 Solution

Here the target $C$ of $g$ is the unit circle. The boundary loop $g(e^{2\pi it})$ extends through the disk, so the radial homotopy $g(s{e}^{2\pi it})$ contracts it to a constant loop. Thus its winding number is zero. On the boundary put $h(z)=g(z)/z^n$, also circle-valued. Adding argument lifts gives $w(h)=0-n=-n$. If $z^n+g(z)$ never vanished, $h$ would avoid $-1$, which is the only negative real point of the unit circle. Part (iii) would give $w(h)=0$, contradicting $n\ne 0$. Therefore there is a boundary point with $z^n+g(z)=0$. The circle-valued hypothesis is essential; it is not a claim about arbitrary complex-valued $g$.