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Multiplication of the square root of complex numbers

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1

$begingroup$

$$
sqrt-21+20i.sqrt-21-20i=pm(2+5i).pm(2-5i)=pm29
$$
But why it is not
$$
sqrt-21+20i.sqrt-21-20i=sqrt(-21+20i)(-21-20i)=sqrtz\
=sqrt441+400=sqrt841=29
$$

Does this has something to do with $sqrt-asqrt-b=-sqrtabneqsqrtab,;a,binmathbbR_+$ ?

Where does all these rules coming from ?

complex-numbers radicals

share|cite|improve this question

edited Mar 17 at 12:10

José Carlos Santos

170k23132238

asked Mar 17 at 11:24

ss1729ss1729

2,04911124

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $sqrt841=pm29$
  $endgroup$
  – InsideOut
  Mar 17 at 11:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @InsideOut for real numbers $sqrty=sqrtx^2=|x|$ we mean the +ve square root, right?
  $endgroup$
  – ss1729
  Mar 17 at 12:38
  
  

  
  
  
  

add a comment | 

1

$begingroup$

$$
sqrt-21+20i.sqrt-21-20i=pm(2+5i).pm(2-5i)=pm29
$$
But why it is not
$$
sqrt-21+20i.sqrt-21-20i=sqrt(-21+20i)(-21-20i)=sqrtz\
=sqrt441+400=sqrt841=29
$$

Does this has something to do with $sqrt-asqrt-b=-sqrtabneqsqrtab,;a,binmathbbR_+$ ?

Where does all these rules coming from ?

complex-numbers radicals

share|cite|improve this question

edited Mar 17 at 12:10

José Carlos Santos

170k23132238

asked Mar 17 at 11:24

ss1729ss1729

2,04911124

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $sqrt841=pm29$
  $endgroup$
  – InsideOut
  Mar 17 at 11:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @InsideOut for real numbers $sqrty=sqrtx^2=|x|$ we mean the +ve square root, right?
  $endgroup$
  – ss1729
  Mar 17 at 12:38
  
  

  
  
  
  

add a comment | 

$begingroup$

$$
sqrt-21+20i.sqrt-21-20i=pm(2+5i).pm(2-5i)=pm29
$$
But why it is not
$$
sqrt-21+20i.sqrt-21-20i=sqrt(-21+20i)(-21-20i)=sqrtz\
=sqrt441+400=sqrt841=29
$$

Does this has something to do with $sqrt-asqrt-b=-sqrtabneqsqrtab,;a,binmathbbR_+$ ?

Where does all these rules coming from ?

complex-numbers radicals

share|cite|improve this question

edited Mar 17 at 12:10

José Carlos Santos

170k23132238

asked Mar 17 at 11:24

ss1729ss1729

2,04911124

$endgroup$

share|cite|improve this question

edited Mar 17 at 12:10

José Carlos Santos

170k23132238

asked Mar 17 at 11:24

ss1729ss1729

2,04911124

share|cite|improve this question

edited Mar 17 at 12:10

José Carlos Santos

170k23132238

asked Mar 17 at 11:24

ss1729ss1729

2,04911124

edited Mar 17 at 12:10

José Carlos Santos

170k23132238

edited Mar 17 at 12:10

José Carlos Santos

170k23132238

asked Mar 17 at 11:24

ss1729ss1729

2,04911124

asked Mar 17 at 11:24

ss1729ss1729

2,04911124

3 Answers
3

active

oldest

votes

4

$begingroup$

In the context of complex numbers, it is not a good idea to write $sqrt z$, since every complex number (other than $0$) has two square roots. So, unless you specify which square root you have in mind, the expression $sqrt z$ is ambiguous. And all you can say about the square roots of a product is that each of them can be obtaind multiplying some square root of the first factor by some square root of the second one.

share|cite|improve this answer

edited Mar 17 at 13:07

answered Mar 17 at 11:48

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

add a comment | 

0

$begingroup$

By definition, $sqrt(x) = y$ means that $y^2 = x$, so both $y$ and $-y$ would work. However, in case of real numbers, square root is conventionally defined as a non-negative root.

In your case $sqrt(29^2) = sqrt((-29)^2) = 29$, so both roots are valid.

share|cite|improve this answer

answered Mar 17 at 12:04

vladzvladz

541211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $sqrtx=yimplies x,yinmathbbR_+$ because $sqrtx$ we mean the +ve(principal) square root
  $endgroup$
  – ss1729
  Mar 17 at 12:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $841$ has two square roots $+29$ and $-29$ I agree, but $sqrt841$ we mean the +ve square root.
  $endgroup$
  – ss1729
  Mar 17 at 12:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Only in case of real numbers. With complex, square root typically means “any root”
  $endgroup$
  – vladz
  Mar 17 at 12:55
  

  
  
  
  

add a comment | 

0

$begingroup$

Square root of any non-zero complex number is two valued function since $(-z)^2=z^2$. In the case of the positive real numbers (and only in this case) both roots are real numbers. The positive of them is termed principal square root and usually is denoted as $sqrt x$. To avoid misunderstanding it is therefore advantageous to use $x^frac12$ instead of $sqrt x$.

