Limit of Sequence is Determined by Three Limits This is a problem from Polya and Latta's book Complex Variables.
Suppose there are three noncollinear points $a,b,c$ in complex plane and
\begin{align*}
\lim \limits_{n \rightarrow \infty} |z_n-a| \\
\lim \limits_{n \rightarrow \infty} |z_n-b|\\
\lim \limits_{n \rightarrow \infty} |z_n-c|
\end{align*}
exist.
Prove that $\lim \limits_{n \rightarrow \infty} z_n$ also exists.
If one of the given limits is $0$ we know the limit exists. I found a difficulty when none of the limits are $0$. Can anyone give some hint? Thank you.
I am also curious if the three points are collinear, can we have an example when $(z_n)$ does not converge?
 A: Here is the sketch of a proof. You need to fill in some details.

Let $r_a, r_b, r_c$ be the three given limits, which we may assume to be strictly positive.
We have $|z_n - a| + |z_n - b| \geq |a - b|$, which implies $r_a + r_b \geq |a - b|$; and $|r_a - r_b| \leq |a - b|$ for the same reason.
This means that, if you draw two circles $C_a, C_b$ such that $C_a$ (resp. $C_b$) is centered at $a$ (resp. $b$) and has radius $r_a$ (resp. $r_b$), then the two circles intersect at one or two points. Note that noncolinear implies that $a \neq b$.
If they only intersect at one point $P$, then $P$ is the limit $\lim z_n$. To see this, for each $0 < \epsilon < \min(r_a, r_b)$, draw two annuli $A_a(\epsilon), A_b(\epsilon)$ such that $A_a(\epsilon) = \{z \in \Bbb C: r_a - \epsilon < |z - a| < r_a + \epsilon\}$ and similarly for $A_b(\epsilon)$.
When $\epsilon$ tends to $0$, the intersection $I(\epsilon)$ of the two annuli "shrinks" to the point $P$. That is, the number $\max\{|z - P|: z \in I(\epsilon)\}$ tends to $0$ when $\epsilon$ tends to $0$. This is clear from the geometric picture, but you may need to do some calculation to get an analytic proof.
This shows that the limit $\lim z_n$ is equal to $P$, because for each $\epsilon$, the numbers $z_n$ eventually all belong to both $A_a(\epsilon)$ and $A_b(\epsilon)$, hence belong to the intersection $I(\epsilon)$.
Now suppose that the two circles $C_a, C_b$ intersect at two points $P, Q$. We define the annali $A_a(\epsilon)$ and $A_b(\epsilon)$ as before. For sufficiently small $\epsilon$, the intersection $I(\epsilon)$ will consist of two connected components, $I_P(\epsilon)$ and $I_Q(\epsilon)$, such that $I_P(\epsilon)$ "shrinks" to $P$, and $I_Q(\epsilon)$ "shrinks" to $Q$.
It remains to show that, for sufficiently small $\epsilon$, the sequence $(z_n)_n$ cannot appear infinitely many times in $I_P(\epsilon)$ and infintely many times in $I_Q(\epsilon)$.
Suppose it is not the case. Then there is a subsequence $(z_{n_{1, i}})_i$ that converges to $P$ and a subsequence $(z_{n_{2, i}})_i$ that converges to $Q$. But this would imply $|P - c| = \lim|z_{n_{1, i}} - c| = \lim|z_{n_{2, i}} - c| = |Q - c|$, which is impossible because $a, b, c$ are noncolinear.

When we drop the noncolinear condition, then there are counter examples.
Take $a, b, c$ to be any real numbers. Define $z_n = i$ or $-i$, depending on whether $n$ is even or odd.
