Let $f\in \mathbb{R}[x]$ be a non-constant polynomial and $a,b\in\mathbb{R}$ be s.t. $b\not= 0$.

Then —I think so— (*) The equation, in $x\in\mathbb{C}$, $f(x)-a-ib=0$ admits at least one simple root.

For example, one can prove (using the Gröbner basis theory) that $(x-a_1-ib_1)^2(x-a_2-ib_2)^2\cdots (x-a_6-ib_6)^2$, where $a_i,b_i\in\mathbb{R}$, $b_i\ne 0$, is never in the form $f(x)-a-ib$.

Yet $f(x)-a-ib$ may have multiple roots:


Question. Is the above statement (*) true?

  • 2
    $\begingroup$ What is the question? $\endgroup$
    – YCor
    Commented Jun 25 at 14:29
  • 1
    $\begingroup$ ($a$ is unnecessary since you can replace $f$ with $f-a$) $\endgroup$
    – YCor
    Commented Jun 25 at 14:30
  • $\begingroup$ ($b$ is unnecessary since you can replace $f-a$ by $(f-a)/b$, hence $b=1$ is enough) $\endgroup$ Commented Jun 25 at 14:32
  • 1
    $\begingroup$ @YCor Jochen does not say that you can replace $b$ by $0$, but that you can replace $b$ by $1$. $\endgroup$ Commented Jun 25 at 14:33
  • $\begingroup$ @AlekseiKulikov I know, I erased it before you replied. Sorry. $\endgroup$
    – YCor
    Commented Jun 25 at 14:34

1 Answer 1


Yes. Let $g$ be a polynomial. If $g$ has no simple roots then the degree of $\gcd(g, g')$ is at least half the degree of $g$. (Each root of $g$ with multiplicity $m$ is a root of $\gcd(g, g')$ with multiplicity $m-1$ and $m-1 \geq m/2$ for $m\geq 2$.)

If $f(x)-a-bi$ has no simple roots then so does $f(x)-a+bi$. Both these polynomials have derivative $f'$ so $\gcd ( f-a -bi, f') \gcd(f-a+bi,f')$ has degree at least $\deg f$.

But $f-a-bi$ and $f-a+bi$ are coprime since their difference $2bi$ is invertible. It follows that $\gcd ( f-a -bi, f') \gcd(f-a+bi,f')=\gcd ( (f-a -bi)(f-a+bi),f') $ divides $f'$, hence (because $f$ is nonconstant and thus $f'\neq 0$) $f'$ has degree at least $\deg f$, a contradiction.

  • 2
    $\begingroup$ This is unexpectedly elegant! $\endgroup$ Commented Jun 25 at 14:42
  • $\begingroup$ Nice (note that the last gcd argument is incorrect if $f'=0$). $\endgroup$
    – YCor
    Commented Jun 25 at 14:43
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    $\begingroup$ Biblical simplicity. Bravissimo. $\endgroup$
    – loup blanc
    Commented Jun 25 at 15:08
  • $\begingroup$ The question was asked and answered before: mathoverflow.net/q/81732 $\endgroup$
    – GH from MO
    Commented Jun 26 at 0:50

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