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Is $e^{qz}=(e^z)^q$ true with $z$ complex and $q\in\mathbb{Q}$?

I've already proved it for $n\in\mathbb{Z}$, but what about for $q\in\mathbb{Q}$? Is it true?

My proof for $n\in\mathbb{Z}$: \begin{equation*} \begin{aligned} (e^z)^n=(e^xe^{iy})^n&=(e^x(\cos(y)+i\sin(y)))^n\\ &=e^{nx}(\cos(y)+i\sin(y))^n\\ &\overset{{\star}}{=}e^{nx}(\cos(ny)+i\sin(ny))\\ &=e^{nx}e^{iny}=e^{n(x+iy)}\\ &=e^{nz}. \end{aligned} \end{equation*} $\star$ By Moivre's Theorem for $|z|=1$.

Manatee
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    LHS has a unique value but RHS has infinitely many values. – Kavi Rama Murthy Jul 16 '21 at 23:18
  • @KaviRamaMurthy I think I understand. But if I have $$\sqrt[n]{|z^m|}\exp{\left(i\frac{1}{n} {\arg(z^m)}\right)},$$ with $\gcd(m,n)=1$, so can I conclude that $$\sqrt[n]{|z^m|}\exp{\left(i\frac{1}{n} {\arg(z^m)}\right)}=\sqrt[n]{|z^m|}\exp{\left(i{\arg(z^m)}\right)}^{\frac{1}{n}}?$$ – Manatee Jul 16 '21 at 23:22
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  • The standard definitions of $e^z$ as single-valued (some authors write $\exp(z)$ to emphasise this choice) are unfortunately inconsistent with the general multi-valued definition of $w^z$ (where $w$ & $z$ are complex and $w$ nonzero): e.g., by the latter, $e^{\frac12}$ equals $\pm\sqrt{e}.$ So, if you rewrite your proof for $n\in\mathbb Z$ by replacing all the instances of $e$ with $\exp,$ then your result is indeed correct.
    • – ryang Jul 19 '21 at 03:25
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  • However, when you replace the integer $n$ in your claim with rational $q,$ by our previous discussion, since $e^z$ here is just a general non-zero complex number, the new result becomes not generally true.
    • – ryang Jul 19 '21 at 03:31