I am pretty sure it's duplicate but on the other hand it's hard to find by a title.
I am having hard time showing that every matrix in $SU(2)$ is of the form $\begin{pmatrix}z & -\overline w \\ w & \overline z\end{pmatrix}$ for some $z, w \in \mathbb{C}$ such that $|z|^2 + |w|^2 =1$. In most of the sources (including answers here) it is stated as something well-known. It is indeed well-known but I have never seen a proof and couldn't come up with one.
On
Note that $$\text{SU}(2):={\{A\in M(2,\mathbb{C}): \langle Ax,Ay\rangle,\; \forall x,y\in\mathbb{C}^2,\;\text{det}(A)=1}\}$$ where $\langle x,y\rangle:=x_1\bar{y_1}+x_2\bar{y_2}.$ Then $$\text{SU}(2):={\{A\in M(2,\mathbb{C}): AA^*=A^*A=I_2,\;\text{det}(A)=1}\}$$ where $A^*=\overline{(A^{t})}.$
So for all $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in\text{SU}(2)$ It is true that $A^{-1}=A^{*}\;\text{and}\;\text{det}(A)=1$:
$A^{-1}=\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}$ and $A^{*}=\begin{pmatrix}\overline{a} & \overline{c} \\ \overline{b} & \overline{d}\end{pmatrix}$ hence $A^{-1}=A^*$ implies that $$d=\overline{a}\;\text{and}\; b=-\overline{c}.$$ Then $A=\begin{pmatrix}a & -\overline{c} \\ c & \overline{a}\end{pmatrix}.$
Let $\begin{bmatrix}x_1 & x_2\\x_3 & x_4\end{bmatrix}\in SU(2),$ then $x_1\cdot x_4-x_3\cdot x_2=1$ because determinant is $1$(this you did not mention) and $$\begin{bmatrix}\overline{x_1} & \overline{x_3}\\\overline{x_2} & \overline{x_4}\end{bmatrix}=\begin{bmatrix}x_4 & -x_2\\-x_3 & x_1\end{bmatrix}$$ so $x_1=\overline{x_4}$ and $\overline{x_2}=-x_3.$ Notice that the other $2$ equations are just manipulations of these two. Check that left hand side is the inverse of the matrix and the left hand side is the adjoint.