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$ |1+z+z^2 +…+z^n| geq |z|^n $ is valid [closed]

5

1

$begingroup$

Let $z=a+ib$ , where $a>0$ & $b $ are integers, then $ |1+z+z^2 +...+z^n| geq |z|^n $ is always holds...

I tried to prove the because this seems to be true... But in vain

complex-analysis inequality complex-numbers

share|cite|improve this question

asked Mar 26 at 8:59

user561527user561527

595

$endgroup$

closed as off-topic by uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會 Mar 26 at 13:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why does it seem to be true? For which integers did you verify it to be true? Do you have an example of equality?
  $endgroup$
  – uniquesolution
  Mar 26 at 9:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I put several integers... That doesn't contradict this inequality
  $endgroup$
  – user561527
  Mar 26 at 9:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It‘s definitely true for $n = 2$, I‘ve just verified that.
  $endgroup$
  – Kezer
  Mar 26 at 10:28
  

  
  
  
  

add a comment | 

5

1

$begingroup$

Let $z=a+ib$ , where $a>0$ & $b $ are integers, then $ |1+z+z^2 +...+z^n| geq |z|^n $ is always holds...

I tried to prove the because this seems to be true... But in vain

complex-analysis inequality complex-numbers

share|cite|improve this question

asked Mar 26 at 8:59

user561527user561527

595

$endgroup$

closed as off-topic by uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會 Mar 26 at 13:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why does it seem to be true? For which integers did you verify it to be true? Do you have an example of equality?
  $endgroup$
  – uniquesolution
  Mar 26 at 9:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I put several integers... That doesn't contradict this inequality
  $endgroup$
  – user561527
  Mar 26 at 9:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It‘s definitely true for $n = 2$, I‘ve just verified that.
  $endgroup$
  – Kezer
  Mar 26 at 10:28
  

  
  
  
  

add a comment | 

$begingroup$

Let $z=a+ib$ , where $a>0$ & $b $ are integers, then $ |1+z+z^2 +...+z^n| geq |z|^n $ is always holds...

I tried to prove the because this seems to be true... But in vain

complex-analysis inequality complex-numbers

share|cite|improve this question

asked Mar 26 at 8:59

user561527user561527

595

$endgroup$

share|cite|improve this question

asked Mar 26 at 8:59

user561527user561527

595

share|cite|improve this question

asked Mar 26 at 8:59

user561527user561527

595

asked Mar 26 at 8:59

user561527user561527

595

asked Mar 26 at 8:59

user561527user561527

595

1 Answer
1

active

oldest

votes

3

$begingroup$

The inequality is true. For $n=0,1$ the inequality is clear, hence let $n geq 2$. The case $z=1$ is also clear, so let $z neq 1$.

Note that $1+z+dots + z^n = fracz^n+1-1z-1$. So the given inequality is equivalent to
$$ left|z - frac1z^n right| geq |z-1| iff left|z - frac1z^n right|^2 geq |z-1|^2.$$
Now adhere to $|z|^2 = z overlinez$. So equivalently we have to prove
$$ z overlinez - left(fraczoverlinez^n + fracoverlinezz^n right) + frac1z^n overlinez^n geq zoverlinez - (z + overlinez) + 1.$$ This is equivalent to
$$2 operatornameRe(z) + frac1 - left(z^n+1 + overlinez^n+1 right) geq 1 iff 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 1. $$ But the estimation $$ 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 2 operatornameRe(z) + frac1 - frac2 overset?geq 1 $$ almost finishes. Note that $operatornameRe(z) geq 1$ by assumption. As $z neq 1$ by assumption, $frac2 leq 1$, so again, we're done.

Our estimations were somewhat crude, so this inequality can be strengthened considerably with a more detailed analysis.

share|cite|improve this answer

edited Mar 26 at 11:26

answered Mar 26 at 11:04

KezerKezer

1,405621

$endgroup$

add a comment | 

3

$begingroup$

The inequality is true. For $n=0,1$ the inequality is clear, hence let $n geq 2$. The case $z=1$ is also clear, so let $z neq 1$.

