# Thread: Complex Analysis, uniformly continous function

1. ## Complex Analysis, uniformly continous function

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

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

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 $\displaystyle f: D\to \mathbb{C}$ is uniformly continous on $\displaystyle D$ if for every $\displaystyle \epsilon > 0$ there is $\displaystyle \delta >0$ so that if $\displaystyle |x-y| < \delta \text{ and }x,y\in D \implies |f(x)-f(y)| < \epsilon$. Note, if $\displaystyle f$ is uniformly continous on $\displaystyle D$ then certainly $\displaystyle f$ is continous on $\displaystyle D$.

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

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

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