$g:\mathbb{C}\rightarrow \mathbb{C}\quad g(z) \equiv 5 \quad z=x+iy $
What is the line integral over the closed curve: $x^2+y^2=9$
Intuitively I would say I add up a bunch of (5*infitesimal) terms and I would get $2r\pi*hight\rightarrow 2*3*5\pi$
but according to Cauchy's Integral theorem it has to be 0 so some terms must be negative and since the bunch of fives cannot be then the minuses must be among the infinitesimals. Even if that's true why isn't that logic applicable to the same function but and curve only considering reals where the actual solution is $2*3*5\pi $ $f: \mathbb{R}^2\rightarrow \mathbb{R} \quad f(x,y) \equiv 5 $
Basically by giving no imaginary part to g(z)=5 I kind of described the same funtion right? (I know I'm wrong but..)
Is it because of the $i$-s getting realised as minus sign?
Where are the minus signs coming from?
$\oint 5 dr\\ r = 3\cos t + i \sin t\\ dr = -3 \sin t + i \cos t\\ \int_0^{2\pi} -15 \sin t + 15i \cos t dt$
Now rather than mechanically integrate this and getting $0.$ Lets think about what is going on.
Half of your infinitessimals have a real compontent that is moving from left to right, and half have a real component that is moving right to left. And simillarly for the imaginary component half are in the $+i$ direction and half are in the $-i$ direction.
Does this help?