Existence of distinct roots of a shifted complex polynomial

Question

Question: Let $f(x)\in \mathbb{C}[x]$. Does there always exist some $\alpha \in \mathbb{C}$ such that $g(x):=f(x)-\alpha$ has distinct roots?

My intuition leads me to believe this is true. For a simple example, if $f(x)=x^n$ then we can set $\alpha=1$ so that the roots of $g(x)$ are the $n$th roots of unity. Proceeding via contradiction, if there were some $f(x)$ such that for all $\alpha\in \mathbb{C}$, $g(x)$ does not have distinct roots, what goes wrong? Taking this route, such a $g(x)$ would be inseparable, so for any $\alpha$ we see that $g(x)$ would always have a common root with its derivative. Can anyone find a contradiction? Is there a simple direct proof? Is the statement incorrect and can someone provide a counterexample?

Answer 1

Yes, such $\alpha$ exists. Note that if $w$ is a double root for the polynomial $g(z)=f(z)-\alpha$ then $g(z)=(z-w)^2q(z)$ and $g'(z)=(z-w) (2q(z)+(z-w)q'(z))$ which implies that $g'(w)=f'(w)=0$. Hence just take $\alpha$ such that $g(w)=f(w)-\alpha\not=0$ when $f'(w)=0$, that is $$\alpha\in \mathbb{C}\setminus\{f(w) : f'(w)=0\}$$ (we are assuming that the degree of $f$ is greater than one).

Answer 2

Suppose $f$ has distinct roots $\lambda_k$ of multiplicity $m_k$. Let $B_k$ be balls centered on $\lambda_k$ whose radii are small enough so the closures do not intersect.

Furthermore, we may choose the radii small enough such that if $m>1$, then $f'$ has exactly $m-1$ zeros in $B_k$.

Continuity of zeros shows that there is some $K>0$ such that if $|\alpha| < K$, then $p_\alpha(x)=f(x)-\alpha$ will have exactly $m_k$ roots in $B_k$.

Suppose $\lambda_k$ is one of these roots with multiplicity $m_k>1$. Note that $p_\alpha'=f'$ has exactly $m_k-1$ roots in $B_k$ and these are $\lambda_k,...,\lambda_k$ ($m-1$ times).

Now suppose $\alpha$ is such that $0 < |\alpha| < K$, and let $z_1,...,z_m$ be the roots of $p_\alpha$ in $B_k$. I claim that the $z_i$ are distinct. Since $p_\alpha(\lambda_k) = f(\lambda_k) -\alpha \neq 0$ we see that $z_i \neq \lambda_k$ and hence $p_\alpha'(z_i) \neq 0$.

Hence there is some $K>0$ such that for any $\alpha$ satisfying $0 < |\alpha| < K$, then $p_\alpha$ has exactly $m_k$ distinct roots in $B_k$.

By further restricting $K$, we can repeat the process for any other root of multiplicity greater than one until we find a $K'>0$ such that for any $\alpha$ satisfying $0 < |\alpha| < K'$, then $p_\alpha$ has distinct roots.