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 $(\frac{1+i}{1-i})^n = 1$

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

$\Big(\frac{|1+i|}{|1-i|}\Big)^n = |1|$

$\Big(\frac{\sqrt{2}}{\sqrt{2}}\Big)^n = 1$

$1^n = 1$

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?

  • 16
    $\begingroup$ 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
    May 12, 2019 at 0:37
  • 3
    $\begingroup$ Whenever you have $a = b$ (whatever things $a$, $b$ might be!), and a function applicable on $a$ and $b$, you always have: $a = b \Rightarrow f(a) = f(b)$. Typically, this comes from basic rules in your logic, e.g. first-order logic with natural deduction. $\endgroup$
    – ComFreek
    May 12, 2019 at 7:11
  • 1
    $\begingroup$ One way to approach this is to grok that the modulo operation "destroys" information. Say, if you take modulo10 and get 2 -- is it 2, 12, 22, or perhaps 222? Similarily x2 destroys information (the sign) hence the +/- value -- how could you compensate for it with modulo operation? (You could define algebras where the modulo operation does not destroy information, but that is something completely different) $\endgroup$
    – ghellquist
    May 12, 2019 at 20:27
  • $\begingroup$ Question: What is the least non-negative value of $x$ such that $x = 2$? That's right, it's $2$. But if we multiply both sides by $0$, as in $0x = 0\cdot 2$, then $0$ also works, why? $\endgroup$
    – user253751
    May 12, 2019 at 22:42

2 Answers 2


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 ${\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)}^{n}$, 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.

  • $\begingroup$ It is a many to one map $\endgroup$
    – jimjim
    May 12, 2019 at 0:41
  • $\begingroup$ 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
    May 12, 2019 at 0:41
  • 2
    $\begingroup$ @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. $\frac{i+1}{i-1} \neq 1$. 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
    May 12, 2019 at 0:47

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 solution to see if it satisfies the original equation.


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