Exercise 23 (page 15) in Golan - The Linear Algebra a Beginning Graduate Student Ought to Know (3rd edition).
Let $0 \neq z_0 \in \mathbb{C}$ satisfy the condition $|z_0| < 2$. Show that there are precisely two complex numbers, $z_1$ and $z_2$, satisfying $|z_1| + |z_2| = 1$ and $z_1 + z_2 = z_0$.
Well.. ok so $z_2 = z_0 - z_1$ and let me set $z = z_1$ which is my real unknown. I need to find all the $z \in \mathbb{C}$ such that $|z| + |z-z_0| = 1$.
The problem is that $|z| + |z-z_0| \geq |z_0|$ for every $z \in \mathbb{C}$. So if $|z_0|>1$ it seems to me that the problem has no solutions at all. Also, if $|z_0|\leq1$, it seems to me that there are an infinity of solutions, not two.
What am I missing?