Constructing a Bounded Non-Convergent Sequence in Complex Variables

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
desaila
Messages
25
Reaction score
0
I can't think of how to title the problem I'm having, but this is what the course is called. Complex being imaginary numbers, ie z = a + ic where i is the sqrt of -1.

So here is the question that I have no idea where to start with:

Construct a sequence {zn} which is bounded and for which the successive
terms get increasingly closer, but which is not convergent. In other words,
{zn} must satisfy:
(i) For some B > 0, |zn| < B for every n = 1, 2,...
(ii) For every n, |zn+2 - zn+1| < |zn+1 - zn|.
(iii) {zn} diverges.
Note that the inequality in (ii) is strict. Make sure to prove that your
sequence satisfies all three parts.


n is a subscript of z.

Thanks in advance.
 
Physics news on Phys.org
Thanks. On the same homework assignment, I have another problem but this is more with the question itself. What does this mean: limit of Arg z, as z approaches zero?
 
?? Just what it says! Or is the question rather "what does Arg z mean"?

Any complex number z= x+ iy can be written in polar form: [itex]z= r (cos(\theta)+ i sin(\theta)[/itex] or simply as [itex]z= r e^{i\theta}[/itex]. In either case [itex]Arg z= \theta[/itex].

Take a look at z= x+ ix. What is Arg z for all x? What is its limit as z (and so x) goes to 0?
Now look at z= x- ix. What is Arg z for all x? What is its limit as z (and so x) goes to 0?