$a^m$ homogeneous implies $a$ homogeneous?

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Let $A = \bigoplus_{r \in \mathbb{Z}/n\mathbb{Z}} A_r$ be an integral $\mathbb{Z}/n\mathbb{Z}$-graded ring of characteristic $0$, and $a \in A$.

Assume there exists an integer $m$ such that $a^m$ is homogeneous, does this imply that $a$ is homogeneous?

Statement is wrong for characteristic p: Consider $A := \mathbb{F}_p[X]$ with the $\mathbb{Z}/p\mathbb{Z}$-grading induced by letting $X$ be of weight $1$. Then $(1+X)$ is not homogeneous, but $(1+X)^p = 1 + X^p$ is.

Comments: The rings I care about are also regular, Noetherian and integrally closed, just in case this helps.

Thoughts so far: Nothing useful to be honest. I tried to write $a$ as the sum of its homogeneous parts, but expanding $a^m = (a_0 + \ldots + a_{n-1})^m$ didn't help me.

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Consider $\mathbb{C}$ as a $\mathbb{Z}/2\mathbb{Z}$-graded ring with $\mathbb{R}$ in degree $0$ and $i\mathbb{R}$ in degree $1$. Then $1+i$ is not homogeneous, but $(1+i)^2=2i$ is.