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
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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
$begingroup$
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
$Big(fracBig)^n = |1|$
$Big(fracsqrt2sqrt2Big)^n = 1$
$1^n = 1$
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
edited 2 hours ago
Ryan Shesler
42911
asked 2 hours 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.
$endgroup$
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
$Big(fracBig)^n = |1|$
$Big(fracsqrt2sqrt2Big)^n = 1$
$1^n = 1$
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
edited 2 hours ago
Ryan Shesler
42911
asked 2 hours 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.
$endgroup$
4
$begingroup$
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
2 hours 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
$Big(fracBig)^n = |1|$
$Big(fracsqrt2sqrt2Big)^n = 1$
$1^n = 1$
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
edited 2 hours ago
Ryan Shesler
42911
asked 2 hours 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.
$endgroup$
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
$Big(fracBig)^n = |1|$
$Big(fracsqrt2sqrt2Big)^n = 1$
$1^n = 1$
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
complex-numbers
edited 2 hours ago
Ryan Shesler
42911
asked 2 hours 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 2 hours ago
Ryan Shesler
42911
asked 2 hours 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 2 hours ago
Ryan Shesler
42911
edited 2 hours ago
Ryan Shesler
42911
edited 2 hours ago
Ryan Shesler
42911
42911
asked 2 hours 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 2 hours ago
user231094user231094
111
asked 2 hours ago
user231094user231094
111
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.
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.
4
$begingroup$
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
2 hours ago
add a comment |
4
$begingroup$
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
2 hours ago
4
4
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 2 hours ago
Morgan RodgersMorgan Rodgers
9,97731440
$endgroup$
$begingroup$
It is a many to one map
$endgroup$
– Arjang
2 hours 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
2 hours 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
2 hours ago
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.
answered 2 hours ago
Lee MosherLee Mosher
53k33892
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 2 hours ago
Morgan RodgersMorgan Rodgers
9,97731440
$endgroup$
$begingroup$
It is a many to one map
$endgroup$
– Arjang
2 hours 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
2 hours 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
2 hours 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.
answered 2 hours ago
Morgan RodgersMorgan Rodgers
9,97731440
$endgroup$
$begingroup$
It is a many to one map
$endgroup$
– Arjang
2 hours 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
2 hours 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
2 hours 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.
answered 2 hours ago
Morgan RodgersMorgan Rodgers
9,97731440
$endgroup$
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.
answered 2 hours ago
Morgan RodgersMorgan Rodgers
9,97731440
edited 2 hours ago
answered 2 hours ago
Morgan RodgersMorgan Rodgers
9,97731440
answered 2 hours ago
Morgan RodgersMorgan Rodgers
9,97731440
answered 2 hours ago
Morgan RodgersMorgan Rodgers
9,97731440
9,97731440
$begingroup$
It is a many to one map
$endgroup$
– Arjang
2 hours 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
2 hours 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
2 hours ago
add a comment |
$begingroup$
It is a many to one map
$endgroup$
– Arjang
2 hours 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
2 hours 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
2 hours ago
$begingroup$
It is a many to one map
$endgroup$
– Arjang
2 hours ago
$begingroup$
It is a many to one map
$endgroup$
– Arjang
2 hours 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
2 hours 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
2 hours ago
1
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
2 hours ago
$begingroup$
@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
2 hours ago
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.
answered 2 hours 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.
answered 2 hours 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.
answered 2 hours ago
Lee MosherLee Mosher
53k33892
$endgroup$
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.
answered 2 hours ago
Lee MosherLee Mosher
53k33892
answered 2 hours ago
Lee MosherLee Mosher
53k33892
answered 2 hours ago
Lee MosherLee Mosher
53k33892
answered 2 hours ago
Lee MosherLee Mosher
53k33892
53k33892
add a comment |
add a comment |
user231094 is a new contributor. Be nice, and check out our Code of Conduct.
user231094 is a new contributor. Be nice, and check out our Code of Conduct.
user231094 is a new contributor. Be nice, and check out our Code of Conduct.
user231094 is a new contributor. Be nice, and check out our Code of Conduct.
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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.
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– jawheele
2 hours ago