# Sequence of complex numbers

1. Mar 12, 2012

### gestalt

Let <zn> be a sequence complex numbers for which Im(zn) is bounded below.
Prove <e^(i*zn)> has a convergent subsequence.

My question on this is what possible help could the boundedness of the Im(zn) to this proof and what theorem might be of help?

2. Mar 12, 2012

### micromass

Staff Emeritus
Do you know anything about compactness or Bolzano-Weierstrass??

3. Mar 12, 2012

### mathwonk

do you know how e^z behaves geometrically?

4. Mar 13, 2012

### gestalt

Yes, I do know the B-W theorem. I looked at it but it applies to bounded sequences. This sequence is only bounded below and only on the Im part. I suppose since you asked you see a way it is still applicable.

It was my understanding e^z forms a circle in the complex. Is this what you mean?

Last edited: Mar 13, 2012
5. Mar 13, 2012

### micromass

Staff Emeritus
Try to show that the sequence is bounded. Try to show that there is a constant C such that

$$|e^{iz_n}|\leq C$$

6. Mar 13, 2012

### gestalt

Edit: The latex was not loading but now I see what you mean. I apologize if this sounds ignorant but I did not think e^(i*zn) was bounded. If you are saying I should show it is then I assume it can be.
When I think of this sequence I must be seeing it all wrong. could you give any explanation of how it behaves. If it helps I see nothing but a circle when I think of it. What attribute am I missing?

Last edited: Mar 13, 2012
7. Mar 13, 2012

### mathwonk

circles are bounded, right?