Expressing $*(z_1\wedge \bar z_2)$ without Hodge star operator

Question

I have been stuck on a computation for hours, and still cannot figure out where is the mistake:

For $2$-dimensional complex vector space, let $z_1,z_2$ be the basis. I want to compute $*(z_1\wedge \bar z_2)$. I am pretty sure the answer is $-z_2\wedge\bar z_1$. But if I write $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ then $x_1,y_1,x_2,y_2$ is the standard basis. Then by the formula here,
we have
\begin{equation}
*z_1\wedge \bar z_2\\
=*(x_1+iy_1)\wedge(x_2-iy_2)\\
=*(x_1\wedge x_2-ix_1\wedge y_2+i y_1\wedge x_2+y_1\wedge y_2)\\
=-y_1\wedge y_2-iy_1\wedge x_2+ix_1\wedge y_2-x_1\wedge x_2\\
=-(x_1+iy_1)\wedge (x_2-iy_2)\\
=-z_1\wedge \bar z_2
\end{equation}

which I also think is correct. So where is my mistake?

Answer 1

For the formula $(\alpha,\alpha)=\int \alpha\wedge * \alpha$ to be valid for complex $\alpha$ we need a complex conjugation in the definition of $*\alpha$. Doing that we get $$*(dz_1 \wedge d\bar z_2) = - d\bar z_1 \wedge dz_2 = dz_2 \wedge d\bar z_1.$$