Note that $1+z+dots + z^n = fracz^n+1-1z-1$. So the given inequality is equivalent to
$$ left|z - frac1z^n right| geq |z-1| iff left|z - frac1z^n right|^2 geq |z-1|^2.$$
Now adhere to $|z|^2 = z overlinez$. So equivalently we have to prove
$$ z overlinez - left(fraczoverlinez^n + fracoverlinezz^n right) + frac1z^n overlinez^n geq zoverlinez - (z + overlinez) + 1.$$ This is equivalent to
$$2 operatornameRe(z) + frac1 - left(z^n+1 + overlinez^n+1 right) geq 1 iff 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 1. $$ But the estimation $$ 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 2 operatornameRe(z) + frac1 - frac2 overset?geq 1 $$ almost finishes. Note that $operatornameRe(z) geq 1$ by assumption. As $z neq 1$ by assumption, $frac2 leq 1$, so again, we're done.

Our estimations were somewhat crude, so this inequality can be strengthened considerably with a more detailed analysis.

share|cite|improve this answer

edited Mar 26 at 11:26

answered Mar 26 at 11:04

KezerKezer

1,405621

$endgroup$

add a comment | 

3

$begingroup$

The inequality is true. For $n=0,1$ the inequality is clear, hence let $n geq 2$. The case $z=1$ is also clear, so let $z neq 1$.

Note that $1+z+dots + z^n = fracz^n+1-1z-1$. So the given inequality is equivalent to
$$ left|z - frac1z^n right| geq |z-1| iff left|z - frac1z^n right|^2 geq |z-1|^2.$$
Now adhere to $|z|^2 = z overlinez$. So equivalently we have to prove
$$ z overlinez - left(fraczoverlinez^n + fracoverlinezz^n right) + frac1z^n overlinez^n geq zoverlinez - (z + overlinez) + 1.$$ This is equivalent to
$$2 operatornameRe(z) + frac1 - left(z^n+1 + overlinez^n+1 right) geq 1 iff 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 1. $$ But the estimation $$ 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 2 operatornameRe(z) + frac1 - frac2 overset?geq 1 $$ almost finishes. Note that $operatornameRe(z) geq 1$ by assumption. As $z neq 1$ by assumption, $frac2 leq 1$, so again, we're done.

Our estimations were somewhat crude, so this inequality can be strengthened considerably with a more detailed analysis.

share|cite|improve this answer

edited Mar 26 at 11:26

answered Mar 26 at 11:04

KezerKezer

1,405621

$endgroup$

add a comment | 

$begingroup$

The inequality is true. For $n=0,1$ the inequality is clear, hence let $n geq 2$. The case $z=1$ is also clear, so let $z neq 1$.

Note that $1+z+dots + z^n = fracz^n+1-1z-1$. So the given inequality is equivalent to
$$ left|z - frac1z^n right| geq |z-1| iff left|z - frac1z^n right|^2 geq |z-1|^2.$$
Now adhere to $|z|^2 = z overlinez$. So equivalently we have to prove
$$ z overlinez - left(fraczoverlinez^n + fracoverlinezz^n right) + frac1z^n overlinez^n geq zoverlinez - (z + overlinez) + 1.$$ This is equivalent to
$$2 operatornameRe(z) + frac1 - left(z^n+1 + overlinez^n+1 right) geq 1 iff 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 1. $$ But the estimation $$ 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 2 operatornameRe(z) + frac1 - frac2 overset?geq 1 $$ almost finishes. Note that $operatornameRe(z) geq 1$ by assumption. As $z neq 1$ by assumption, $frac2 leq 1$, so again, we're done.

Our estimations were somewhat crude, so this inequality can be strengthened considerably with a more detailed analysis.

share|cite|improve this answer

edited Mar 26 at 11:26

answered Mar 26 at 11:04

KezerKezer

1,405621

$endgroup$

share|cite|improve this answer

edited Mar 26 at 11:26

answered Mar 26 at 11:04

KezerKezer

1,405621

answered Mar 26 at 11:04

KezerKezer

1,405621

answered Mar 26 at 11:04

KezerKezer

1,405621