Investigate the change of arg(sqrt[n]z-a) as z traces a…

Investigate the change of $\arg(\sqrt[n]{z-a})$ as $z$ traces a closed Jordan curve around $a$ once in the positive direction. What kind of point is $z=a$ for $w=\sqrt[n]{z-a}$? Is $z=\infty$ a critical point of this function?

Part 1 · Change in argument around $a$

As $z$ traces one full positive circuit of a Jordan curve enclosing $a$ and returns to its start, the point $z-a$ makes exactly one full revolution about the origin:

$$ \Delta[\arg(z-a)]=2\pi \qquad\Longrightarrow\qquad \Delta[\arg w]=\Delta\left[\frac1n\arg(z-a)\right]=\frac{2\pi}n $$

Since $\frac{2\pi}n$ is not a multiple of $2\pi$ (for $n\ge2$), $w$ does not return to its initial value after one circuit; it takes exactly $n$ circuits for $w$ to return, since $n\cdot\frac{2\pi}n=2\pi$.

Part 2 · Nature of the point $z=a$

The point $z=a$ is a branch point of order $n-1$: the function $w=\sqrt[n]{z-a}$ has exactly $n$ distinct branches (sheets), and $a$ is where all $n$ sheets meet. Although $w(a)=0$ is well-defined as a value, no single-valued continuous branch of $w$ exists in any punctured neighborhood of $a$.

Part 3 · Behavior at infinity

Substitute $\zeta=1/z$ (so $z\to\infty\Leftrightarrow\zeta\to0$):

$$ w=\sqrt[n]{\frac1\zeta-a}=\sqrt[n]{\frac{1-a\zeta}\zeta}=\zeta^{-1/n}(1-a\zeta)^{1/n} $$

For $\zeta\to0$ (i.e. $z\to\infty$): $\ w\approx\zeta^{-1/n}=z^{1/n}$. As $\zeta$ traces one small loop around $0$ (i.e. $z$ traces one large loop around $\infty$), $\arg\zeta$ changes by $2\pi$, so:

$$ \Delta[\arg w]=-\frac1n\Delta[\arg\zeta]=-\frac{2\pi}n\neq0 $$

so $w$ again fails to return to its original value — $z=\infty$ is also a branch point.