I am reading "Calculus of Several Variables 3rd Edition" by Serge Lang.
Let $G(x,y) := (\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})$.
Then, $\phi(x,y):=\arg(x+iy)$ is a potential function for the vector field $G$ on the plane from which the shaded region has been deleted. (Please see the picture below.)
I proved the above fact.
But I am not sure that I am right or not.
Is my answer right?
My solution is here:
If $0\leq\arg(x+iy)<\frac{\pi}{2}$, then $\phi(x,y)=\arctan(\frac{y}{x})$.
$\nabla \phi = G(x,y)$.
If $\arg(x+iy)=\frac{\pi}{2}$, then $\phi(x,y)=\frac{\pi}{2}$.
$\lim_{h\to0, x>0} \frac{\arg(h+iy)-\arg(0+iy)}{h} = \lim_{h\to0, h>0} \frac{\arctan(\frac{y}{h})-\frac{\pi}{2}}{h} = \frac{-1}{y}$ by L'Hospital's rule. $\lim_{h\to0, h<0} \frac{\arg(h+iy)-\arg(0+iy)}{h} = \lim_{h\to0, h>0} \frac{\arctan(\frac{y}{h})+\pi-\frac{\pi}{2}}{h} = \frac{-1}{y}$ by L'Hospital's rule.
$\lim_{h\to0} \frac{\arg(0+i(y+h))-\arg(0+iy)}{h} = \lim_{h\to0} \frac{\frac{\pi}{2}-\frac{\pi}{2}}{h} = 0$.
So, $\nabla \phi(0,y) = G(0,y)$.
If $\frac{\pi}{2}<\arg(x+iy)<\frac{3\pi}{2}$, then $\phi(x,y)=\arctan(\frac{y}{x})+\pi$.
$\nabla \phi = G(x,y)$.
If $\arg(x+iy)=\frac{3\pi}{2}$, then $\phi(x,y)=\frac{3\pi}{2}$.
$\lim_{h\to0, x>0} \frac{\arg(h+iy)-\arg(0+iy)}{h} = \lim_{h\to0, h>0} \frac{\arctan(\frac{y}{h})+2\pi-\frac{3\pi}{2}}{h} = \frac{-1}{y}$ by L'Hospital's rule. $\lim_{h\to0, h<0} \frac{\arg(h+iy)-\arg(0+iy)}{h} = \lim_{h\to0, h>0} \frac{\arctan(\frac{y}{h})+\pi-\frac{3\pi}{2}}{h} = \frac{-1}{y}$ by L'Hospital's rule.
$\lim_{h\to0} \frac{\arg(0+i(y+h))-\arg(0+iy)}{h} = \lim_{h\to0} \frac{\frac{3\pi}{2}-\frac{3\pi}{2}}{h} = 0$.
So, $\nabla \phi(0,y) = G(0,y)$.
If $\frac{3\pi}{2}\leq\arg(x+iy)\leq 2\pi-c$, then $\phi(x,y)=\arctan(\frac{y}{x})+2\pi$.
$\nabla \phi = G(x,y)$.
