There are given two vectors describing quantic state:
$$ x= \begin{pmatrix} e^{j^{30^\circ}}\\ 1+2j \end{pmatrix} $$
$$ y= \begin{pmatrix} 3+j\\ e^{j^{60^\circ}} \end{pmatrix} $$ How to normalize them so i'm able to calculate tensor product like: $$ \hat x \otimes \hat y $$
It's confusing because of the "e" component which is obscure for me.
Following on from your comments, I believe that what you want to compute is the Tensor product:
$$\hat{x}\otimes \hat{y}$$
Where $\hat{x}$ and $\hat{y}$ are unit vectors (i.e. $\|\hat{x}\| = \|\hat{y}\|=1$), defined by:
$$\hat{x} = \frac{\vec{x}}{\|x\|}, \qquad \hat{y}=\frac{\vec{y}}{\|y\|}$$
We note that we can define the norm $\|\cdot\|$ in $\mathbb{C}^{2}$ using the following:
$$\|\vec{x}\| = \sqrt{\left|x_{1}\right|^{2}+\left|x_{2}\right|^{2}}$$
So we can calculate, (where $j$ is the imaginary unit $j = \sqrt{-1}$):
$$\|\vec{x}\|=\sqrt{\left|e^{j\cdot 30^{\circ}}\right|^{2}+|1+2j|^{2}}=\sqrt{1+5}=\sqrt{6}$$
Where we have used Euler's Formula: $e^{j\theta} = \cos(\theta) + j\sin(\theta)$, thus giving us $\left|e^{j\theta}\right| = 1$ for all $\theta$. So we can write our normalized $\vec{x}$ vector:
$$\hat{x}=\frac{\vec{x}}{\|\vec{x}\|}=\frac{1}{\sqrt{6}}\begin{pmatrix}e^{j \cdot 30^{\circ}} \\ 1+2j\end{pmatrix}$$
And for $\vec{y}$:
$$\|\vec{y}\|=\sqrt{|3+j|^{2}+\left|e^{j \cdot 60^{\circ}}\right|^{2}}=\sqrt{11} \implies \hat{y} = \frac{1}{\sqrt{11}}\begin{pmatrix}3+j \\ e^{j \cdot 60^{\circ}}\end{pmatrix}$$
We therefore have that:
$$\begin{align*}\hat{x}\otimes \hat{y} &= \frac{1}{\sqrt{66}}\begin{pmatrix}e^{j \cdot 30^{\circ}} \\ 1+2j\end{pmatrix}\begin{pmatrix}3-j & e^{-j \cdot 60^{\circ}}\end{pmatrix} \\ &= \frac{1}{\sqrt{66}}\begin{pmatrix}(3-j)e^{j \cdot 30^{\circ}} & e^{j \cdot 30^{\circ}} \\ 5+5j & (1+2j)e^{-j \cdot 60^{\circ}}\end{pmatrix}\end{align*}$$
Where we have used that in $\mathbb{C}^{n}$ we have:
$$\vec{x} \otimes \vec{y} = \vec{x}\vec{y}^{\dagger}$$