Complex number inequality on a disk

Question

Let $z_0 \in D(0,1) \subseteq \mathbb{C}$ and $\epsilon_0>0$ such that $\bar D(z_0, \epsilon_0) \subset D(0,1)$. Show that $\forall z \in \bar D(z_0, \epsilon_0), \forall t \in [0,1]: |1-zt|\ge 1-\epsilon_0-|z_0|>0$

I have been struggling with this question the past few days and don't know how to approach it although it seems easy. For the second part of the inequality, I don't think this is a rigorous proof, I noticed that for the closed disk to be a subset of the open disk, we require its radius to be strictly smaller than $1-|z_0|$, or else the borders of the disks touch and we would have a common point for the two disks on the border $x^2+y^2=1$, but the disk $D(0,1)$ is open and so that point could not belong in that disk and so the disk $\bar D(z_0, \epsilon_0) \nsubseteq D(0,1)$, which contradicts our hypothesis. I would like some help on how to formalize this statement. Now for the first part I'm stuck. I tried to use the triangle inequality: $|1-zt|+|z_0| \ge |1+z_0-zt|$ and so it would be suffiecient to prove that $|1+z_0-zt| \ge 1-\epsilon_0, \forall t \in [0,1]$ but even after substituting $z=a+bi$ for some $a,b \in \mathbb{R}$ the expression under the root is complicated. I tried the Cauchy-Schwarz inequality but I probably didn't apply it right. I can see how this statement holds from the image that I linked. I would appreciate some hints on how to progress further by myself and a way to formalize my proof of the second part (I think I have an analytic geometric proof on my mind but it requires some cases)

Answer 1

Utilizing the triangle inequality is the correct approach. Using cartesian coordinates $a+ib$ is not necessary.

You have $$ |z| \le |z_0 + z-z_0| \le |z_0| + |z-z_0| \le |z_0| + \epsilon_0 $$ and therefore $$ |1-zt| \ge 1 - |zt| = 1 - t|z| \ge 1-|z| \ge 1 - |z_0| - \epsilon_0 \, . $$ And if we write $z_0 = |z_0| e^{i \alpha}$ with $\alpha \in \Bbb R$ then $$ (|z_0|+ \epsilon_0)e^{i \alpha} = z_0 + \epsilon_0 e^{i \alpha} \in \bar D(z_0, \epsilon_0) \subset D(0,1) $$ so that necessarily $|z_0|+ \epsilon_0 < 1$.