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.
Best Answer
Hint:) As an alternative way you can show that this function is differentiable in the unit disk.