# Delta epsilon proof of continuity complex analysis

• xokaitt
I think I need someone who has done this before to help me out.In summary, the function F is continuous at every z0 if given an epsilon > 0, there exists a delta > 0 such that for all z_0 in C, |z-z_0|< delta implies that's |F(z) - F(z_0)|< epsilon.f

## Homework Statement

show that the function
F:C$$\rightarrow$$C
z $$\rightarrow$$ z+|z|

is continuous for every z0$$\in$$ C

2. Proof
F is continuous at every z0$$\in$$ C if given an \epsilon > 0 , there exists a $$\delta$$ > 0 such that $$\forall$$ z 0 $$\in$$ C, |z-z 0|< $$\delta$$ implies |F(z)-F(z0)|< $$\epsilon$$.

I know basically how to do this, if someone could just help me with the theoretical steps. First we suppose we are given an epsilon that works? then we have to relate epsilon and delta to find a delta (in terms of epsilon) that works...? then once we have the epsilon and delta we plug back into verify?
or do i have the concept backwards?

It's not that you are given some specific epsilon by someone else that "works", its basically saying that for ANY given epsilon > 0, there exists a delta > 0 such that for all z_0 in C, |z-z_0|< delta implies that's |F(z) - F(z_0)|< epsilon.

Since its ANY epsilon > 0, they inequality should still hold if I made epsilon as close to zero as I wanted. So the concept is that : If F is continuous, then there must be SOME value of delta I can pick, ie get z_0 close enough to z, so that epsilon is as small as I wanted, ie F(z_0) and F(z) are as close as I want them to be.

maybe i should have been more specific ... I understand the geometrical concept of an epsilon neighborhood. What i do not understand however, is how to go about the proof. Formally speaking, what is the correct way to construct a delta-epsilon proof and how do I begin?

i have been manipulating the properties of the modulus for a little while now and have a long string of complicated inequalities ... but I'm just confusing myself.