= Solution
The argument principle says $(2\pi i)^{-1}\int_\gamma f\prime/f$ equals zeros minus poles. Applying it to the homotopy $f+tg$ proves Rouché when $|g|\lt |f|$ on the boundary. On $|z|=1$, $15z$ dominates $z^6+1$, so there is one zero inside; on $|z|=2$, $z^6$ dominates $15z+1$, so there are six. Hence the annulus contains five. For the <limit> theorem, if nonconstant $f$ took the same value at two points, use disjoint small circles and Rouché on $f_n-f_n(a)$ to force a second preimage, contradicting injectivity.
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