Let $\mathbb S$ be the collection of all univalent and analytic functions defined on unit disc $\mathbb D$ such that for each $f \in \mathbb S,$ we have $f(0)=0, f'(0)=1.$ Let $P$ be the subsets of $\mathbb S$ consisting polynomials. I want to check if $P$ is dense in $S?$
The elements of $P$ are in the form of $p_n(z)=z + c_2z^2+c_3z^3+ \cdots + c_n z^n.$ It is not difficult to see that if $p_n$ converges uniformly to a nonconstant function then the limiting function is also univalent. Also another observation is $p_n(z)$ will not have any roots in $\mathbb D$ as $0$ is already a root.
Let $f \in \mathbb S.$ Then $f$ has a power series representation $f(z) = z + \sum_{n\ge 2}a_n z^n .$ Since, partial sums uniformly converge to $f$, if we are able to prove that partial sums are univalent then it will imply that $P$ is dense in $\mathbb S.$ But how to show that partial sums are univalent? Or are there other polynomials which will work?
By the argument prnciple if $f(z)=z+\sum_{n \ge 2} a_n z^n$ is univalent in the unit disc and $P_m(z)=z+\sum_{2 \le k \le m}a_kz^k$ are its partial sums, then for every $r<1$ there is $n_r$ st $P_n$ univalent for $n >n_r$
(otherwise taking two sequences in $|z| \le r$ on which infinitely many $P_n$ are equal, we can find limit points of same subsequences of both which are in $|z| \le r$ and either $f$ is equal there if the limit points are distinct or $f'(w)=0$ if the limit points coincide and are $w$)
But then picking $r_n \to 1$ eg $r_n =1-1/n$ and $P_{m_n}$ univalent on $|z| \le r_n$ as above, we can normalize them to $Q_n(z)=P_{m_n}(r_nz)/r_n$ which are polynomials univalent in the unit disc and in $S$ and clearly have same normal limit as $P_{m_n}$ so they converge normally to $f$, showing that univalent polynomials are dense in $S$
However, there is much more that is true here, namely that in $S$ there is a uniform lower bound $$|\frac {f(z)-f(w)}{z-w}| \ge \frac{1-r}{(1+r)^3}, |z|=|w|=r, f \in S$$ which follows from the Grunsky-Lebedev inequalities and the standard growth theorem.
If $P_n$ as above fails to be univalent in some disc $|z| \le r$, then by standard stuff (argument principle again) there are $|z|=|w| \le r, P_n(z)=P_n(w)$ but one can easily obtain bounds for the tail of $|\frac {f(z)-f(w)}{z-w}|$ using the simple Littlewood estimate $|a_n| < en$ (or of course the De Branges - Bieberbach theorem $|a_n| \le n$ but that is difficult while Littlewood estimate is much simpler) ensuring that those tails are smaller than the lower bound above for $n \ge 1- cn \log n$ for some effective constant that depends on which $|a_n|$ bound we use, so $P_n$ is univalent in $|r| < 1- cn \log n$ uniformly in $f \in S$.