Proving a function is a Nondegenerate bilinear form

Question

Let $V$ be a finite dimensional inner product vector space above $\Bbb C$. Let $f \in V\times V \to \Bbb R $ defined by

$$f(u,v)=Im \langle u,v \rangle$$ (Here $Im(a+bi)=b$ the imaginary part of a complex number).

I managed to prove that $f$ is a skew symmetric bilinear form, and I'm trying to prove it is also a nondegenarate, but pretty stuck. If it wasn't, then for some $0 \neq v\in V$ I have $\langle u,v \rangle \in \Bbb R$ for every $u \in V$. I feel like I'm close to a contradiction but I'm missing something. Any ideas?

Answer 1

If $\langle u , v \rangle \in \mathbb{R} \setminus {0}$ then $\langle iu , v \rangle = i \langle u , v \rangle \in \mathbb{ C} \setminus \mathbb{R}$.

Observe that you cannot have $\langle u , v \rangle =0$ for all $u \in V$, because $\langle \cdot, \, \cdot \rangle$ is non-degenerate.