What kind of points are z=0 and z=∞ for w=sqrt[n]z^k , where…
What kind of points are $z=0$ and $z=\infty$ for $w=\sqrt[n]{z^k}$, where $k$ is a prime number, $k\neq0$?
Explanation of Method
Write $w=z^{k/n}=r^{k/n}e^{ik\theta/n}$ and track $\arg w$ as $z$ circles the origin once; the reduced fraction $k'/n'=k/n$ (after dividing by $d=\gcd(k,n)$) determines exactly how many circuits are needed for $w$ to return to its original value. Repeat at infinity via $\zeta=1/z$.
Step 1 · One circuit around $z=0$
Let $z=\varepsilon e^{i\phi}$, $\phi:0\to2\pi$. Then $\arg(z^k)=k\phi$ changes by $2\pi k$, so:
$$ \Delta[\arg w]=\frac{2\pi k}n $$
Step 2 · Reduce the fraction
Let $d=\gcd(k,n)$ and write $k=dk',\ n=dn'$ with $\gcd(k',n')=1$, so $w=z^{k/n}=z^{k'/n'}$ (in lowest terms). Since $k$ is prime, $d$ is either $1$ (if $k\nmid n$) or $k$ (if $k\mid n$).
Step 3 · Number of circuits to return
With $k'/n'$ in lowest terms, $w$ returns to its original value only after $n'=n/d$ complete circuits around $z=0$ (since $n'\cdot\frac{2\pi k'}{n'}=2\pi k'$, a multiple of $2\pi$, and no smaller multiple works because $\gcd(k',n')=1$). Hence:
$$ \boxed{z=0 \text{ is a branch point of order } \frac nd-1} $$
Step 4 · Behavior at infinity
Substitute $\zeta=1/z$: $\ w=\zeta^{-k/n}$. As $\zeta$ circles $0$ once ($z$ circling $\infty$ once), $\arg w$ changes by $-\frac{2\pi k}n$ — the same magnitude as at the origin. By the identical argument:
$$ \boxed{z=\infty \text{ is also a branch point of order } \frac nd-1} $$
Applying $k$ prime
Since $k$ is prime, $d=\gcd(k,n)\in\{1,k\}$:
| Condition | $d$ | Branch order at $0$ and $\infty$ |
|---|---|---|
| $k\nmid n$ (typical case) | $1$ | $n-1$ |
| $k\mid n,\ n=k$ | $k$ | $0$ — not a branch point ($w=z$) |
| $k\mid n,\ n=mk,\ m>1$ | $k$ | $m-1$ |
Result
Only in the degenerate case $n=k$ does the exponent reduce to an integer ($w=z$), eliminating the branching entirely.