Prove a complex set defined by inequalities is open

[skriabin]

Homework Statement

Prove that the set D = {z in C; |z^2 - 1| < 1} is open

Homework Equations

The Attempt at a Solution

I have to show that for any z in D, there exists r > 0 s.t. the nbhd N(z,r) is contained in D. Let w in N(z,r) => |z - w| < r. Need to show |w^2 - 1| < 1 for some r > 0.

I tried |w^2 - 1| = |(w^2 - z^2) + (z^2 - 1)| <= |w - z||w + z| + |z^2 - 1|,

but since we have |z^2 - 1| < 1 already, it doesn't seem to help much. Not sure what else to try. Thanks for any help!

 

[Office_Shredder]

I tried |w^2 - 1| = |(w^2 - z^2) + (z^2 - 1)| <= |w - z||w + z| + |z^2 - 1|,

but since we have |z^2 - 1| < 1 already, it doesn't seem to help much. Not sure what else to try. Thanks for any help!

So that looks pretty good...for a given z, let |z²-1| be equal to 1-e for some (probably small) e. |w²-1| <= |w-z||w+z| + 1 - e < 1 if e>|w-z||w+z|

When |w-z| is really small, what can you say about |w+z|, and then what can you say about comparing e and |w-z|

 

[skriabin]

hmm |w + z| then approaches 2|z| ? Giving then e > 2|z||w - z|? and?

 

[Office_Shredder]

You know what e is. You know what z is. Your goal is to find how small |w-z| is. What can you do with the inequality?

Also note that |w+z| is not going to be exactly 2|z|, you should allow a little wiggle room (for example, say |w+z|<2.1|z| then when you pick r, pick it to be either what's required from your calculation, or what's required to make |w+z|<2.1|z|, whichever is smaller)

 

[skriabin]

But can't we choose r = |w - z| < e/|w + z| instead of r < e/2.1|z| ? Then we would have the inequality satisfied no?

 

[Office_Shredder]

But can't we choose r = |w - z| < e/|w + z| instead of r < e/2.1|z| ? Then we would have the inequality satisfied no?

You can't define r in terms of w, since w is picked based on what r is!

 

[skriabin]

oops meant r = e/|w + z|. sry about that.

 

[skriabin]

You can't define r in terms of w, since w is picked based on what r is!

But if we pick a w based on what r is, how can we prove any w in the nbhd is in D?

 

[skriabin]

Maybe I misunderstood. I will check again next morning. I'm off to bed. Thanks for the help.