Note after reviewing answers. This question illustrates in a non-trivial way how the choice of how to compute the square root in the complex plane can make real difference to the answer you get.
In this question finding all the solutions came down to solving:
$$ab-1=\sqrt{a^2+b^2+1}$$ subject to $$(a+1)(b+1)=1$$
The condition implies $a^2+b^2+1=(ab-1)^2$, but this leaves open the possibility of $\sqrt {a^2+b^2+1}=1-ab$ rather than $ab-1$.
The solution and discussion on the original question was getting rather long, and dealt mainly with real roots, so I've extracted this sub-question on complex roots.
Could someone please explain what constraints can be put on the complex solutions of this equation - the values of $a$ and $b$ - which would be consistent with the square root operation. It would help if answers were elementary and didn't assume a prior knowledge of complex square roots.
Answer
Every nonzero complex number $z$ has two "square roots". You might call one $\sqrt{z}$ and the other $-\sqrt{z}$, but which is which? A particular choice (for all $z$ in some region of the complex plane) is called a branch of the function $\sqrt{z}$. There is no single branch that is continuous everywhere: if you start at some point with a particular choice of $\sqrt{z}$ and travel in a loop around $0$, keeping $\sqrt{z}$ continuous along your path, when you come back to the start $\sqrt{z}$ will have $-$ its original value. A curve $\Gamma$ in whose complement your branch is continuous is called a branch cut.
One popular choice, called the "principal branch", is to make the real part of $\sqrt{z}$ always $\ge 0$. The branch cut in this case is the negative real axis.
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