I have to integrate three functions over the quarter-circle from $z=4$ to $z=4i$.
The functions are: $z^2$, $|z|^2$, and $\bar{z}$.
I'm trying to parameterize them because the circle $C(t) = 4\cos t + 4i\sin t$ on $0 \leq t \leq \pi/2$
Then $x = \sqrt{16-y^2}$ and $y = t$ so $C(t) = \sqrt{16-t^2}+it$ and $C'(t) = \frac{-t}{\sqrt{16-t^2}}+i$
Then $\int_{C} z^2 dz$ becomes $\int_{0}^{4} C(t)^2 C'(t)dt$ for the first one, right? But how do I do the next two?
I think that the easiest way it to do this.
$C(t) = 4\cos t + 4i\sin t$ $f(z) = z^2\\ f(C(t)) = 16(\cos^2 t -\sin^2 t) + 16i(2\sin t\cos t)\\ f(C(t))C'(t) = (16(\cos^2 t -\sin^2 t) + 16i(2\sin t\cos t))(-4\sin t + i4\cos t) = 64(-3\cos^2 t\sin t + \sin^3t) + 64i(-3\sin^2 t\cos t +\cos^3 t)\\ $
$\int_0^\frac {\pi}{2} 64(-3\cos^2 t\sin t + \sin^3t) \ dt + i\int_0^\frac {\pi}{2} 64(-3\sin^2 t\cos t +\cos^3 t)\ dt$
$f(z) = |z|^2\\ f(C(t)) = 16(\cos^2 t + \sin^2 t)\\ f(C(t))C'(t) = (16(\cos^2 t +\sin^2 t)(-4\sin t + i4\cos t) = 64(-\cos^2 t\sin t - \sin^3 t) + 64i(\cos^3 +\cos t\sin^2 t)$
$f(z) = \bar z\\ f(C(t)) = 4(\cos t - \sin t)\\ f(C(t))C'(t) = \cdots$