Can someone please help me setup
a) $\int _\Gamma z dz$
b) $\int _\Gamma \bar z dz$
and given the admissible parametrization of $\Gamma$
$\Gamma_1 : z_1: 2 + i(t - 1) ; 1 \leq t \leq 2$
and $\Gamma_2 : z_2: (3 - t)(2 + i) ; 2 \leq t \leq 3$
Is it $\int _\Gamma z dz = \int_{1}^{2} 2 + i(t - 1)dt + \int_{2}^{3} (3 - t)(2 + i) dt$ ?
For the part b) I don't know even how to start.
I would really appreciate if you anyone can set up the problem. Thank you
I think that what could help you is the expression $$\int_\Gamma f(z) dz = \int_a^b f(\Gamma(t))\Gamma'(t)dt.$$
For example, in the first case, your $F(z) = z$, $\Gamma(t) = 2 + i(t-1)$, $a = 1$, $b = 2$ and $\Gamma'(t) = i$. So you get that $$\int_\Gamma f(z) dz = \int_1^2 (2 + i(t-1))idt. = (2it - \frac{t^2}{2} + t)\big|_{t=a}^b = -\frac{1}{2} + 2i$$
For the second part, you will need to calculate $\overline{\Gamma(t)}$, but it shouldn't be too difficult.