Let $\mathrm{Sp}(2m;\mathbb{C})=\{X\in\mathrm{GL}(2m;\mathbb{C});X^t\Omega X=\Omega\}$, where $\Omega=\begin{bmatrix}0& I_m\\ -I_m& 0\end{bmatrix}$, $I_m$ is $m\times m$ identity matrix.
The compact symplectic group is defined by $\mathrm{Sp}(m)=\mathrm{Sp}(2m;\mathbb{C})\cap U(2m)$. The following statement is from Kobayashi's Differential Geometry of Complex Vector Bundles pp.45.
By simple calculation we obtain $$\mathrm{Sp}(m)=\left\{\begin{bmatrix} A& B\\-\bar{B}&\bar{A}\end{bmatrix};\bar{A}^tA+B^t\bar{B}=I_m,\bar{A}^tB=B^t\bar{A}\right\}.$$
However I cannot obtain this form via computation. If we write $X\in\mathrm{Sp}(m)$ as $X=\begin{bmatrix} A& B\\C&D\end{bmatrix}$, then from $X^t\Omega X=\Omega$ and $\bar{X}^tX=I$, we obtain $$A^tC=C^tA\quad B^tD=D^t B\quad A^tD=C^tB+I_m\quad D^tA=B^tC+I_m$$ $$ \bar{A}^tB=-\bar{C}^tD\quad\bar{B}^tA=-\bar{D}^tC\quad\bar{A}^tA+\bar{C}^tC=I_m\quad\bar{B}^tB+\bar{D}^tD=I_m.$$ The third and fourth equations are equivalent, fifth and sixth equations are equivalent. How to deduce $C=-\bar{B}$ and $D=\bar{A}$?
I appreciate it if you can give a proof or a counterexample.
I did the following calculation:
$$\begin{array}{rl} &~&\overline{(C+\bar{B})}^t(C+\bar{B})+\overline{(\bar{D}-A)}^t(\bar{D}-A)\\ &=&(\bar{C}^t+B^t)(C+\bar{B})+(D^t-\bar{A}^t)(\bar{D}-A)\\ &=&\bar{C}^tC+B^t\bar{B}+\bar{C}^t\bar{B}+B^tC+D^t\bar{D}+\bar{A}^tA-(\bar{A}^t\bar{D}+D^tA)\\ &=&\bar{C}^tC+\bar{A}^tA+B^t\bar{B}+D^t\bar{D}+\overline{C^tB} + B^tC-(\bar{A}^t\bar{D}+D^tA)\\ &=&2I_m+\overline{C^tB} + B^tC-(\bar{A}^t\bar{D}+D^tA)\qquad\text{by eq 7 and 8}\\ &=&2I_m+\overline{(A^tD-I_m)} + (D^tA-I_m)-(\bar{A}^t\bar{D}+D^tA)\qquad\text{by eq 3 and 4}\\ &=&0 \end{array}$$
Hence we have $$0=\text{Tr}\left(\overline{(C+\bar{B})}^t(C+\bar{B})+\overline{(\bar{D}-A)}^t(\bar{D}-A)\right)\\ =\text{Tr}\left(\overline{(C+\bar{B})}^t(C+\bar{B})\right)+\text{Tr}\left(\overline{(\bar{D}-A)}^t(\bar{D}-A)\right)\\ =\Vert C+\bar{B}\Vert^2+\Vert \bar{D}-A\Vert^2$$ and the result follows.
I hope I didn't do any mistake!