Let $X=\{[z_0,z_1,z_2,z_3]\ \big{|}\ [z_0,z_1,z_2,z_3]\in\mathbb{C}P^3,z_0^4+z_1^4+z_2^4+z_3^4=0\}$. $c_1(X)$ is the first Chern class of $X$. Prove that $c_1(X)=0$.
$\textbf{My try}$:
It's easy to show that $X$ is a submanifold of $\mathbb{C}P^3$. Let $D$ be a connection on $X$. $\Theta=D\circ D$ is a curvature. Then $c_1(X)=\frac{\sqrt{-1}}{2\pi}tr(\Theta)$. So I need to prove that $tr(\Theta)=0$. Let $h$ be the Hermitian metric. $X_0=\{[z_0,z_1,z_2,z_3]\ \big{|}\ [z_0,z_1,z_2,z_3]\in X,z_0\ne 0\}\ $. On $X_0\ $ $h([z_0,z_1,z_2,z_3])=1+\frac{|z_1|^2}{|z_0|^2}+\frac{|z_2|^2}{|z_0|^2}+\frac{|z_3|^2}{|z_0|^2}=1+\xi_1^2+\xi_2^2+\xi_3^2$. Let $D$ be the Chern connection. Then $\Theta=d\theta\ $ where $\theta=h^{-1}\partial h$. On $X_0$, $\xi_1^4+\xi_2^4+\xi_3^4=-1$. But I can't cauculate $tr(\Theta)$.
Any ideas?