Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to solve the equation $$z^n = 1.$$

Taking $\log$ on both sides I get $n\log(z) = \log(1) = 0$.

$\implies$ $n = 0$ or $\log(z) = 0$

$\implies$ $n = 0$ or $z = 1$.

But I clearly missed out $(-1)^{\text{even numbers}}$ which is equal to $1$.

How do I solve this equation algebraically?

share|cite|improve this question
The moment you take logarithm on both sides, you implicitly assume that it is defined. So, if you want to get such solutions, you need to refrain from taking logarithm because it is not defined in the first place – user21436 Dec 27 '11 at 20:10
You can't apply logarythms like this if $z$ might be a negative number, at least, not until you've gotten to the complex numbers, where $log$ becomes a much trickier animal. – Thomas Andrews Dec 27 '11 at 20:12
@KS I am interested in an algebraic solution. thats all. – Ajay George Dec 27 '11 at 20:12
In particular, you can't say $\log (z^n) = n \log z$ when $\log z$ is not well-defined. – Thomas Andrews Dec 27 '11 at 20:12
it's not clear what you mean by "algebraic," but one approach may be to factor the polynomial $z^n-1$. In particular, for $n=2m$ even, $z^{2m}-1 = (z^2-1)(1+z^2+...+z^{2m-2})$ and clearly $1+z^2+..+z^{2m-2}$ is always non-zero, so the only (real) roots are the roots of $z^2-1$. On the other hand, for $n$ odd, $z^n-1$ can be shown to be strictly increasing, so it can have at most one real root. – Thomas Andrews Dec 27 '11 at 20:19
up vote 3 down vote accepted

You can't take the logarithm of a negative number, unless you consider the multivalued complex logarithm.

If you are willing to expand to complex numbers in that manner, then you can take the log of both sides. $\log(1) = 2\pi i k$, $k \in \mathbb{N}$, so then you're solving for $n \log(z) = 2 \pi i k $, which gives $\log z = 2\pi i\frac{k}{n}$, or $z = e^{2 \pi i\frac{k}{n}}$, which describes all of the roots of unity.

share|cite|improve this answer
This is the best answer. +1 for roots of unity. Thanks Dustan. – Ajay George Dec 27 '11 at 20:37

Working in the reals, take the $n$th root of both sides.

If $n$ is even then you'd write $(z^n)^{1/n}=|z|$.

For example $z^2=1 \iff (z^2)^{1/2}=1^{1/2} \iff |z|=1$.

For odd powers, you could say, for example: $z^3=1 \iff (z^3)^{1/3}=1^{1/3} \iff z =1$.

share|cite|improve this answer

If you want to do this properly you need complex numbers. The logarithm is a multivalued function; $z^n = 1$ is equivalent not to $n \log z = \log 1$ but to $n \log z = 2 \pi i m$ where $m$ is an arbitrary integer. If $n \ne 0$ this says $\log z = (2 \pi i m)/n$ and $z = e^{2 \pi i m/n}$. In particular with $n = 2 m$, $z = e^{\pi i} = -1$.

share|cite|improve this answer

First of all, note that, apart from $z=1$, all other answers will be complex. For the equation

$z^n=1$, we use the theorems $e^{ix}=cosx + i sin x$ and that $(cosx + i sin x)^{n}=cos(nx)+ sin(nx)$.

Then, $z^n=1$ implies $r^n(cos(nx)+ sin(nx))=1$, where $z=re^{ix}$. Hence, $r=1$ and

$cos(nx)+ isin(nx)=1$. Thus, $cos(nx)=1$ and $sin(nx)=0$, equating real and imaginary parts separately.

That gives the answer $z=cos(\frac{2\pi}{k})+isin(\frac{2\pi}{k}), k=1,2 \cdots (n-1)$

share|cite|improve this answer

To solve it algebraically, I'd say:

  • For even values of $n$, $z=1$ or $z=-1$ are the solutions.
  • For odd values of $n$, $z=1$ is the answer.
  • And if $n=0$, $\forall z \in \mathbb{R}$ would be valid as an answer.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.