"Any matrix $A\in SU(2)$ can be written as $\begin{pmatrix}z & -\overline w \\ w & \overline z\end{pmatrix}$ for some $w,z\in\mathbb C$ with $|w|^2+|z|^2=1$. Moreover, every matrix of this form is diagonalizable, that is, there exists $B\in SO(2)$ such that $BAB^{-1}$ is a diagonal matrix."
I have read this statement but I do not understand why any matrix of $SU(2)$ is diagonalizable? and why the matrix that we will find will be in $SO(2)$ and not in $SU(2)$? can not it be with complex entries?
On
Unitary matrices are normal, and therefore diagonalizable using a unitary $B$.
EDIT: Look up normal matrix
However, it is not true that $B$ is in $SO(2)$: that must be a misprint for $SU(2)$. A rotation by an angle not a multiple of $\pi$ is a real member of $SU(2)$ with non-real eigenvalues, and therefore the $B$ can't be real.
For example, try $$A = \pmatrix{0 & -1\cr 1 & 0\cr}$$
where with $$B = \frac{1}{\sqrt{2}} \pmatrix{1 & i\cr i & 1\cr}$$ we have $$BAB^{-1} = \pmatrix{i & 0\cr 0 & -i}$$
The way over normal matrices is really the way to go here. However, you can also see the result by a simple investigation into the characteristic polynomial.
Take $$A=\begin{pmatrix}z &-\bar w\\w &\bar z\end{pmatrix},$$ then the characteristic polynomial of $A$ is given by $$p_A(\lambda)=(z-\lambda)(\bar z-\lambda)+w\bar w=z\bar z-\bar z\lambda-z\lambda+\lambda^2+w\bar w$$ Now, for $z\in\mathbb C$, you have $z\bar z=|z|^2$, and thus as $|z|^2+|w|^2=1$:
$$p_A(\lambda)=\lambda^2-(z+\bar z)\lambda+1$$
Now, $\mathfrak R(z)=\frac{z+\bar z}{2}$ and thus
$$p_A(\lambda)=\lambda^2-2\mathfrak{R}(z)\lambda+1$$. You can now see that $p_A$ has two complex roots by looking at the discriminant of $p_A$:
$$(\frac{-2\mathfrak{R}(z)}{2})^2-1=(\mathfrak R(z))^2-1$$
and as $|z|^2<1$, you have $(\mathfrak R(z))^2<1$ and thus $(\mathfrak R(z))^2-1<0$. Thus, the matrix $A$ has two complex eigenvalues and thus can be diagonalized.