Is $\delta\left(a+bi\right)=\delta\left(a-bi\right)$?

I wonder whether Dirac Delta (as defined via Fourier transform) is symmetric against the real axis.

We can write Delta function as

$$\delta(z) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{itz}\, dt=\delta\left(a+bi\right)=\frac1{2\pi}\int_{-\infty}^{+\infty}e^{-bx}\cos ax\, dx+\frac{i}{2\pi}\int_{-\infty}^{+\infty}e^{-bx}\sin ax\, dx.$$

The second integral is always zero (using Abel regularization), the first integral does not depend on the sign of $b$. So, $\delta\left(a+bi\right)$ should be equal to $\delta\left(a-bi\right)$.

But this contradicts the fact that $$\int_{-\infty}^\infty \delta(t+bi)f(t)dt=f(-bi)$$

which depends on the sign of $b$.

I have asked this on Math.Stackexchange, but received no answers.