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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!

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So you're given two complex-conjugate poles $p=1+i$ and $p^*=1-i$. Your denominator polynomial is then given by

$$D(z)=(z-p)(z-p^*)$$

I trust that you can do the multiplication (for general $p$) and verify that the result is purely real-valued for any pair of complex-conjugate poles.

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Thank you very much for your fast reply!

In the meantime, I came across a really embarassing knowledge gap: $$i^2 = -1$$ (yeah, I'm that blank).

That way, I can transform $$(z-1-i)*(z-1+i)$$ to $$z^2-z+zi-z+1-i-zi+i-i^2$$ and further to $$z^2-2z+2$$

Since transfer function $H(z)$:

$$ H(z) = \frac{Y(z)}{X(z)} $$

I conclude the difference equation should be

$$ y[n] = b_0 x[n] - a_1 y[n-1] - 2 * a_2 y[n-2] + 2 \tag{2} $$

This leads me to a result and even more to read about. I am sorry, but apparently I didn't understand how to solve for 'general' $$p$$ yet. Thank you very much again for your time and help!

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    $\begingroup$ also, in the electrical engineering subculture, we use "$i$" for current in the time domain, like "$i(t)$" or "$v(t)\,=\,R\,i(t)$" so then to reduce confusion (among EEs, because this increases confusion for everyone else), we use the letter "$j$" for the imaginary unit. or $$ j^2 = -1 $$ $\endgroup$ – robert bristow-johnson Jun 26 '18 at 23:30

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