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Is sign function continuous in complex plane? [closed]

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1

$begingroup$

Is sign function continuous in the complex plane? Is it continuous everywhere except at $z = 0$ ? Is it continuous in a circle of unit radius and argument having $exp(-itheta)$ ?

complex-analysis complex-numbers

share|cite|improve this question

edited Mar 25 at 19:50

J. W. Tanner

5,0751520

asked Mar 25 at 17:06

Jewel JohnsonJewel Johnson

11

$endgroup$

closed as off-topic by rschwieb, Thomas Shelby, Arnaud D., user21820, Erick Wong Mar 25 at 18:43

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." – rschwieb, Thomas Shelby, Arnaud D., user21820, Erick Wong

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

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  $begingroup$
  How do you define $sign(x + iy)$? Do you define it as $sign(x)$?
  $endgroup$
  – Siddharth Bhat
  Mar 25 at 17:09
  

  
  
  
  


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  $begingroup$
  You can define is as sgn(z) = |z|/z
  $endgroup$
  – Jewel Johnson
  Mar 25 at 17:19
  

  
  
  
  

add a comment | 

-1

1

$begingroup$

Is sign function continuous in the complex plane? Is it continuous everywhere except at $z = 0$ ? Is it continuous in a circle of unit radius and argument having $exp(-itheta)$ ?

complex-analysis complex-numbers

share|cite|improve this question

edited Mar 25 at 19:50

J. W. Tanner

5,0751520

asked Mar 25 at 17:06

Jewel JohnsonJewel Johnson

11

$endgroup$

closed as off-topic by rschwieb, Thomas Shelby, Arnaud D., user21820, Erick Wong Mar 25 at 18:43

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." – rschwieb, Thomas Shelby, Arnaud D., user21820, Erick Wong

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  How do you define $sign(x + iy)$? Do you define it as $sign(x)$?
  $endgroup$
  – Siddharth Bhat
  Mar 25 at 17:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can define is as sgn(z) = |z|/z
  $endgroup$
  – Jewel Johnson
  Mar 25 at 17:19
  

  
  
  
  

add a comment | 

$begingroup$

Is sign function continuous in the complex plane? Is it continuous everywhere except at $z = 0$ ? Is it continuous in a circle of unit radius and argument having $exp(-itheta)$ ?

complex-analysis complex-numbers

share|cite|improve this question

edited Mar 25 at 19:50

J. W. Tanner

5,0751520

asked Mar 25 at 17:06

Jewel JohnsonJewel Johnson

11

$endgroup$

share|cite|improve this question

edited Mar 25 at 19:50

J. W. Tanner

5,0751520

asked Mar 25 at 17:06

Jewel JohnsonJewel Johnson

11

share|cite|improve this question

edited Mar 25 at 19:50

J. W. Tanner

5,0751520

asked Mar 25 at 17:06

Jewel JohnsonJewel Johnson

11

edited Mar 25 at 19:50

J. W. Tanner

5,0751520

edited Mar 25 at 19:50

J. W. Tanner

5,0751520

asked Mar 25 at 17:06

Jewel JohnsonJewel Johnson

11

asked Mar 25 at 17:06

Jewel JohnsonJewel Johnson

11

1 Answer
1

active

oldest

votes

2

$begingroup$

$frac{|z|}{z}$ is continuous in $mathbb{C} setminus {0}$ as it is the quotient of continuous functions. However, $frac{|z|}{z}$ is obviously not defined in $z = 0$ (hence not continuous there either).

By the way, it is more common to use $frac{z}{|z|}$ instead of $frac{|z|}{z}$ because then you have $z = |z| cdot frac{z}{|z|}$ just like for the reals.

share|cite|improve this answer

edited Mar 25 at 17:29

answered Mar 25 at 17:22

KlausKlaus

2,955214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think this is what you are saying, but I thought it was at first, so to avoid confusion: $frac{|z|}{z} neq frac{z}{|z|}$. For example $frac{i}{|i|} = i neq -i = frac{|i|}{i}$
  $endgroup$
  – Callus
  Mar 25 at 17:27
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Callus Thanks, I improved the wording.
  $endgroup$
  – Klaus
  Mar 25 at 17:30
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$frac{|z|}{z}$ is continuous in $mathbb{C} setminus {0}$ as it is the quotient of continuous functions. However, $frac{|z|}{z}$ is obviously not defined in $z = 0$ (hence not continuous there either).

By the way, it is more common to use $frac{z}{|z|}$ instead of $frac{|z|}{z}$ because then you have $z = |z| cdot frac{z}{|z|}$ just like for the reals.

share|cite|improve this answer

edited Mar 25 at 17:29

answered Mar 25 at 17:22

KlausKlaus

2,955214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think this is what you are saying, but I thought it was at first, so to avoid confusion: $frac{|z|}{z} neq frac{z}{|z|}$. For example $frac{i}{|i|} = i neq -i = frac{|i|}{i}$
  $endgroup$
  – Callus
  Mar 25 at 17:27
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Callus Thanks, I improved the wording.
  $endgroup$
  – Klaus
  Mar 25 at 17:30
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$frac{|z|}{z}$ is continuous in $mathbb{C} setminus {0}$ as it is the quotient of continuous functions. However, $frac{|z|}{z}$ is obviously not defined in $z = 0$ (hence not continuous there either).

By the way, it is more common to use $frac{z}{|z|}$ instead of $frac{|z|}{z}$ because then you have $z = |z| cdot frac{z}{|z|}$ just like for the reals.

share|cite|improve this answer

edited Mar 25 at 17:29

answered Mar 25 at 17:22

KlausKlaus

2,955214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think this is what you are saying, but I thought it was at first, so to avoid confusion: $frac{|z|}{z} neq frac{z}{|z|}$. For example $frac{i}{|i|} = i neq -i = frac{|i|}{i}$
  $endgroup$
  – Callus
  Mar 25 at 17:27
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Callus Thanks, I improved the wording.
  $endgroup$
  – Klaus
  Mar 25 at 17:30
  
  
  

  
  
  
  

add a comment | 

$begingroup$

$frac{|z|}{z}$ is continuous in $mathbb{C} setminus {0}$ as it is the quotient of continuous functions. However, $frac{|z|}{z}$ is obviously not defined in $z = 0$ (hence not continuous there either).

By the way, it is more common to use $frac{z}{|z|}$ instead of $frac{|z|}{z}$ because then you have $z = |z| cdot frac{z}{|z|}$ just like for the reals.

share|cite|improve this answer

edited Mar 25 at 17:29

answered Mar 25 at 17:22

KlausKlaus

2,955214

$endgroup$

share|cite|improve this answer

edited Mar 25 at 17:29

answered Mar 25 at 17:22

KlausKlaus

2,955214

answered Mar 25 at 17:22

KlausKlaus

2,955214

answered Mar 25 at 17:22

KlausKlaus

2,955214