Complex analysis computing the integral

Question

I=$\int_\gamma \operatorname{Im}(z)\mathrm dz$

$\gamma$ is the interval between $\omega_1=0$ and $\omega_2=1+2\mathrm i$

How can I compute the integral above? I have no process about this question. I`m sorry about it.

Answer 1

Just apply the definition: If $D\subseteq\mathbb C$, $\gamma:[a,b]\to D$ is the parametrization of a smooth curve and $f:D\to\mathbb C$, then

$$\int_\gamma f(z)\mathrm dz:=\int_a^b \gamma'(t) f(\gamma(t))\mathrm dt.$$

In your case, $\gamma:[0,1]\to\mathbb C,~\gamma(t)=(1+2\mathrm i)t$ is a suitable parametrization, and $f(z)=\operatorname{Im}z$. Plugging everything in:

$$\int_\gamma\operatorname{Im}z\mathrm dz=\int_0^1(1+2\mathrm i)\operatorname{Im}((1+2\mathrm i)t)\mathrm dt=\int_0^1(1+2\mathrm i)2t\mathrm dt.$$

I think you can do the rest by yourself.