Assume $G = \{z \in \mathbb{C}: \vert z \vert < 1 \quad and \quad Re(z) + Im(z) > 1 \}$. Construct a path $ \gamma: [a,b] \rightarrow \mathbb{C} $ with $\gamma([a,b])=\partial G$ and calculate $\int_\gamma Im(z) dz $ and $\int_\gamma Re(z)dz$. Furthermore calculate and interprete $\int_\gamma \bar{z}dz$.
Hint: Use the Leibniz sector formula.
My problem is that I'm not sure if I'm understanding this exercise right because I didn't use the sector formula for my result. In my opinion $G$ is a circular segment in the first quadrant of the complex plane. Therefore I can easily construct the path:
$\gamma_1: [0,1] \rightarrow \mathbb{C}, t \mapsto t+ i(1-t)$
$\gamma_2: [0,\pi/2] \rightarrow \mathbb{C}, t \mapsto \cos(t)+ i\sin(t)$
With these paths the results of my integrals are (results edited):
$\int_\gamma Im(z) dz = \frac{1}{2}(1-\frac{\pi}{2})$ and $\int_\gamma Re(z)dz = \frac{i}{2}(\frac{\pi}{2}-1)$
Since $\bar{z}=Re(z)-iIm(z)$ I can write: $\int_\gamma \bar{z}dz = \int_\gamma Re(z)-iIm(z)dz = \int_\gamma Re(z) dz - i\int_\gamma Im(z) dz=i(\frac{\pi}{2}-1)$
The name "Leibnitz Sector Formula" is not one I am familiar with (almost certainly know the formula, but not this name for it). And attempts to search for it online turned up only a few mentions, none of which actually stated what it was. So I am unable to address the point of the hint.
You are correct about the boundary. But I do not get the same answer you did. For $\gamma_1$, note that $\gamma_1'(t)= 1-i$, so $$\int_{\gamma_1}Re(z)dz = \int_0^1(t)(1-i)dt$$
$\gamma_2'(t) = -\sin t + i\cos t$, so $$\int_{\gamma_2}Re(z)dz = \int_0^{\pi/2}\cos t(-\sin t + i\cos t)dt$$ These do not simplify to $\frac {\sqrt 2}2 + 1$