Remember that if $a\in\mathbb C\setminus\{0\}$, then $\phi\in\mathbb R$ is called argument of $a$ if $$a=|a|e^{{\rm i}\phi}\tag1.$$ Let $$\operatorname{Arg}a:=\left\{\phi\in\mathbb R:\phi\text{ is an argument of }a\right\}$$ and $\arg a$ denote the unique intersection point of $\operatorname{Arg}a$ and $(-\pi,\pi]$.
Question 1: Are we able to show that $\operatorname{Arg}a^n=n\operatorname{Arg}a$?
It should hold, since we should have $$\phi\in\operatorname{Arg}a\Leftrightarrow a=\left|a\right|e^{{\rm i}\phi}\Leftrightarrow a^n=\left|a^n\right|e^{{\rm i}n\phi}\Leftrightarrow n\phi\in\operatorname{Arg}a^n\tag2.$$
Now, if $n\in\mathbb N$, remember that $z\in\mathbb C$ is called $n$th root of $a$ if $$z^n=a\tag3.$$ Let $$\sqrt[n]a:=\left\{z\in\mathbb C:z\text{ is a }n\text{th root of }a\right\}.$$
Question 2: Let $E$ be a normed $\mathbb R$-vector space and $g_i:E\to\mathbb C\setminus\{0\}$ with
- $g_i$ is continuous;
- $g_i(0)=1$.
If $g_1^n=g_2^n$, are we able to infer that $g_1=g_2$?
I know that it holds that whenever $f_i:E\to\mathbb C$ is continuous and $e^{f_1}=e^{f_2}$, then $$\exists k\in\mathbb Z:f_1-f_2=2k\pi\tag4.$$ So, if $f_1(x_0)=f_2(x_0)$ for some $x_0\in E$, then $f_1=f_2$. Maybe we can reduce the desired claim to this situation.
On the other hand, it should hold $$\sqrt[n]a=\left|a\right|^{\frac1n}e^{{\rm i}\frac{\operatorname{Arg}a}n}=\left|a\right|^{\frac1n}\left\{e^{{\rm i}\frac{\arg a+2k\pi}n}:k\in\{0,\ldots,n-1\}\right\}\tag5.$$ So, unless I'm missing something and if the answer to Question 1 is positive, it should hold $$\sqrt[n]{a^n}=\left|a\right|e^{{\rm i}\frac{\operatorname{Arg}a^n}n}=\left|a\right|e^{{\rm i}\frac{n\operatorname{Arg}a}n}=\left|a\right|e^{{\rm i}\operatorname{Arg}a}=a;\tag6$$ from which the desired claim should follow even without assuming that $g_i$ satisfies (1.) and (2.). Am I missing something?
Martin R has already commented on the first question.
Answer fo quesion 2): Any normed linear space is (path) connected. So the range of $\frac {g_1} {g_2} $ is a connected set contained in the $n-$th roots of unity. Hence, $g_1=cg_2 $ wher $c$ is an $n-$th root of unity. Since $g_1(0)=g_2(0)$ we get $c=1$.