If $a_nb_n\to 1$ and $a_n+b_n\to2$, do $a_n,b_n\to1$?
Hint. $|a_n - b_n| = \sqrt{(a_n+b_n)^2-4a_n b_n}$ converges to zero. From this, both
$$\max\{a_n, b_n\} = \frac{(a_n+b_n)+|a_n-b_n|}{2}, \qquad \min\{a_n, b_n\} = \frac{(a_n+b_n)-|a_n-b_n|}{2}$$
converge to $1$. Can you now finish the proof by applying the squeezing lemma?
The same conclusion holds for complex case but we may need a slightly more lengthy solution. First, we still know that
$$|a_n - b_n| = |(a_n+b_n)^2-4a_n b_n|^{1/2}$$
converges to $0$. Then by the triangle inequality,
$$|a_n|, |b_n| \leq \frac{|a_n+b_n|+|a_n-b_n|}{2}$$
and hence both $(a_n)$ and $(b_n)$ are bounded. Now for any subsequene of $(a_n, b_n)$, we can extract a further subsequence, say $(a_{n_k}, b_{n_k})$ such that both components converge. Then by the relation $a_{n_k} + b_{n_k} \to 2$ and $|a_{n_k} -b_{n_k}|\to 0$, it follows that both $(a_{n_k})$ and $(b_{n_k})$ converge to $1$. This implies that the original sequence $(a_n, b_n)$ also converges to $(1,1)$ by the following well-known trick:
Lemma. For a metric space $(X, d)$ and for a sequence $(x_n)$ in $M$, the followings are equivalent:
$(x_n)$ converges to $x\in M$.
Any subsequence of $(x_n)$ has a further subsequence that converges to $x$.
Consider $$c_n=(a_n-1)^2+(b_n-1)^2.$$ Then $$c_n=(a_n+b_n)^2-2a_nb_n-2(a_n+b_n)+2\to2^2-2-4+2=0.$$ As $0\le(a_n-1)^2\le c_n$ then $a_n-1\to0$, and $a_n\to1$. Similarly $b_n\to1$.