How to find difference equation, given poles and zeros?
We have a zero at $z=0$ and poles at $z=1+i$ and $z=1-i$

Therefore the Transfer function is:

$$H(z) = k \dfrac{z}{[z-(1+i)][z-(1-i)]}$$

$$\Rightarrow H(z) = k\dfrac{z}{(z-1)^2+1}=\dfrac{kz}{z^2-2z+2}$$

$$\Rightarrow \dfrac{Y(z)}{X(z)} = \dfrac{kz}{z^2-2z+2}$$

$$\therefore Y(z)[z^2-2z+2] = kzX(z)$$

This gives:
$$z^2Y(z)-2zY(z)+2Y(z) =  kzX(z)$$
Taking inverse Z transform:
$$y(n+2)-2y(n+1)+2y(n) = kx(n+1)$$