= Solution
There is a small domain issue in the printed notation. If $\mathbb R^3/\mathbb R^2$ means a <quotient vector space> by the horizontal plane, the displayed formula does not descend to that <quotient vector space>: $(0,0,0)$ and $(1,0,0)$ represent the same class but give different values. The intended construction is <stereographic projection>, restricted to the unit <sphere> with the north pole removed. Interpreting the slash as removal of the plane $z=1$ also supplies a suitable ambient domain. The <holomorphic stereographic atlas of the sphere> is obtained as follows.
Write $N=(0,0,1)$ and $S=(0,0,-1)$. On $U_N=S^2\setminus\{N\}$ use $\zeta=(x+iy)/(1-z)$. Its inverse, with $\zeta=u+iv$, is
$$
x=\frac{2u}{1+u^2+v^2},\qquad
y=\frac{2v}{1+u^2+v^2},\qquad
z=\frac{u^2+v^2-1}{1+u^2+v^2}.
$$
These formulas give a smooth <manifold chart> from $U_N$ onto $\mathbb C$. On $U_S=S^2\setminus\{S\}$ choose the second <manifold chart>
$$
\eta=\frac{x-iy}{1+z}.
$$
Its inverse is $x=2\operatorname{Re}\eta/(1+|\eta|^2)$, $y=-2\operatorname{Im}\eta/(1+|\eta|^2)$ and $z=(1-|\eta|^2)/(1+|\eta|^2)$, so this is also a smooth <manifold chart> onto $\mathbb C$. The conjugation in this second <stereographic projection> is essential. On the overlap, $x^2+y^2=1-z^2$, so
$$
\zeta\eta=\frac{x^2+y^2}{1-z^2}=1,\qquad
\boxed{\eta=\zeta^{-1}}.
$$
Both directions of this transition are <holomorphic maps> on $\mathbb C^\times$, with nonzero derivative. The two <manifold charts> cover the <sphere>, hence define a <holomorphic atlas>, giving precisely the <Riemann sphere>. If one instead used $x+iy$ in both <manifold charts>, the transition would be $1/\bar\zeta$ and would not be <holomorphic>.
Orient the <sphere> by this <holomorphic atlas>. The given <volume form> is smooth at infinity: replacing $\zeta$ by $1/\eta$ in its <exterior product> gives the same expression
$$
\Omega=\frac{i\,d\eta\wedge d\bar\eta}{(1+|\eta|^2)^2}.
$$
Moreover, $i\,d\zeta\wedge d\bar\zeta=2\,du\wedge dv$, so the normalization of this <volume form> is
$$
\int_{S^2}\Omega
=2\int_0^{2\pi}\int_0^\infty\frac{r}{(1+r^2)^2}\,dr\,d\theta
=2\pi.
$$
Thus the printed <volume form> has half the area of the standard round unit <sphere>; replacing its integral by $4\pi$ would introduce an erroneous factor of two.
For $k\ge1$, the <holomorphic map> $f(\zeta)=\zeta^k$ extends over infinity, since the target reciprocal coordinate is $\eta^k$ when the source reciprocal coordinate is $\eta$. Its <pullback of a differential form> is
$$
f^*\Omega=\frac{i\,k^2|\zeta|^{2k-2}\,d\zeta\wedge d\bar\zeta}
{(1+|\zeta|^{2k})^2}.
$$
Using $s=r^{2k}$ gives
$$
\int_{S^2}f^*\Omega
=4\pi k^2\int_0^\infty\frac{r^{2k-1}}{(1+r^{2k})^2}\,dr
=2\pi k\int_0^\infty\frac{ds}{(1+s)^2}
=2\pi k.
$$
The <degree of a map between oriented manifolds> is therefore
$$
\boxed{\deg f=\frac{\int_{S^2}f^*\Omega}{\int_{S^2}\Omega}=k}.
$$
For $k=0$, the formula on the finite <manifold chart> is the constant $1$. Its unique continuous extension is also $1$ at infinity, rather than an undefined expression $\infty^0$. Its <pullback of a differential form> is zero and its <degree of a map between oriented manifolds> is \b[zero].
For the preimage calculation when $k\ge1$, choose a <regular value> $w\in\mathbb C^\times$. There are exactly $k$ distinct roots of $\zeta^k=w$. At each root the real <Jacobian determinant> is $|k\zeta^{k-1}|^2>0$, so every local contribution to the <degree of a map between oriented manifolds> is $+1$. Their sum is $k$, agreeing with the integral. The exceptional values $0$ and infinity are avoided because they are branch values when $k>1$. For the constant map, any $w\ne1$ is a <regular value> with no preimages, giving the same answer zero. This establishes the <degree of a power map of the Riemann sphere> for every allowed $k$.
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