A quadratic form is a map $q: Q(q) \times Q(q) \rightarrow \mathbb{C}$, where $Q(q)$ is a dense linear subset of the Hilbert space $H$. If $q(\phi,\psi)=\overline{q(\psi,\phi)}$, then we say q is symmetric. If $q(\phi,\phi)\geq -M||\phi||^2$ for some $M$, we say $q$ is semibounded.
If $q$ is semibounded, then it is automatically symmetric if $H$ is complex.
Anyone could give my a hint for the proof?
I have made an attempt to solve this: Let $\phi,\psi \in Q(q)$. We know that $q(\phi+\psi,\phi+\psi)\in \mathbb{R}$ and $q(\phi+i\psi,\phi+i\psi)\in \mathbb{R}$. $$q(\phi+\psi,\phi+\psi)=q(\phi,\phi)+q(\psi,\psi)+q(\psi,\phi)+q(\phi,\psi)(1)$$ $$q(\phi+i\psi,\phi+i\psi)=q(\phi,\phi)+q(\psi,\psi)-iq(\psi,\phi)+iq(\phi,\psi) \;(2)$$
From (2), we know that $q(\psi,\phi)=q(\phi,\psi)$. From (1), $q(\psi,\phi)+q(\phi,\psi)$ is real. Thus, $q(\phi,\psi)$ and $q(\phi,\psi)$ are real. $q(\psi,\phi)=\overline{q(\phi,\psi)}$