Given question
Evaluate $\displaystyle\int_C(z^2+3z)\,\mathbb dz$ along
- The circle $|z|=2$ from $(2,0)$ to $(0,2)$ in a counterclockwise direction.
- The straight line from $(2,0)$ to $(0,2)$.
- The straight line from $(2,0)$ to $(2,2)$ and then $(2,2)$ to $(0,2)$.
Actually all of those cases give me the answer $-\dfrac{44}{3}-\dfrac{8}{3}i$, and sorry i can't give the whole of my answer in detail, cz it'll consume a lot of time. But, i'm pretty sure i didn't make a mistake cz i did it carefully.
Then i tried to checked the function with substituting $z=x+iy$, it satisfied the C-R equation.
Could i state, in the same initial point and end point, if the path of the curve that is the function is holomorphic, then it gives the same result no matter which path that i choose?
Really? Is it just a nice coincidence or that is so?
Note that, $g:z\mapsto\frac{z^3}{3}+\frac{3z^2}{2},z\in\Bbb C$ is an antiderivative of $f:z\mapsto z^2+3z,z\in \Bbb C$ i.e. $g'=f$. Hence for any smooth parametrization $\gamma:[0,1]\to\Bbb C$ of $C$ we have, $$\int_Cf(z)dz=\int_0^1f(\gamma(t))\cdot\gamma'(t)dt$$$$=\int_0^1g'(\gamma(t))\cdot\gamma'(t)dt$$$$=\int_0^1\frac{d}{dt}\big(g(\gamma(t))\big)dt$$$$=g(\gamma(1))-g(\gamma(0))$$$$=g(0,2)-g(2,0).$$
So as long as you can find antiderivative of an analytic function defined on an open set, the integral over any path of this function only depends on the two end points of the path, irrespective of how the path actually looks like.