If $p(z)$ is a monic polynomial then $p(z)+b=(z-z_1)(z-z_2)\cdots (z-z_n)$

Question

I need some help with this problem:

If $p(z)$ is a monic polynomial of degree $n$ then there is a $b\in\mathbb{C}$ such that $p(z)+b=(z-z_1)(z-z_2)\cdots (z-z_n)$ where $z_1,z_2,\dots,z_n$ are simple roots.

$p(z)\in\mathbb{C}[z]$

Any hint would be appreciated.

Answer 1

Hint: A polynomial $q(z)$ has only simple roots iff it has no roots in common with its derivative $q'(z)$. What is the derivative of $p(z) + b$? Could you choose $b$ such that $p(z)+b$ and its derivative have no roots in common?

Answer 2

It is sufficient to take $b$ as: $$ b = 1+\max_{\xi:p'(\xi)=0}\| p(\xi) \| $$ since, in such a way, $p'(\nu)=0$ implies $p(\nu)+b\neq 0$, so all the roots of $p(x)+b$ are simple.