# Thread: Complex Analysis, uniformly continous function

1. ## Complex Analysis, uniformly continous function

Prove that $f(z)=z^3$ is uniformly continuous in the region { ${z\in C \mid \mid z \mid < 2 }$}

I do not understand what is the meaning of "uniformly continuous"... I read on wikipedia, but I still do not understand "uniformly continuous" in complex numbers. What theory should I use? How should I start in this case?

2. Originally Posted by andreas
Prove that $f(z)=z^3$ is uniformly continuous in the region { ${z\in C \mid \mid z \mid < 2 }$}

I do not understand what is the meaning of "uniformly continuous"... I read on wikipedia, but I still do not understand "uniformly continuous" in complex numbers. What theory should I use? How should I start in this case?
A function $f: D\to \mathbb{C}$ is uniformly continous on $D$ if for every $\epsilon > 0$ there is $\delta >0$ so that if $|x-y| < \delta \text{ and }x,y\in D \implies |f(x)-f(y)| < \epsilon$. Note, if $f$ is uniformly continous on $D$ then certainly $f$ is continous on $D$.

An easy way to prove uniform continuity here is to note that $f$ on $|z|\leq 1$ is a continous function. Continous functions on compact sets are uniformly continous on the set (this is a theorem). It follows that $f$ is uniformly continous on $|z| \leq 1$. But if it is uniformly continous on $|z|\leq 1$ then it is surly uniformly continous on $|z|<1$ for it is a subset.

3. Originally Posted by andreas
Prove that $f(z)=z^3$ is uniformly continuous in the region { ${z\in C \mid \mid z \mid < 2 }$}
I do not understand what is the meaning of "uniformly continuous"
The meaning of "uniformly continuous on $C = \left\{ {z:\left| z \right| < 2} \right\}$ means that: $\left( {\forall \varepsilon > 0} \right)\left( {\exists \delta > 0} \right)\left( {\forall z_0 \in C} \right)\left[ {\left| {z - z_0 } \right| < \delta \, \Rightarrow \,\left| {f(z) - f(z_0 )} \right| < \varepsilon } \right]$.
In other words it is the usual $\varepsilon \backslash \delta$ but it works for any point in the set.
That is, we do not have to first pick a point.

Note that $z_0 \in C\, \Rightarrow \,\left| {z^3 - z_0 ^3 } \right| = \left| {z - z_0 } \right|\left| {z^2 + zz_0 + z_0 ^2 } \right| < \left| {z - z_0 } \right|(12)$.
Thus if $\varepsilon > 0 \text{ let }\delta = \frac{\varepsilon }{{12}}$.