I need help parametrizing. I never really understood how to parametrize. I know that
$C_1:$ $0$ to $1$ $\Rightarrow$ $z= ?$
$C_2:$ $1$ to $i$ $\Rightarrow$ $z= ?$
and $C_3:$ $i$ to $0$ $\Rightarrow$ $z= ?$.
I don't understand how to get $z$. Could someone explain what my $z$'s are and how you obtain them.
On
Identifying $\;C\cong\Bbb R^2\;$, the line from point $\;A(x_1,y_1)\;$ to point $\;B(x_2,y_2)\;$ can easily be parametrized as
$$r(t):=tB+(1-t)A=\left((1-t)x_1+tx_2\,,\,\,(1-t)y_2+ty_1\right)\;,\;\;0\le t\le1$$
For example, from $\;1\to i\;$ is line from $\;(1,0)\to (0,1)\;$ , so you can parametrize
$$r(t):=t(0,1)+(1-t)(1-0)=(1-t,t)\;,\;\;t\in[0,1]\;\implies$$
if
$$f(z)=e^{\pi \overline z}\;,\;\;\text{then}\;f(r(t))=e^{\pi(1-t-it)}=e^{\pi(1-t)}\left(\cos\pi t,\, i\sin\pi t\right)\;,\;0\le t\le1\;,\;\;\text{and}$$
$$\;r'(t)=(-1,1)\implies f(r(t))\cdot r'(t)=-e^{\pi(1-t)}\cos t+ie^{\pi(1-t)}\sin t$$
and now just integrate
$$\int_0^1\left[-e^{\pi(1-t)}\cos t+ie^{\pi(1-t)}\sin t\right]dt$$
Just define:$$C(t)=\begin{cases}t&\text{ if }t\in[0,1]\\2-t+(t-1)i&\text{ if }t\in[1,2]\\(3-t)i&\text{ if }t\in[2,3].\end{cases}$$Indeed, if I want to go from a point $A$ to a point $B$, I use the parametrization $t\mapsto(1-t)A+tB$ ($t\in[0,1]$). That's how I got the first part, in order to go from $0$ to $1$. Then, when $t$ goes from $1$ to $2$, $t-1$ goes from $0$ to $1$, and so$$\bigl(1-(t-1)\bigr)+(t-1)i=(2-t)+(t-1)i$$goes from $1$ to $i$. And a similar apporach allows me to construct the third branch.