What is the limit of the complex sequence z_n = [(1+i)/sqrt(3)]^n?

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Homework Help Overview

The discussion revolves around finding the limit of the complex sequence z_n = [(1+i)/sqrt(3)]^n as n approaches infinity. Participants are exploring the behavior of this sequence in the context of complex analysis.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Some participants express confusion regarding the limit being zero, questioning how this conclusion is reached given that (1+i)^n appears unbounded. Others explore the absolute value of (1+i) and its implications for the limit of z_n.

Discussion Status

The discussion is active, with participants examining the relationship between the absolute value of the sequence and its limit. There is a recognition that the limit of |z_n| approaching zero would imply the limit of z_n itself approaches zero, although some caution against assuming this relationship universally.

Contextual Notes

Participants are grappling with the implications of the absolute value of complex sequences and the conditions under which limits can be inferred. There is an acknowledgment of the complexity of convergence in the context of complex numbers.

DotKite
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Homework Statement



find the limit z_n = [(1+i)/sqrt(3)]^n as n -> ∞.

Homework Equations


3. The Attempt at a Solution

Apparently the limit is zero (via back of the book), but I have no clue how they got that answer.

(1 + i)^n seems to be unbounded, thus i do not see how z_n can go to zero I am lost.
 
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DotKite said:

Homework Statement



find the limit z_n = [(1+i)/sqrt(3)]^n as n -> ∞.

Homework Equations


3. The Attempt at a Solution

Apparently the limit is zero (via back of the book), but I have no clue how they got that answer.

(1 + i)^n seems to be unbounded, thus i do not see how z_n can go to zero I am lost.


(1+i)^n is unbounded. But its absolute value is |1+i|^n. What's that??
 
Last edited:
Dick said:
(1+i)^n is unbounded. But its absolute value is |1+i|^n. What's that??

|1+i|^n = [sqrt(2)]^n?
 
DotKite said:
|1+i|^n = [sqrt(2)]^n?

Sure, so can you show the limit of |z_n| is 0? That would show the limit of z_n is also 0.
 
Dick said:
Sure, so can you show the limit of |z_n| is 0? That would show the limit of z_n is also 0.

that is not generally true,

take |(-1)^n + i/n| which converges to 1

{(-1)^n + i/n} does not converge.
 
DotKite said:
that is not generally true,

take |(-1)^n + i/n| which converges to 1

{(-1)^n + i/n} does not converge.
But it is true if the limit is zero. ##|z_n| \rightarrow 0## if and only if ##z_n \rightarrow 0##.
 
oh i see
 

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