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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?

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    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 place2011-12-27
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    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.2011-12-27
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    @KS I am interested in an algebraic solution. thats all.2011-12-27
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    In particular, you can't say $\log (z^n) = n \log z$ when $\log z$ is not well-defined.2011-12-27
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    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.2011-12-27
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    @ThomasAndrews this is a brilliant answer. +1 for avoiding complex numbers.2011-12-27

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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.
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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: $z^3=1 \iff (z^3)^{1/3}=1^{1/3} \iff z =1$.

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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.

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    This is the best answer. +1 for roots of unity. Thanks Dustan.2011-12-27
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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 $n = 2 m$, $z = e^{\pi i} = -1$.

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