Prove that the function $1/z$ is uniformly continuous on $\frac{1}{3}\le\left|z\right|<1$, is the function uniformly continuous on $0<\left|z\right|\le1$?
My definition of uniform continuity is:Assume $S \subseteq \mathbb C$ and $f:S \to \mathbb C$ is a complex function, then $f$ is uniformly continuous on $S$ if for every positive real number $\epsilon$ there exists a positive real number $\delta$ such that for all $z_1,z_2 \in S$ if $\left|z_{1}-z_{2}\right|<\delta$ then $\left|f(z_{1})-f(z_{2})\right|<\epsilon$.
For $f(z)=1/z$ we have that $$\left|f(z_{1})-f(z_{2})\right|=\left|\frac{z_{1}-z_{2}}{z_{1}z_{2}}\right|<\frac{ \delta}{\left|z_{1}\right|\left|z_{2}\right|}$$
So taking $\delta=\left|z_{1}\right|\left|z_{2}\right|\epsilon/2$ shows that for all $\epsilon>0$ and $z_1,z_2 \in S$ if $\left|z_{1}-z_{2}\right|<\left|z_{1}\right|\left|z_{2}\right|\epsilon/2$ then $$\left|\frac{1}{z_{1}}-\frac{1}{z_{2}}\right|\le\frac{2\left|z_{1}-z_{2}\right|}{\left|z_{1}\right|\left|z_{2}\right|}<\epsilon$$
However this shows that the function is continuous at $\mathbb C$, and I think there should be some errors.
You cannot make $\delta$ depend on $z_1$ and $z_2$. Use the fact that $|\frac 1 {z_1}-\frac 1 {z_2}|=\frac {|z_1-z_2|} {|z_1||z_2|}\leq 9|z_1-z_2|$ so we can take $\delta= \epsilon /9$.
The function is not uniformly continuous on $\{z: 0<|z| \leq 1\}$. To prove this note that $|f(\frac 1 n)-f(\frac 1 {n+1})|=1$ for all $n$. Write down the definition of uniform continuity and show that the condition for uniform continuity fails when $z_1=\frac 1n, z_2=\frac 1 {n+1}$ with $n$ sufficiently large.