How to show the fact that the set of all orthogonal matrices over $\mathbb C$ is compact
By an orthogonal matrix over $\mathbb C$ I mean a matrix $A$ satisfying $AA^T=I$ and here $A^T=(a_{ji})$ where $A=(a_{ij})$
It is not the same as unitary matrix where in unitary matrix we take transpose and then conjugate or vice versa
I know that set of all orthogonal matrices over $\mathbb R$ is compact.
I think the closedness of the set will follow from the same arguements as in the above case. But the boundedness part not sure
It is not bounded, unless you are working with $1\times 1$ matrices.
The only complex, $1\times 1$ orthogonal matrix are $(1)$ and $(-1)$.
In $2\times 2$ (and by extension you get $n\times n$) you can consider the family, where $\lambda \geq 1$ is a real parameter, $$ A_\lambda = \begin{pmatrix} \lambda & i \sqrt{\lambda^2 - 1} \\ - i \sqrt{\lambda^2 - 1} & \lambda \end{pmatrix} $$ which you can easily check to be orthogonal. But any matrix norm of $A_\lambda$ would tell you it is of size $\approx \lambda$.