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?
Best Answer
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.