How to parametrize a complex curve I just started Complex Analysis course in my Uni! We are at the part of Complex Line Integrals! There is something there saying something about the Parametrization of line integrals and i wanted to know how you can do that! I read the description in my slides but i can't figure it out how to do it!
Can someone help me understand how to do them! Ideally by providing some examples! If you can't can you help me solve this?
C segment from 1 + i to 3
f(z)=iz
 A: To calculate an integral over a complex line froma definition (and not using theorems like the Cauchy's theorem) you need a parametrization of the curve.
A paramterization of a straight line from $z_1$ to $z_2$ is
$$ z(t) = z_1 + t (z_2-z_1), \qquad t\in[0,1]$$
Another useful curve (not in your specific problem, just in general) is an arc of a circle. It can be parametrized as
$$ z(t) = z_0 + R e^{it}$$
when going counterclockwise or
$$ z(t) = z_0 + R e^{-it}$$
when going clockwise, where $z_0$ is the center of the circle, $R$ is the radius, and you need to choose the range of $t$ to fit the initial and final points of the arc.
One you have the parametrization, the integral is defined as
$$ \int_\gamma f(z) dz : = \int_{t_1}^{t_2} f(z(t)) \frac{dz}{dt} dt = \int_{t_1}^{t_2} {\rm Re}\big( f(z(t)) \frac{dz}{dt} \big)  dt + i \int_{t_1}^{t_2} {\rm Im}\big( f(z(t)) \frac{dz}{dt} \big) dt$$
If a curve is more complex, but it can be decomposed into straight lines and arcs, you write the integral as sum of the integrals over each segment and calculate them separately. If it involves a more difficult curve, you need a parametrization of this curve or use some theorems about integrals over complex curves.
Taking your specific example, you need a straight line from $1+i$ to $3$. It can be parametrized as $$ z(t) = 1+i+ t (2-i), \qquad t\in[0,1]$$
Therefore we have
$$ \int_\gamma f(z)dz = \int_0^1 iz(t) \frac{dz}{dt} dt = \int_0^1 i\big(1+i+ t (2-i) \big)(2-i) dt$$
A: To parameterize the line segment, do it just like you would do it in $\mathbb{R}^2$: $$z(t) = (1-t)(1+i)+3t,\ 0\leq t\leq 1$$
By definition, $\mathrm{d}z=z'(t)\mathrm{d}t= (-1-i+3)\mathrm{d} t$, so the integral is $$\int_0^1 i((1-t)(1+i)+3t))(2-i)\mathrm{d}t$$
