Showing matrices in $SU(2)$ are of form $\begin{pmatrix} a & -b^* \\ b & a^*\end{pmatrix}$ Matrices $A$ in the special unitary group $SU(2)$ have determinant $\operatorname{det}(A) = 1$ and satisfy $AA^\dagger = I$.
I want to show that $A$ is of the form $\begin{pmatrix} a & -b^* \\ b & a^*\end{pmatrix}$ with complex numbers $a,b$ such that $|a|^2+|b|^2 = 1$.

To this end, we put $A:= \begin{pmatrix} r & s \\ t & u\end{pmatrix}$ and impose the two properties.
This yields \begin{align}\operatorname{det}(A) &= ru-st \\ &= 1 \ ,\end{align}
and 
\begin{align}
AA^\dagger &= \begin{pmatrix} r & s \\ t & u\end{pmatrix}  \begin{pmatrix} r^* & t^* \\ s^* & u^* \end{pmatrix} \\&= \begin{pmatrix} |r|^2+|s|^2 & rt^* +su^* \\ tr^*+us^* & |t|^2 + |u|^2\end{pmatrix} \\
&= \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \ .\\
\end{align}
The latter gives rise to 
\begin{align}
 |r|^2+|s|^2 &= 1 \\
&= |t|^2+|u|^2 \ ,
\end{align}
and 
\begin{align}
tr^*+us^* &= 0 \\
&= rt^*+su^* \ .
\end{align}

At this point, I don't know how to proceed. Any hints would be appreciated.

@Omnomnomnom's remark 
\begin{align}
A A^\dagger &= \begin{pmatrix} |r|^2+|s|^2 & rt^* +su^* \\ tr^*+us^* & |t|^2 + |u|^2\end{pmatrix} \\
&= \begin{pmatrix} |r|^2+|t|^2 & sr^* +ut^* \\ rs^*+tu^* & |s|^2 + |u|^2\end{pmatrix} = A^\dagger A \ ,
\end{align}
gives rise to 
$$
|t|^2 = |s|^2 \\
|r|^2 = |u|^2
$$
and 
$$
AA^\dagger :\begin{pmatrix}
rt^* +su^* = sr^* +ut^* \\
tr^*+us^* = rs^*+tu^*  
\end{pmatrix}: A^\dagger A $$ 

At this point, I'm looking in to find a relation between $t,s$ and $r,u$ respectively. 

The condition $A^{\ast}A=I$ says that $A$ has orthonormal columns.
Suppose the first column is $v=[\begin{smallmatrix}a\\b\end{smallmatrix}]$. It must have unit norm, so $|a|^2+|b|^2=1$. What can the second column be? It must be orthogonal to the first, which means it must be in the complex one-dimensional orthogonal complement. Thus, if $w$ is orthogonal to $v$, then the possibilities for the second column are $\lambda w$ for $\lambda\in\mathbb{C}$. Since $\det[v~\lambda w]=\lambda\det[v~w]$, only one value of $\lambda$ will make the determinant $1$, hence the second column is unique. So it suffices to check $w=[-b ~~ a]^{\ast}$ works, which is natural to check because in ${\rm SO}(2)$ the second column would be $[-b~~a]^T$.
