Proving Inequalities for Complex Analysis Limits

[Metahominid]

Homework Statement

I'm not very good with LaTeX and the reference button seems to broken.
So

Assume lim h(z) = 1+i, as z->w, prove there exists a delta, d>0
s.t. 0<|z-w|<d -> (2^.5)/2 < |h(z)| < 3(2^.5)/2

Homework Equations

The Attempt at a Solution

Kinda been running in circles but from assumption
there exists d>0 s.t. 0<|z-w|<d -> |h(z) - (1+i)| < e (e > 0, epsilon)

So
|h(z)| = |h(z) - (1+i) + (1+i)| <= |h(z) - (1+i)| + |(1+i)|
therefore |h(z)| < e + |1+i| = e + (2^.5)

This seems alright so far but I feel like there is a much better way so
continuing for the other part of the inequality doesn't seem right

 

[micromass]

Your proof isn't finished yet. You've got:

|h(z)| < e + |1+i| = e + (2^.5)

But how did you choose e to obtain $$|h(z)|<\frac{3\sqrt{2}}{2}$$?

The proof looks good. But what I would do is show in general that a function $$f:D\rightarrow \mathbb{R}$$ with $$\lim_{x\rightarrow a}{f(x)}=c$$ satisfies that there exists a $$\delta$$ such that $$c-\epsilon<f(x)<c+\epsilon$$ for all $$0<|x-a|<\delta$$.
The thing you have to prove follows form this with f(x)=|h(x)|.