Find a sequence of complex polynomials with certain properties. (Hardy spaces over unit circle) 
Let $\lambda\in \Bbb S^1$. Find a sequence of complex polynomials $p_n(z)$ such that for any $c>0$ the following inequality does not hold:
  $$|p_n(\lambda)|\le c\cdot \|p_n\|$$
  where $$\|p_n\|=\left(\dfrac{1}{2\pi}\int\limits_0^{2\pi}{\left|p_n(e^{i\theta})\right|^2d\theta}\right)^{1/2}.$$

I was asked  to show this by finding a sequence $p_n(z)$ such that $\|p_n\|\to 0$ but $|p_n(\lambda)|\nrightarrow 0$. For this I first consider $p_n(z)=\left(\dfrac{z}{\lambda}\right)^n$, which did not help. Also I tried to find some polynomials like $\dfrac{1}{n}\left(\dfrac{z}{\lambda}\right)^n$  which did not help. Any help is appreciated. 

For the sake of simplicity I assume $\lambda=1$. The general case can be generated by rotating the arguments. Consider 
\begin{align*}
q_n(z):=\prod_{k=1}^{n-1}\left(z-\exp\left(\frac{2\pi ik}{n}\right)\right).
\end{align*}
and define $f_n(t):=q_n(e^{it})$. Then the Fourier Coefficients $c_k$ of $f_n$ are  $c_k=1$ for $0\leq k<n$ and $c_k=0$ otherwise. In particular, $q_n(1)=n$. By Parceval's Identity we have
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}|f_n(t)|^2dt=\sum_{k\in\mathbb Z}|c_k|^2=n.
\end{align*}
Thus, if we define $p_n(z):=q_n(z)/n$, then
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}|p_n(e^{it})|^2dt=\frac{1}{n}\to0,
\end{align*}
but $p_n(1)=1$ for each $n\in\mathbb N$.
