Let $N$ be a $2× 2$ complex matrix such that $N^2=0$. how could I show $N=0$, or $N$ is similar over the matrix. Let $N$ be a $2× 2$ complex matrix such that $N^2=0$. how could I show $N=0$, or $N$ is similar over $\mathbb{C}$ to \begin{bmatrix}0 & 0\\1 & 0\end{bmatrix} 

Elementarily, assume $$N = \begin{bmatrix}a & b\\c & d\end{bmatrix} $$
Then
$$0=N^2 = \begin{bmatrix}a^2+bc & b(a+d)\\c(a+d) & d^2+bc\end{bmatrix} $$
Now if $b=0$, you get 
$$0=N^2 = \begin{bmatrix}a^2 & 0\\c(a+d) & d^2\end{bmatrix} $$
so that $a=d=0$, and the result follows.
Otherwise, you must conclude $a=-d$ so 
$$0=N^2 = \begin{bmatrix}a^2+bc & 0\\0 & a^2+bc\end{bmatrix} $$
giving that $c=-\frac{a^2}{b}$ so the original matrix is
$$N = \begin{bmatrix}a & b\\-\frac{a^2}{b} & -a\end{bmatrix} $$
and then observe that
$$\begin{bmatrix}0 & 1\\\frac{1}{b} & -\frac{a}{b}\end{bmatrix}
  \begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}
  \begin{bmatrix}a & b\\1 & 0\end{bmatrix} = N$$
