How to prove $\mu(f,z) = \mu(f,\bar{z})$? Let $\mu(f,z)$ denote the multiplicity of the zero $z$ of the polynomial $f\in \mathbb{R}[t]$.

How to prove that $\mu(f,z) = \mu(f,\bar{z})$?

There is one hint: Show by induction that $\forall k\in\mathbb{N}: [\mu(f,z) \geq k \Longrightarrow  \mu(f,\bar{z}) \geq k]$. However, I do not really understand how to prove the hint or how it helps.
 A: Let $f(t) \in \mathbb R[t]$ be a real polynomial. Let $z$ be it's root, so that $f(z)=0$. Then,
$$f(\bar z) = \overline{f(z)} = \bar 0 = 0$$
(the first equality is due to $f$ having real coefficients).
So, $\bar z$ is also a root of $f$. This means that we can decompose $f$ as
$$f(t) = (t-z)(t-\bar z)g(t) = (t^2 -2\operatorname{Re} z + |z|^2)g(t) = h_z(t)g(t).$$
Note that $h_z(t)$ has real coefficients, thus $g(t)$ is real, too, and
$$\deg g = \deg f - 2.$$
$h_z$ has roots $z$ and $\bar z$, each with multiplicity $1$ (assuming $z \neq \bar z$). Now, proceed with induction: if $g$ has $z$ and $\bar z$ with the same multiplicities, then so does $f$.
A: For a polynomial $p: \>t\mapsto p(t)$ denote by $\bar p$ the polynomial obtained by conjugating the coefficients of $p$. Let $z\in{\mathbb C}$ be a zero of multiplicity $m$ of the given $\ f: \>t\mapsto f(t)$. Then $f(t)=(t-z)^m g(t)$.  According to the rules governing polynomial multiplication we therefore have
$$f(t)=\bar f(t)=(t-\bar z)^m \bar g(t)\ .$$
It follows that the number $\bar z$ is a zero of multiplicity $m^*\geq m$. By symmetry we get $m^*=m$.
