I am trying to factor $(a^2 - 2ab + b^2 + 1)$ into $n$ terms for $a$ and $b$. Let $z_a$ and $z_b$ be any real or complex-valued number. I am trying to factor the following expression into separate terms for $z_a$ and $z_b$:
$$z_a^2 - 2z_az_b + z_b^2 +1 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space (1)$$
A concrete example would be the attempt to rewrite this expression into two quadratic polynomials with coefficients that can be real or imaginary. Let $\xi$ be a real valued coefficient of some kind (possibly denoting the branches at intervals of $\frac{2m\pi}{n}:m \in \mathbb{Z}(0,...,n-1)$). I have thought about expressing this as some relation of complex conjugate pairs, IE:
$$z_a^2 - 2z_az_b + z_b^2 + 1  = (z_a \pm e^{\xi i})(z_b \pm e^{\xi i)})$$
To be honest though, I'm very confused about what my options are for factoring $(1)$, and if it is even possible. Any help would be massively appreciated.
EDIT: I tried to follow the comment from @Suzu Hirose and took my quadratic formula coefficients $(a=1),(b=0),(c=0)$ coefficients from $$1(z_a - z_b)^2 + 0(z_a - z_b)^1 + 1(z_a - z_b)^0$$
I then inserted into the formula of
$$ \frac{-0 \pm\sqrt{0^2-4(1)(1)}}{2} = \pm \frac{2i}{2}=\pm{i}$$
I'm not sure how I would proceed from here though.

Write
$$a^2-2ab+(b^2+1)=0\\
\Delta_a=b^2-b^2-1=-1=i^2\\
a_1=b+i,\quad a_2=b-i$$
Then applying Fundamental theorem of algebra we have,
$$a^2-2ab+(b^2+1)=(a-b-i)(a-b+i).$$

Note: You can also write
$$a^2-2ab+b^2+1=b^2-2ab+(a^2+1)=0.$$
