Proving complex function $f(z) = \frac1{1-z}$ is continuous on open disk 
Prove that $f(z) = \frac1{1-z}$ is continuous on the open disk $\mathbb{D}_1(0) = \{z \in \mathbb{C}: |z|<1 \}$

I've been having a lot of trouble with this one. Here's my attempt thus far:
Fix $\epsilon > 0$, and let $z_0$ be an arbitrary element of $\mathbb{D}_1(0)$
$$|f(z) - f(z_0)| = |\frac1{1-z} - \frac1{1-z_0}| = \frac{|z-z_0|}{|1-z_0||1-z|} \le  \frac{|z-z_0|}{(|1-|z_0||)(|1-|z||)} \space\space \text{ by the reverse triangle inequality}$$
Let $$\frac1{|1-|z||} < \frac1{|1-|z_0||} \implies |1 - |z_0|| < |1-|z||$$
Which implies that $|z_0|$ is closer to $1$ than $|z|$, and since both must be less than $|1|$, we have $|z| < |z_0|$. In this case, choosing $\delta = \epsilon|1-|z_0||^2$:
$$\frac{|z-z_0|}{(|1-|z_0||)(|1-|z||)} < \frac{|z-z_0|}{|1-|z_0||^2} < \epsilon$$
However, when $|z_0| < |z|$, I'm not sure how to handle this, as I end up with $\delta$ in terms of $|z|$, which is not permitted.
I also tried writing $f(z) = \frac{1 - \bar{z}}{1 - 2\Re(z) + |z|^2}$, and converting to polar coordinates, but these ended up being harder to work with than the original representation.
 A: Hint:) As an alternative way you can show that this function is differentiable in the unit disk.
A: I believe it is simpler to prove that $f(z) = 1/z$ is continuous in $\mathbb C \setminus\{0\}$ and then use composition to obtain the function you want. 
Lets prove $f(x) = 1/z$ is continuous at $p\not = 0$.
$$\left|\frac{1}{z}-\frac{1}{p}\right| = \frac{|z-p|}{|zp|}.$$
For each $\epsilon$ take a $\delta>0$ such that $\delta < |p|/2$.
If $|z-p|<\delta$ then we have that $|z-p|<|p|/2$. Since $|z|\geq |p|-|z-p|$ and $|z-p|<|p|/2$ we obtain $|z|>|p|/2$ and therefore
$$\frac{|z-p|}{|zp|} \leq 2\frac{|z-p|}{|p|^2}.$$
Now we take $\delta < (|p|^2/2) \epsilon$ and by the above inequality we conclude that
$$\frac{|z-p|}{|zp|} <\epsilon.$$
So, if $\epsilon > 0$ then we can take a $\delta>0$ such that $\delta<|p|/2$ and $\delta < (|p|^2/2)\epsilon$ and we obtain
$$|z-p|<\delta \implies \left|\frac{1}{z} - \frac{1}{p}\right|<\epsilon.$$
Observation: The composition to obtain the function you want is the following: $f(z) = 1/z$ and $g(z) = 1-z$. The function you described is $f \circ g: \mathbb D_1(0) \to \mathbb C$ and it is continuous because $f,g$ are continuous.
