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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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    @ThomasAndrews this is a brilliant answer. +1 for avoiding complex numbers.2011-12-27

5 Answers 5

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