Let $X:=\begin{bmatrix} c_1 & c_2 \\ c_3 & -c_1 \end{bmatrix}$ be any $2\times 2$-traceless matrix with complex entries. I want to find a curve on $SL(2,\mathbb{C})$ passing through the identity element $I:=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ such that its tangent vector at $I$ is $X$.
Suppose such a curve $\gamma : (-\epsilon,\epsilon) \to SL(2,\mathbb{C})$ defined by $\gamma(t)=\begin{bmatrix} \gamma_1(t) & \gamma_2(t) \\ \gamma_3(t) & \gamma_4(t) \end{bmatrix} $ exists for some $\epsilon>0$. Then we must have $$\gamma'(0)=\begin{bmatrix} c_1 & c_2 \\ c_3 & -c_1 \end{bmatrix}\iff \gamma_1'(0)=c_1 , \gamma_2'(0)=c_2, \gamma_3'(0)=c_3, \gamma_4'(0)=-c_1$$ $$\gamma_1(t)\gamma_4(t)-\gamma_2(t)\gamma_3(t)=1$$ $$\gamma(0)=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\iff \text{$1=\gamma_1(0)=\gamma_4(0)$ and $0=\gamma_2(0)=\gamma_3(0)$}$$ Our guess is that $\gamma(t)=\begin{bmatrix} 1+c_1t & c_2 t \\ c_3t & 1-c_1 t \end{bmatrix}$. But we forgot something here because we have that $\det\gamma(t)=1-(c_1^2+c_2c_3)t^2\neq 1$ for all $c_1,c_2,c_3\in\mathbb{C}$ and $t\in (-\epsilon,\epsilon)$. For remedy we want to find $\lambda$ such that $\det(\lambda\gamma(t))=1$. Hence $\lambda=\frac{1}{\sqrt{\det(\gamma(t))}}=:\lambda(t)$. So we may define a differentiable curve $A(t):=\lambda(t)\gamma(t)$ on $SL(2,\mathbb{C})$ for all $t\in(\epsilon,\epsilon)$. Now it remains to check that the curve $A$ satisfies the other conditions $A(0)=I$ and $A'(0)=\begin{bmatrix} c_1 & c_2 \\ c_3 & -c_1 \end{bmatrix}$. First is clear. For the second, we have $$A'(t)=\lambda'(t)\gamma(t)+\lambda(t)\gamma'(t)\implies A'(0)=\lambda'(0)\gamma(0)+\lambda(0)\gamma'(0) $$ Note that $$\lambda'(t)=-\frac{1}{2}\det(\gamma(t))^{-3/2}(-2(c_1^2+c_2c_3)t)\implies \lambda'(0)=0$$ So $A'(0)=\gamma'(0)=\begin{bmatrix} c_1 & c_2 \\ c_3 & -c_1 \end{bmatrix}$, as was to be shown.
My question:
Is it true that $A(t)=\exp{\Big(t\begin{bmatrix} c_1 & c_2 \\ c_3 & -c_1 \end{bmatrix}\Big)}$? If it is so, how can we show this?
Surely you may take $A(t)=e^{tX}$, but there might be other choices as well. For instance, suppose $X=\pmatrix{1/\epsilon&0\\ 0&-1/\epsilon}$. Then $A(t):=\pmatrix{\frac{t+\epsilon}\epsilon&0\\ 0&\frac\epsilon{t+\epsilon}}$ is well-defined on $(-\epsilon,\epsilon),\,\det A(t)=1,\,A(0)=I$ and $A'(0)=X$, but $e^{tX}=\pmatrix{e^{t/\epsilon}&0\\ 0&e^{-t/\epsilon}}\ne A(t)$.
However, if you also want $A$ to be a one-paramter subgroup, it is well-known that $A$ has to be a matrix exponential. In that case, the answer to your question is affirmative.