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Is taking modulus on both sides of an equation valid?

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Is taking modulus on both sides of an equation valid?

Square root of a squared number changes sign, which to apply first?Why is it valid to multiply both sides of an equation by its complex conjugate?Will there be two square roots for a Complex number?Taking Mod on both sides, mathematically correct?Solving and graphing all values of $z$.When do you take into account the +2kpi for complex numbers arguments in complex equations$z^2 = sqrt3+ 3i$ (complex equation)Range of sum of complex numberssolve $(x+iy)^2= a+ ib$Question on the logic of a proof involving complex numbers

2

$begingroup$

This might look like a copy of another question, but what I'm about to propose here is new. There's this question,

Find the least positive integral value of n for which $(frac1+i1-i)^n = 1$

While solving, if we multiply what is within the bracket, by the conjugate of denominator and divide by the same thing, we get $i$ in the bracket, that means, the question boils down to

$i ^ n= 1$

We know that the least positive value $n$ can have is $4$, for $i^n$ to be $1$.
Done.
Now,
IF I were to solve it by taking mod on both sides of the given equation,
I would get

$Big(frac1+iBig)^n = |1|$

$Big(fracsqrt2sqrt2Big)^n = 1$

$1^n = 1$

NOTE that the least positive value of $n$ changes from $4$ to $1$.

Why is it so?
I read an answer on stack exchange, that it is valid to do anything with the equation until it maintains the equality. I didn't destroy the equality, so why does the answer vary?

Is there any restriction as to where to use the "taking-mod-both-sides" thing?

complex-numbers

share|cite|improve this question

edited 44 mins ago

Ryan Shesler

40711

asked 53 mins ago

user231094user231094

111

New contributor

user231094 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  If $a=b$, then $|a|=|b|$. In this sense, it is "valid" to take the modulus of both sides. It is not necessarily the case, however, that if $|a|=|b|$, then $a=b$. Assuming this latter statement is what is leading you to incorrectly conclude that $n=1$ is a solution to the original problem.
  $endgroup$
  – jawheele
  42 mins ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

This might look like a copy of another question, but what I'm about to propose here is new. There's this question,

Find the least positive integral value of n for which $(frac1+i1-i)^n = 1$

While solving, if we multiply what is within the bracket, by the conjugate of denominator and divide by the same thing, we get $i$ in the bracket, that means, the question boils down to

$i ^ n= 1$

We know that the least positive value $n$ can have is $4$, for $i^n$ to be $1$.
Done.
Now,
IF I were to solve it by taking mod on both sides of the given equation,
I would get

$Big(frac1+iBig)^n = |1|$

$Big(fracsqrt2sqrt2Big)^n = 1$

$1^n = 1$

NOTE that the least positive value of $n$ changes from $4$ to $1$.

Why is it so?
I read an answer on stack exchange, that it is valid to do anything with the equation until it maintains the equality. I didn't destroy the equality, so why does the answer vary?

Is there any restriction as to where to use the "taking-mod-both-sides" thing?

complex-numbers

share|cite|improve this question

edited 44 mins ago

Ryan Shesler

40711

asked 53 mins ago

user231094user231094

111

New contributor

user231094 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  If $a=b$, then $|a|=|b|$. In this sense, it is "valid" to take the modulus of both sides. It is not necessarily the case, however, that if $|a|=|b|$, then $a=b$. Assuming this latter statement is what is leading you to incorrectly conclude that $n=1$ is a solution to the original problem.
  $endgroup$
  – jawheele
  42 mins ago
  
  

  
  
  
  

add a comment | 

$begingroup$

This might look like a copy of another question, but what I'm about to propose here is new. There's this question,

Find the least positive integral value of n for which $(frac1+i1-i)^n = 1$

While solving, if we multiply what is within the bracket, by the conjugate of denominator and divide by the same thing, we get $i$ in the bracket, that means, the question boils down to

$i ^ n= 1$

We know that the least positive value $n$ can have is $4$, for $i^n$ to be $1$.
Done.
Now,
IF I were to solve it by taking mod on both sides of the given equation,
I would get

$Big(frac1+iBig)^n = |1|$

$Big(fracsqrt2sqrt2Big)^n = 1$

$1^n = 1$

NOTE that the least positive value of $n$ changes from $4$ to $1$.

