Do All nth Roots Of Complex Numbers Only Yield Complex Solutions, That Is, No Purely Real Solutions?

Question

For Example, if i had the nth root of (a+ib), are its solutions always in the form of (x + iy) such that y can't be zero [in other words, the nth root of a complex number always yields another complex number, that is, never any purely real numbers].

Answer 1

There are complex numbers, where the n.th root of this number is allways purely real like 0. Don't forget that purely real numbers are complex numbers too! $$ \begin{align*} \mathbb{R} &\in \mathbb{C}\\ \mathbb{R} &\in \left\{a + b \cdot \mathrm{i} ~ \mid ~ a, ~b \in \mathbb{R} \land i^{2} = -1 \right\}\\ \end{align*} $$

Aka if we found a purely real number a with $$ \sqrt[n]{a} \in \mathbb{R} $$ for every n, we would find a complex number z with $$ \sqrt[n]{z_{z \in \mathbb{C}}} \in \mathbb{R} $$ for every n too.

A number for which this applies is zero. The n.th root of Zero (a complex Number) is allways Zero (if n is real and bigger then 0) (a purely real number). $$ \begin{align*} \sqrt[n]{z_{z \in \mathbb{C}}} \in \mathbb{R}\\ \sqrt[n]{0} = 0 \in \mathbb{R}\\ \sqrt[1010,2233]{0} = 0 \in \mathbb{R}\\ \end{align*} $$

Answer 2

I believe what you are trying to ask is:

If $z \in \mathbb{C} \setminus \mathbb{R}$, can $z^{\frac{1}{n}} \in \mathbb{R}$ for $n \in \{1,2,3,...\}$?

Suppose that this is true, then let $\omega = z^{\frac{1}{n}}$ and $\omega \in \mathbb{R}$. We know any complex number can be written as $r(\cos(\theta)+i\sin(\theta))$ for some constant $r \in \mathbb{R}$; thus, $z = \omega^{n}=r^{n}(\cos(\theta)+i\sin(\theta))^{n}$. Applying De Moivre's formula will give us $z = r^{n}(\cos(n\theta)+i\sin(n\theta))$. Therefore, it is evident that if $\omega \in \mathbb{R}$, then $z \in \mathbb{R}$ (because $\sin(n\theta)$ must be $0$). This contradicts our initial assumption. So the answer to the revised question is no, it cannot.