Show that the points ± i are branch points of the function…

Show that the points $\pm i$ are branch points of the function $w=(z^2+1)^{1/3}$.

Step 1 · Locate the zeros of $z^2+1$

$$ z^2+1=0 \ \Longrightarrow\ z=\pm i $$

Step 2 · Circuit around $z=i$

Let $z=i+\varepsilon e^{i\phi}$ with $\varepsilon$ small and $\phi:0\to2\pi$ (one loop around $i$). Then:

$$ z^2+1=(i+\varepsilon e^{i\phi})^2+1=\underbrace{i^2+1}_{=0}+2i\varepsilon e^{i\phi}+O(\varepsilon^2)=2i\varepsilon e^{i\phi}+O(\varepsilon^2) $$

so for small $\varepsilon$: $\ \arg(z^2+1)\approx\phi+\frac\pi2$, which increases by exactly $2\pi$ as $\phi$ sweeps $0\to2\pi$.

Step 3 · Effect on $w$

Since $w=(z^2+1)^{1/3}=|z^2+1|^{1/3}e^{i\arg(z^2+1)/3}$, one full loop around $i$ changes $\arg w$ by only $\frac13(2\pi)=\frac{2\pi}3$ — not a multiple of $2\pi$, so $w$ does not return to its starting value.

Step 4 · Circuit around $z=-i$

Similarly, with $z=-i+\varepsilon e^{i\phi}$: $\ z^2+1=-2i\varepsilon e^{i\phi}+O(\varepsilon^2)$, and $\arg(z^2+1)$ again increases by $2\pi$ over one loop, so $\arg w$ again changes by $\frac{2\pi}3$.

Conclusion

Both $z=i$ and $z=-i$ are branch points. Since $\frac13(2\pi)$ is not a multiple of $2\pi$, three complete loops around either point are required before $w$ returns to its original value — each is a branch point of order $2$ (a $3$-valued root has $n-1=2$ sheets meeting there), and the function has three branches overall.

As $z\to\infty$, $z^2+1\sim z^2$ so $w\sim z^{2/3}$, which also fails to return after one large loop — $z=\infty$ is a third branch point of this function.