Why is it so?
I read an answer on stack exchange, that it is valid to do anything with the equation until it maintains the equality. I didn't destroy the equality, so why does the answer vary?

Is there any restriction as to where to use the "taking-mod-both-sides" thing?

complex-numbers

share|cite|improve this question

edited 44 mins ago

Ryan Shesler

40711

asked 53 mins ago

user231094user231094

111

New contributor

user231094 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 44 mins ago

Ryan Shesler

40711

asked 53 mins ago

user231094user231094

111

New contributor

user231094 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 44 mins ago

Ryan Shesler

40711

asked 53 mins ago

user231094user231094

111

New contributor

user231094 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 44 mins ago

Ryan Shesler

40711

edited 44 mins ago

Ryan Shesler

40711

asked 53 mins ago

user231094user231094

111

New contributor

user231094 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 53 mins ago

user231094user231094

111

2 Answers
2

active

oldest

votes

3

$begingroup$

The problem is, "taking-mod-of-both-sides" is valid, but only gives a one way implication. If $z_1 = z_2$ then $|z_1| = |z_2|$. But there are lots of possibilities for $z_3$,$z_4$ with $|z_3| = |z_4|$ and $z_3 neq z_4$, since modulus is not a one-to-one map.

When you look at $left(frac1+mathrmi1-mathrmiright)^n$, this always has modulus $1$ for all values of $n$. But this just means it lies on the unit circle in the complex plane, it does not mean that it is equal to 1.

share|cite|improve this answer

edited 30 mins ago

answered 41 mins ago

Morgan RodgersMorgan Rodgers

9,95731440

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is a many to one map
  $endgroup$
  – Arjang
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So while solving questions relating to this, how am I supposed to know that my taking mod would yield me a valid result? With reference to the quoted example question, of course.
  $endgroup$
  – user231094
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user231094 In order to know, you can explicitly check the result you obtain after taking the modulus. For instance, in the originally posed problem, we can explicitly check that $n=1$ is not a solution, i.e. $fraci+1i-1 neq 1$. Sometimes taking the modulus might help you narrow down the possible solutions, even if it doesn't tell you the solutions exactly. Oftentimes, it's more useful in determining what the solutions aren't.
  $endgroup$
  – jawheele
  32 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I'm sure you have learned that when solving equations, if you square both sides you can introduce spurious solutions. The point is that
$$A = B implies A^2 = B^2
$$
but the converse can fail. That's why we teach algebra students to check their answers back in the original equation.

Taking the modulus on both sides of the equation has the same effect:
$$A = B implies |A| = |B|
$$
but the converse can fail, for example $|i|=|1|$ but $i ne 1$.

When you took the modulus on both sides of your equation, you introduced spurious solutions. So you have to check each answer to see if it satisfies the original equation.

share|cite|improve this answer

answered 31 mins ago

Lee MosherLee Mosher

53k33892

$endgroup$

add a comment | 

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3

$begingroup$

The problem is, "taking-mod-of-both-sides" is valid, but only gives a one way implication. If $z_1 = z_2$ then $|z_1| = |z_2|$. But there are lots of possibilities for $z_3$,$z_4$ with $|z_3| = |z_4|$ and $z_3 neq z_4$, since modulus is not a one-to-one map.

When you look at $left(frac1+mathrmi1-mathrmiright)^n$, this always has modulus $1$ for all values of $n$. But this just means it lies on the unit circle in the complex plane, it does not mean that it is equal to 1.

share|cite|improve this answer

edited 30 mins ago

answered 41 mins ago

Morgan RodgersMorgan Rodgers

9,95731440

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is a many to one map
  $endgroup$
  – Arjang
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So while solving questions relating to this, how am I supposed to know that my taking mod would yield me a valid result? With reference to the quoted example question, of course.
  $endgroup$
  – user231094
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user231094 In order to know, you can explicitly check the result you obtain after taking the modulus. For instance, in the originally posed problem, we can explicitly check that $n=1$ is not a solution, i.e. $fraci+1i-1 neq 1$. Sometimes taking the modulus might help you narrow down the possible solutions, even if it doesn't tell you the solutions exactly. Oftentimes, it's more useful in determining what the solutions aren't.
  $endgroup$
  – jawheele
  32 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

The problem is, "taking-mod-of-both-sides" is valid, but only gives a one way implication. If $z_1 = z_2$ then $|z_1| = |z_2|$. But there are lots of possibilities for $z_3$,$z_4$ with $|z_3| = |z_4|$ and $z_3 neq z_4$, since modulus is not a one-to-one map.

