Let $\lambda$ be the following contour oriented clockwise: the real axis from $0$ to $2$, followed by an arc of the circle $|z|=2$ from $2$ to $-2i$, followed by the imaginary axis from $-2i$ to $0$. Calculate the following:
$\Large \int_{\lambda}\overline z dz$ and $\Large \int_{\lambda}e^zdz$
For the first one:
First path: $z(t)=2t$ when $0\leq t\leq 1$, $z'(t)=2$
Second path: not entirely sure how to denote the arc can anyone clarify?
I thought perhaps $z(t)=2e^{it}$ when $0\leq t\leq \pi$, $z'(t)=2ie^{it}$
Third path: $z(t)=-2i+2it$ for $0 \leq t \leq 1$, $z'(t)=2i$
Input into $\int_{\lambda}\overline z dz$:
$\int_{0}^1(2t)\cdot 2 dt+\int_0^{\pi}2e^{-it}\cdot 2ie^{it}dt+\int_0^1(2i-2it)\cdot 2i dt$
Is this correct?
Second one:
Same paths as before but
$\int_{0}^1e^{2t}\cdot 2 dt+\int_0^{\pi}e^{2e^{it}}\cdot 2ie^{it}dt+\int_0^1e^{(-2i+2it)}\cdot 2i dt$
Again, is this correct and what would I do about the $e^{2e^{it}}$ term if so?
You have tacitly assumed that the "arc of the circle $|z|=2$" is meant to be counterclockwise. That's o.k. with me, but the parametrization of this arc is $$t\mapsto 2e^{it}\qquad\left(0\leq t\leq{3\pi\over2}\right)\ .$$ Otherwise your computation of $\int_\lambda \bar z\>dz$ is correct.
As for the integral $\int_\lambda e^z\>dz$ note that the integrand is analytic in the full plane ${\mathbb C}$; furthermore $\lambda$ is a cycle (a closed curve). It follows that $\int_\lambda e^z\>dz=0$, without any calculation.