# A polynomial with at least a simple root

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?

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

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.
• Nice (note that the last gcd argument is incorrect if $f'=0$).