Quadratic form on $\mathbb{C}^3$ Let 
$$\Phi(x)= \frac{1}{4}x_1^2+\frac{3}{4}i(\bar{x_1}x_2-\bar{x_2}x_1)-\frac{1}{2\sqrt{2}}(\bar{x_1}x_3+\bar{x_3}x_1)+\frac{1}{4}x_2^2-\frac{1}{2\sqrt{2}}i(\bar{x_2}x_3+\bar{x_3}x_2)-\frac{1}{2}x_3^2$$
Show that $\Phi$ is hermitian and find its signature. 
So in order to solve this I was thinking that, since we also have to show the signature, I could write the quadratic form in matrix form. Which is 
$$A=\begin{pmatrix}\frac{1}{4}&\frac{3}{8}i&\frac{-1}{4\sqrt{2}}\\\frac{-3}{8}i&\frac{1}{4}&\frac{-1}{4\sqrt{2}}i\\\frac{-1}{4\sqrt{2}}&\frac{1}{4\sqrt{2}}i&\frac{-1}{2}\end{pmatrix}$$
It is easy to see that the matrix $A$ is hermitian since $A=A^{*}$.
My question is how do I find the signature? Do I have to diagonalise the matrix to see which eigenvalues it has? 

Your matrix is $\ast$-congruent to a diagonal real matrix.
We need not worry about the eigenvalues, those may or may not be nice. 
$$
R =
\left(
\begin{array}{rrr}
  1& -\frac{3i}{2}  & - \sqrt 2  \\
  0& 1  &  - i \sqrt 2 \\
  0& 0 & 1 \\
\end{array}
\right)
$$ 
$$
R^\ast A R =
\left(
\begin{array}{rrr}
 \frac{1}{4} & 0  & 0  \\
  0 & - \frac{5}{16}  & 0  \\
  0 & 0  & 0  \\
\end{array}
\right)
$$ 
Without the bother of finding the eigenvalues, we know that one is positive, one zero, and one negative. Having one of them zero means finding them is not bad after all.
