} \int_C \bar{z}^3\,\mathrm dz &= \int_0^\frac{\pi}{2} \overline{(2 \exp(-it))}^3 (-2i\exp(-it))\,\mathrm dt \\ &= -16i \int_0^\frac{\pi}{2} \exp(2it)\,\mathrm dt \\ &= \left[ -8\exp(2it) \right]_0^\frac{\pi}{2} \\ &= 16. \end{align*}$$ I am new to this and i am wondering if i did this the right way.

Answer 1

Hint. For $R>0$ and $n\in\mathbb Z$ it holds that $$\oint_{|z|=R}\overline{z}^n\,\mathrm dz = \int_0^{2\pi}(R\mathrm e^{-\mathrm it})^n\cdot\mathrm iR\mathrm e^{\mathrm it}\,\mathrm dt = \mathrm iR^{n+1}\int_0^{2\pi}\mathrm e^{-\mathrm i(n-1)t}\,\mathrm dt=2\pi\mathrm i R^2\delta_{n,1}.$$ Can you adapt this for your needs? Just keep in mind to reverse the direction since your $C$ has a different orientation as well as not taking the entire circle as I did. Furthermore keep in mind to take $C'(t)$ as well like Argon mentioned it in the comments.