How to find difference equation, given poles and zeros? I'm given poles at ${1+i}$ and ${1-i}$, and zero at $0$.

I have to find the difference equation and find out whether the system is stable.

Now I found in a similar question, which structure I tried to follow ( How to find system function, H(z) in the z-domain, given zero-pole plot of the system? ), that we can express $H(z)$ as $$H(z)=G_0z^{-1}\frac{(z-z_0)(z-z_1)\ldots}{(z-p_0)(z-p_1)\ldots} $$

It's clear to me that the numerator of H(z) is simply = z, but I can't wrap my head around the denominator.

I tried to go for: (z- (1-i))*(z- (1+i)) - my prof looked over my shoulder, nodded and told me to simply multiply it, adding "there will be no i in the polynomial". Now I really don't find any combination of tricks to achieve this, and I've been stuck for several hours right now. 

I'd very much appreciate a hint!
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:
$$y(n+2)-2y(n+1)+2y(n) = kx(n+1)$$