When you look at $left(frac1+mathrmi1-mathrmiright)^n$, this always has modulus $1$ for all values of $n$. But this just means it lies on the unit circle in the complex plane, it does not mean that it is equal to 1.

share|cite|improve this answer

edited 30 mins ago

answered 41 mins ago

Morgan RodgersMorgan Rodgers

9,95731440

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is a many to one map
  $endgroup$
  – Arjang
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So while solving questions relating to this, how am I supposed to know that my taking mod would yield me a valid result? With reference to the quoted example question, of course.
  $endgroup$
  – user231094
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user231094 In order to know, you can explicitly check the result you obtain after taking the modulus. For instance, in the originally posed problem, we can explicitly check that $n=1$ is not a solution, i.e. $fraci+1i-1 neq 1$. Sometimes taking the modulus might help you narrow down the possible solutions, even if it doesn't tell you the solutions exactly. Oftentimes, it's more useful in determining what the solutions aren't.
  $endgroup$
  – jawheele
  32 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

The problem is, "taking-mod-of-both-sides" is valid, but only gives a one way implication. If $z_1 = z_2$ then $|z_1| = |z_2|$. But there are lots of possibilities for $z_3$,$z_4$ with $|z_3| = |z_4|$ and $z_3 neq z_4$, since modulus is not a one-to-one map.

When you look at $left(frac1+mathrmi1-mathrmiright)^n$, this always has modulus $1$ for all values of $n$. But this just means it lies on the unit circle in the complex plane, it does not mean that it is equal to 1.

share|cite|improve this answer

edited 30 mins ago

answered 41 mins ago

Morgan RodgersMorgan Rodgers

9,95731440

$endgroup$

share|cite|improve this answer

edited 30 mins ago

answered 41 mins ago

Morgan RodgersMorgan Rodgers

9,95731440

answered 41 mins ago

Morgan RodgersMorgan Rodgers

9,95731440

answered 41 mins ago

Morgan RodgersMorgan Rodgers

9,95731440

2

$begingroup$

I'm sure you have learned that when solving equations, if you square both sides you can introduce spurious solutions. The point is that
$$A = B implies A^2 = B^2
$$
but the converse can fail. That's why we teach algebra students to check their answers back in the original equation.

Taking the modulus on both sides of the equation has the same effect:
$$A = B implies |A| = |B|
$$
but the converse can fail, for example $|i|=|1|$ but $i ne 1$.

When you took the modulus on both sides of your equation, you introduced spurious solutions. So you have to check each answer to see if it satisfies the original equation.

share|cite|improve this answer

answered 31 mins ago

Lee MosherLee Mosher

53k33892

$endgroup$

add a comment | 

2

$begingroup$

I'm sure you have learned that when solving equations, if you square both sides you can introduce spurious solutions. The point is that
$$A = B implies A^2 = B^2
$$
but the converse can fail. That's why we teach algebra students to check their answers back in the original equation.

Taking the modulus on both sides of the equation has the same effect:
$$A = B implies |A| = |B|
$$
but the converse can fail, for example $|i|=|1|$ but $i ne 1$.

When you took the modulus on both sides of your equation, you introduced spurious solutions. So you have to check each answer to see if it satisfies the original equation.

share|cite|improve this answer

answered 31 mins ago

Lee MosherLee Mosher

53k33892

$endgroup$

add a comment | 

$begingroup$

I'm sure you have learned that when solving equations, if you square both sides you can introduce spurious solutions. The point is that
$$A = B implies A^2 = B^2
$$
but the converse can fail. That's why we teach algebra students to check their answers back in the original equation.

Taking the modulus on both sides of the equation has the same effect:
$$A = B implies |A| = |B|
$$
but the converse can fail, for example $|i|=|1|$ but $i ne 1$.

When you took the modulus on both sides of your equation, you introduced spurious solutions. So you have to check each answer to see if it satisfies the original equation.

share|cite|improve this answer

answered 31 mins ago

Lee MosherLee Mosher

53k33892

$endgroup$

share|cite|improve this answer

answered 31 mins ago

Lee MosherLee Mosher

53k33892

answered 31 mins ago

Lee MosherLee Mosher

53k33892

answered 31 mins ago

Lee MosherLee Mosher

53k33892