Kernel with complex vectors
Kernels are considered valid if they are positive-definite, so we must have
$$
\sum_{i=1}^n \sum_{j=1}^n c_i c_j K(x_i, x_j) \ge 0
$$
for all $n \in \mathbb{N}$ and $c_1, \dots, c_n \in \mathbb{R}$ and (in your case) all $x^1, \dots, x^n \in \mathbb{R}^2.$

Letting $n=1$ and $c_1 = 1$ and $x^1 = (0, 1)$ shows that $K$ is not positive-definite, since
$$
K\big((0,1), (0,1)\big) = 0\times 0 - 1\times 1 = -1.
$$

The problem with your approach is that you define $\Phi$ on $\mathbb{C}$ but write $K$ as the dot product for real-valued vectors.
The feature map argument for positive definite vectors only works if the product is an inner product on $\Phi$'s codomain.
The dot product for real-valued vectors is not an inner product on $\mathbb{C}^2$ (because $i\cdot i = -1 &lt; 0$).
You would have to use the generalized dot product $x \cdot y = \sum_i x_i \overline{y_i}$, or another inner product on $\mathbb{C}^2$.