Let $f$ be a non-degenerate sesquilinear form on $E$, is it true that for all linear $\varphi$ in $E$, so $\varphi \in E^*$, there exist $z \in E$ such that $$\varphi(x) = f(x,z)$$ and if so, why?
I know that $f$ is non-degenerate if $E^{\perp} = \{0\}$, so for all non zero $z \in E$, we know that $f(x,z) \neq 0$. Does this $z$ exist because we can always modify the value of $f(x,z)$ by choosing a different $z$ and because $f$ is linear with respect to the first variable?
Would be glad if someone would help me to explain this more precisely and better, because it is not so clear for me right now.