Wikipedia states for $z\in\mathbb{C}$ and $n\in\mathbb{N}$
${\displaystyle (z^{n})^{1/n}\neq z}$
Is this correct?
In general, the exponent law
$(x^a)^{1/b}=x^{a/b}$, $a,b\in\mathbb{N}$
for nonnegative reals $x\in\mathbb{R}^+$ doesn't apply to complex numbers. In particular, if $w=a/b$ is a rational number with $a$ and $b$ coprime integers and $w>b$ then $z^w$ has exactly $n$ values (unless $z=0$), so even the equation $(-1)^{1/2}=(-1)^{1/2}$ doesn't make sense (for complex $z=-1$ and $a = 1$) as the square root of each side of the equation has two solutions.
However, for $w\in \mathbb{N}$ and complex $z=r\exp^{i\varphi + i2k\pi}, k \in\mathbb{Z}$ the solutions to
$(z^a)^{1/b} = (r^{a}\exp^{ia(\varphi + 2k_1\pi)})^{1/b}=r^{a/b}\exp^{i{a/b}(\varphi + 2k_1\pi)}=r^{a/b}\exp^{i\varphi a/b}$
and
$(z)^{a/b} = r^{a/b}\exp^{i{a/b}(\varphi + 2k_2\pi)}=r^{a/b}\exp^{i \varphi a/b}$
are unique (i.e., independent of $k_1$ and $k_2$) and identical in real and imaginary part for all $z$. Doesn't this imply that the exponent laws above holds for complex numbers and $w\in\mathbb{N}$?
Best Answer
What you are missing is that you need to keep the argument inside some fixed interval $(x, x + 2\pi]$. Let's choose $(-\pi,\pi]$, i.e. the principal branch of logarithm: $\ln z = \ln|z| + i\arg z$, where $\arg z\in (-\pi,\pi]$.
The crucial thing to notice is that if $z = re^{i\varphi}$ for $\varphi\in(-\pi,\pi]$, then, obviously, $\arg z = \varphi$. However, if $\varphi \neq 0$, you can always choose $n\in\mathbb N$ such that $n\varphi \not\in (-\pi,\pi]$, so even though $z^n = r^n e^{in\varphi}$, $\arg(z^n) \neq n\varphi$!
Now, $(z^n)^{1/n} = \exp(\frac 1n\ln(z^n)) = \exp(\frac 1n(\ln|z^n|+i\arg(z^n))) = |z|\exp(i \frac 1n\arg(z^n)).$ This is equal to $z$ if and only if $\frac 1n\arg(z^n) = \varphi + 2k\pi$, for some integer $k$. We can write this as $\arg(z^n) = n\varphi + 2nk\pi$. Since we chose argument to be inside $(-\pi,\pi]$, we need to have
$$-\pi < n\varphi + 2nk\pi \leq \pi \iff -\pi/n < \varphi + 2k\pi \leq \pi/n\tag{1}.$$ Since $\varphi \in (-\pi,\pi]$, we have $\varphi + 2k\pi\in (-\pi+2k\pi,\pi+2k\pi],$ so $(1)$ can hold only if $k = 0$. We are left with $-\pi/n <\varphi \leq \pi/n$, but this is contradiction with our choice of $n$ to be such that $n\varphi\not\in (-\pi,\pi]$.
What we can conclude is that $(z^n)^{1/n} = z$ if and only if $\arg z \in (-\pi/n,\pi/n]$. This is always satisfied if $\arg z = 0$, i.e. $z$ is a positive real, but obviously not in general.
For example, let $z = i$ and $n = 3$. Then $z^3 = i^3 = -i$ and $$(z^3)^{1/3} = \exp\left(\frac 13\ln(-i)\right) = \exp\left(\frac 13(\ln |-i| + i \arg(-i))\right) = e^{-i\frac{\pi}6}\neq i.$$
Nothing changes essentially by choosing a different branch of logarithm, except that the whole thing becomes more tedious to write down. And, if you don't want to choose a branch and keep logarithm multivalued, then $z$ is always one of the values of $(z^n)^{1/n}$.
To reflect on what went wrong with your calculation, it's circular logic. You are using the exponent law $(e^z)^w = e^{zw}$ to justify the exponent law in a particular case. Just to be clear, $(e^z)^w \neq e^{zw}$ in general. Problematic part boils down to that even though $\exp(\ln z) = z$, it is not true that $\ln (\exp z) = z$ in general since $\exp$ is surjective, but not injective. Just like $\sqrt x^2 = x$, but $\sqrt{x^2} \neq x$ in general.
Concretely, $$(e^{it})^{1/b} = \exp\left(\frac 1b \ln(e^{it})\right) = \exp\left(\frac 1bi(t+2k\pi)\right) = \exp\left(\frac{it}b + \frac{2k\pi}b\right)$$
which is not the same as $e^{it/b}$, unless $k/b$ is integer. Note that $k$ is chosen such that $t + 2k\pi \in (-\pi,\pi]$.