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:

$$x^3/3+x^2/2+x-(-5/12+i\sqrt{3}/4)=1/24(2x+1+2i\sqrt{3})(i\sqrt{3}-2x-1)^2.$$

**Question.** Is the above statement (*) true?

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