And $841^frac12=pm29$.

share|cite|improve this answer

edited Mar 17 at 16:58

answered Mar 17 at 11:48

useruser

5,93011031

$endgroup$

add a comment | 

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4

$begingroup$

In the context of complex numbers, it is not a good idea to write $sqrt z$, since every complex number (other than $0$) has two square roots. So, unless you specify which square root you have in mind, the expression $sqrt z$ is ambiguous. And all you can say about the square roots of a product is that each of them can be obtaind multiplying some square root of the first factor by some square root of the second one.

share|cite|improve this answer

edited Mar 17 at 13:07

answered Mar 17 at 11:48

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

add a comment | 

4

$begingroup$

In the context of complex numbers, it is not a good idea to write $sqrt z$, since every complex number (other than $0$) has two square roots. So, unless you specify which square root you have in mind, the expression $sqrt z$ is ambiguous. And all you can say about the square roots of a product is that each of them can be obtaind multiplying some square root of the first factor by some square root of the second one.

share|cite|improve this answer

edited Mar 17 at 13:07

answered Mar 17 at 11:48

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

add a comment | 

$begingroup$

In the context of complex numbers, it is not a good idea to write $sqrt z$, since every complex number (other than $0$) has two square roots. So, unless you specify which square root you have in mind, the expression $sqrt z$ is ambiguous. And all you can say about the square roots of a product is that each of them can be obtaind multiplying some square root of the first factor by some square root of the second one.

share|cite|improve this answer

edited Mar 17 at 13:07

answered Mar 17 at 11:48

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

share|cite|improve this answer

edited Mar 17 at 13:07

answered Mar 17 at 11:48

José Carlos SantosJosé Carlos Santos

170k23132238

answered Mar 17 at 11:48

José Carlos SantosJosé Carlos Santos

170k23132238

answered Mar 17 at 11:48

José Carlos SantosJosé Carlos Santos

170k23132238

0

$begingroup$

By definition, $sqrt(x) = y$ means that $y^2 = x$, so both $y$ and $-y$ would work. However, in case of real numbers, square root is conventionally defined as a non-negative root.

In your case $sqrt(29^2) = sqrt((-29)^2) = 29$, so both roots are valid.

share|cite|improve this answer

answered Mar 17 at 12:04

vladzvladz

541211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $sqrtx=yimplies x,yinmathbbR_+$ because $sqrtx$ we mean the +ve(principal) square root
  $endgroup$
  – ss1729
  Mar 17 at 12:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $841$ has two square roots $+29$ and $-29$ I agree, but $sqrt841$ we mean the +ve square root.
  $endgroup$
  – ss1729
  Mar 17 at 12:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Only in case of real numbers. With complex, square root typically means “any root”
  $endgroup$
  – vladz
  Mar 17 at 12:55
  

  
  
  
  

add a comment | 

0

$begingroup$

By definition, $sqrt(x) = y$ means that $y^2 = x$, so both $y$ and $-y$ would work. However, in case of real numbers, square root is conventionally defined as a non-negative root.

In your case $sqrt(29^2) = sqrt((-29)^2) = 29$, so both roots are valid.

share|cite|improve this answer

answered Mar 17 at 12:04

vladzvladz

541211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $sqrtx=yimplies x,yinmathbbR_+$ because $sqrtx$ we mean the +ve(principal) square root
  $endgroup$
  – ss1729
  Mar 17 at 12:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $841$ has two square roots $+29$ and $-29$ I agree, but $sqrt841$ we mean the +ve square root.
  $endgroup$
  – ss1729
  Mar 17 at 12:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Only in case of real numbers. With complex, square root typically means “any root”
  $endgroup$
  – vladz
  Mar 17 at 12:55
  

  
  
  
  

add a comment | 

$begingroup$

By definition, $sqrt(x) = y$ means that $y^2 = x$, so both $y$ and $-y$ would work. However, in case of real numbers, square root is conventionally defined as a non-negative root.

In your case $sqrt(29^2) = sqrt((-29)^2) = 29$, so both roots are valid.

share|cite|improve this answer

answered Mar 17 at 12:04

vladzvladz

541211

$endgroup$

share|cite|improve this answer

answered Mar 17 at 12:04

vladzvladz

541211

answered Mar 17 at 12:04

vladzvladz

541211

answered Mar 17 at 12:04

vladzvladz

541211

0

$begingroup$

Square root of any non-zero complex number is two valued function since $(-z)^2=z^2$. In the case of the positive real numbers (and only in this case) both roots are real numbers. The positive of them is termed principal square root and usually is denoted as $sqrt x$. To avoid misunderstanding it is therefore advantageous to use $x^frac12$ instead of $sqrt x$.

And $841^frac12=pm29$.

share|cite|improve this answer

edited Mar 17 at 16:58

answered Mar 17 at 11:48

useruser

5,93011031

$endgroup$

add a comment | 

0

$begingroup$

Square root of any non-zero complex number is two valued function since $(-z)^2=z^2$. In the case of the positive real numbers (and only in this case) both roots are real numbers. The positive of them is termed principal square root and usually is denoted as $sqrt x$. To avoid misunderstanding it is therefore advantageous to use $x^frac12$ instead of $sqrt x$.

And $841^frac12=pm29$.

share|cite|improve this answer

edited Mar 17 at 16:58

answered Mar 17 at 11:48

useruser

5,93011031

$endgroup$

add a comment | 

$begingroup$

Square root of any non-zero complex number is two valued function since $(-z)^2=z^2$. In the case of the positive real numbers (and only in this case) both roots are real numbers. The positive of them is termed principal square root and usually is denoted as $sqrt x$. To avoid misunderstanding it is therefore advantageous to use $x^frac12$ instead of $sqrt x$.

And $841^frac12=pm29$.

share|cite|improve this answer

edited Mar 17 at 16:58

answered Mar 17 at 11:48

useruser

5,93011031

$endgroup$

share|cite|improve this answer

edited Mar 17 at 16:58

answered Mar 17 at 11:48

useruser

5,93011031

answered Mar 17 at 11:48

useruser

5,93011031

answered Mar 17 at 11:48

useruser

5,93011031