Positive definite matrix must be Hermitian

As Marvis says in the comments, the problem reduces to showing that if $V$ is a finite-dimensional complex inner product space and $A : V \to V$ an operator such that $\langle v, Av \rangle = 0$ for all $v$, then $A = 0$. Letting $v$ run across all eigenvectors of $A$, the hypothesis implies that $A$ has all eigenvalues $0$, hence is nilpotent. Suppose by contradiction that $A \neq 0$, hence there exists a vector $v_0$ such that $v_1 = A v_0$ is nonzero. We may assume WLOG that $A v_1 = 0$, hence

$$\langle v_0 + v_1, A(v_0 + v_1) \rangle = \langle v_1, v_1 \rangle \neq 0$$

which is a contradiction. Hence $A = 0$.

Note that the corresponding assertion for real inner product spaces is false; the condition $\langle v, Av \rangle = 0$ is satisfied by all skew-symmetric matrices.

You don't even need positive definiteness (or semi,negative etc definitiveness for that matter)

Only thing you need is that $x^*Ax \in R \; \forall x \in C$ which is ofcourse true when we compare this value with $0$ while defining such matrices.

So let $x^*A x\in R \; \forall x\in C$ (and NOT just $R$)

I will show $A$ is hermitian

$x^*Ax= \langle Ax,x\rangle=\langle x,A^*x\rangle = \overline{\langle A^*x,x\rangle} = \langle A^*x,x \rangle \rightarrow \langle (A-A^*)x,x\rangle = 0 \; \forall x\in C$

Claim: if $\langle Bx,x\rangle = 0 \; \forall x\in C \rightarrow B=0$

Proof: For arbitrary $y\in C$, $\; \langle B(x+ky), x+ky\rangle = \bar{k}\langle Bx,y \rangle + k\langle By,x \rangle$

Now set $k=1$ and $k=\iota$ to get two equations, solve them to get $\langle Bx,y\rangle =0 \; \forall y\in C \rightarrow Bx= 0 \; \forall x\in C \rightarrow B=0$