a sequence of polynomials converges to $0$ I am trying to show that  

there is a sequence $(P_{n})_{n}$ of polynomials such that $P'_{n}(0)=1$ for all $n$, $P'_{n}(z)\rightarrow0$ if $z \in \mathbb{C}^{\times}$ and $P_{n}(z)\rightarrow0$ if $z \in \mathbb{C}$   

but I could not able to do that. I will appreciate for any help.
 A: Let $K_n=\{0\}\cup\{z\in\Bbb C:\frac1n\le |z|\le n\}\setminus(\Bbb R^+\times(0,\frac1n))$. This is an annulus of inner radius $\frac1n$, outer radius $n$, with $0$ added and with a strip of width $\frac1n$ just above the positive real axis deleted. This peculiar sequence of sets was chosen because:


*

*$\bigcup_{n\in\Bbb N}K_n=\Bbb C$

*$K_n$ is compact

*$\Bbb C\setminus K_n$ is connected

*$K_n^\times$ is connected


Now define $f_n$ to be $0$ on a neighborhood of $K_n^\times$ and $1$ on a neighborhood of $0$. This is a holomorphic function, so by Runge's theorem, there is a polynomial $g_n$ that is within $\frac1{2n(n+1)}$ of $f_n$ on $K_n$, so letting $Q_n(x)=g_n(x)-g_n(0)+1$, $Q_n$ is a polynomial that is less than $\frac1{n(n+1)}$ on $K_n^\times$ and such that $Q_n(0)=1$. Thus $Q_n(x)\to0$ pointwise on $\Bbb C^\times$.
Now define $P_n(z)=\int_{-1}^zQ_n(t)\,dt$. Then for any $z\in K_n^\times$, $$|P_n(z)|\le|z+1|\sup_{t\in K_n^\times}|Q_n(t)|\le(n+1)\cdot\frac1{n(n+1)}=\frac1n,$$
so $P_n(z)\to0$ pointwise on $\Bbb C^\times$, and also $P_n'=Q_n$, hence the other properties.
But there is one piece missing, namely $P_n(0)\to0$. To do this, define $R_n(z)=\frac12(P_n(z)-P_n(-z))$. That way, the limits are preserved, the derivative is preserved, and $R_n(0)=0$ for each $n$.
