= Solution
{c}
Suppose, for a contradiction, that $p$ and $q$ have no common <root of a polynomial>. Since the <polynomial ring> $\mathbb C[x]$ has no <zero divisors> and $p,q\ne0$, the relation $ap+bq=0$ with $a,b$ not both zero forces both $a$ and $b$ to be nonzero.
By the <fundamental theorem of algebra>, write
$$
p(x)=c\prod_{j=1}^r(x-\alpha_j)^{m_j},\qquad c\ne0,\quad \sum_{j=1}^r m_j=n,
$$
where the $\alpha_j$ are distinct. For each $j$, $q(\alpha_j)\ne0$. The equality $bq=-ap$ implies that $(x-\alpha_j)^{m_j}$ divides $bq$. The <multiplicity of a root> in <polynomial multiplication> is the sum of its multiplicities in the factors: factor out the largest powers of $x-\alpha_j$ and note that the two remaining factors have nonzero values at $\alpha_j$. Since $q(\alpha_j)\ne0$, $b$ must contain $(x-\alpha_j)^{m_j}$.
Thus $b$ has at least $\sum_jm_j=n$ <roots of a polynomial> counted with <multiplicity of a root>, so $\deg b\geq n$. This contradicts $b\in V_{n-1}$. Consequently
$$
\boxed{\text{there is }\alpha\in\mathbb C\text{ with }p(\alpha)=q(\alpha)=0.}
$$
Counting <multiplicity of a root> is necessary: counting only distinct <roots of a polynomial> would miss repeated factors.
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