Let $u$ be a smooth function on $\mathbb R^n\backslash\{0\}$ homogeneous with degree $\lambda$ (on $\mathbb R^n\backslash\{0\}$). If $\lambda$ is not an integer $\le -n$, then $u$ can be uniquely extended to a tempered distribution homogeneous with degree $\lambda$. Moreover, the Fourier transform of an homogeneous distribution with degree $\lambda$ is an homogeneous distribution of degree $-\lambda-n$.
As a result, the Fourier transform of your $u$ is homogeneous with degree $n-i\tau-n=-i\tau$ when $\tau\in \mathbb R^*$, so is in one dimension a linear combination of $\xi_\pm^{-i\tau}$ which is thus bounded, proving the sought $L^2$ boundedness.
If $\tau =0$, then $u$ is homogeneous of degree $-1$ in one dimension. You need $u$ to be odd for the $L^2$ boundedness to hold. Take for instance (still in one dimension) $u(x)=1/\vert x\vert$, obviously homogeneous with degree $-1$ and smooth on $\mathbb R^*$. The singular integral with kernel $1/\vert x-y\vert$ is not bounded on $L^2$, but the Hilbert transform with kernel $1/(x-y)$ is bounded on $L^2$ (with norm π).