The Lie algebra elements fulfill
$$\mathfrak su(2) =\{X: \text {Tr} (X) = 0 \;, X^\dagger = - X \}.\tag 1$$
$$\mathfrak su(2) = \left\{\begin{bmatrix} ix & y + iz \\ -y+iz & -ix \end{bmatrix}\;x,y,z\in \mathbb R \right\},$$
On the other hand the Lie group is
$$SU(2)=\{A: \det(A)=1 \;, A^\dagger A=1 \}.$$
This implies matrices of the form
$$SU(2) = \left\{\begin{bmatrix}a & b \\ -\bar b & \bar a \end{bmatrix}\;a,b\in \mathbb C,\vert a \vert^2+ \vert b \vert^2=1 \right\},$$
since $A^\dagger A=1 \implies A^\dagger A^{-1},$ and the determinant is $1.$
I guess that to prove $(1)$ either the log or the exponential mapping would have to be used with some Taylor expansion at least to the linear term, but I don't know where to start.
You have$$\mathfrak{su}(2)=\left\{X\in GL(2,\Bbb C)\,\middle|\,(\forall t\in\Bbb R):\exp(tX)\in SU(2)\right\}.$$But, if $X\in GL(2,\Bbb C)$ and $t\in\Bbb R$,\begin{align}\exp(tX)\in SU(2)&\iff\exp(tX)^\dagger\exp(tX)=\operatorname{Id}\text{ and }\det\bigl(\exp(tX)\bigr)=1\\&\iff\exp\left(tX^\dagger\right)=\exp(tX)^{-1}\text{ and }\exp\bigl(\operatorname{tr}(tX)\bigr)=1\\&\iff\exp\left(tX^\dagger\right)=\exp(-tX)\text{ and }\exp\bigl( t(\operatorname{tr}(X)\bigr)=1\end{align}and, since these equalities occur for each $t\in\Bbb R$,$$\left.\frac{\mathrm d}{\mathrm dt}\exp\left(tX^\dagger\right)\right|_{t=0}=\left.\frac{\mathrm d}{\mathrm dt}\exp(-tX)\right|_{t=0}\quad\text{and}\quad\left.\frac{\mathrm d}{\mathrm dt}\exp\bigl( t(\operatorname{tr}(X)\bigr)\right|_{t=0}=0,$$which is the same thing as asserting that $X^\dagger=-X$ and that $\operatorname{tr}(X)=0$.