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I was thinking that it might has to be $m$ and $n$ coprimes, but I don't have a consolidated idea of how I can prove it. Incidentally, how could I prove that it doesn't work for any integers? (is there any counterexample? I was thinking about $z^{\frac{1}{2}}=\pm z$).

So, my first question is, Can we define $z^{\frac{m}{n}}$, where $z\in\mathbb{C}$ and $m,n\in\mathbb{Z}$?

After that, if the answer is "no", can we say something using that fact that i said previously?

PS: I need to prove that statement without using exponential definition of complex numbers. So, what I need to use is:

Find a $z$ that satisfies $z^n=z_0$ with: $$z=\sqrt[n]{|z_0|}\left(\cos\left(\frac{\theta_0+2k\pi}{n}\right)+i\sin\left(\frac{\theta_0+2k\pi}{n}\right)\right),\text{ for all }k\in\mathbb{Z}.$$

Manatee
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2 Answers2

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You first need to write $\frac mn$ such that it is reduced and $n$ is a positive integer. Then write $z^{m/n}=(z^m)^{1/n}$ and apply the given formula with $z_0=z^m$. You get $n$ solutions as long as $z\ne0$ or $m\ge0$.

Parcly Taxel
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  1. Regarding the necessity of the coprime condition, consider that $$\{-1, e^{\pm\frac\pi3}\}=(-1)^{\frac13}=\left(i^6\right)^{\frac13}\neq i^{\frac63}=i^2=-1.$$

  2. Let $z\in\mathbb C\setminus\{0\}, \,m\in\mathbb Z, \,n\in\mathbb Z^+$ such that $\gcd(m,n)=1.$ Then $$z^{\frac mn}:=\exp\left(\frac mn\log(z)\right)\\ =\exp\left(\frac mn\left[\ln|z|+i\arg(z)\right]\right)\\ =\exp\left({\frac mn\ln|z|}\right)\exp\left(i\frac{m\arg(z)}n\right)\\ =|z|^{\frac mn}\exp\left(i\frac{m\mathrm{Arg}(z)+2k\pi}n\right)\\ =\sqrt[n]{|z|^m}\exp\left(i\frac{\mathrm{Arg}(z^m)+2k\pi}n\right)\\ =\sqrt[n]{|z^m|}\exp\left(i\frac{\mathrm{arg}(z^m)}n\right)\\ =\left(z^m\right)^{\frac1n}.$$ In other words, under the given conditions, it is valid to handle $z^{\frac mn}$ by rewriting it as $\left(z^m\right)^{\frac1n}$, then applying De Moivre’s theorem within the parenthesis, then finally taking the $n$th root as described in your final paragraph.

    P.S. The coprime condition ensures that $$\frac{m\arg(z)}n\mod 2\pi$$ does not return extraneous values.

    P.P.S. For $z\in\mathbb C\setminus\{0\}, \,m\in\mathbb Z, \,n\in\mathbb Z^+,$ $$\quad z^{\frac mn}=\left(z^{\frac1n}\right)^m.$$ In this case, the coprime condition is unnecessary.

ryang
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  • In the first item, you said that -1 $\neq$ -1. I didn't get your example. :( – Manatee May 07 '21 at 12:08
  • Furthermore, how can I take the $n$th root of that: $z^{\frac{m}{n}}=(z^m)^{\frac{1}{n}}=\left(|z|^m(\cos(m\theta)+i\sin(m\theta))\right)^{\frac{1}{n}}=???$ – Manatee May 07 '21 at 12:27
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    @Manatee I've just expanded my answer. The answers to both of your questions in the comments can be found within my expanded working. BTW, I use the symbol := to mean that I'm specifying a definition. – ryang May 07 '21 at 13:13
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    Is there some way to do that without using exp, log, ln. Only using my request above? I really like your way to solve this. It is perfect, but I can't use this properties. :(

    Besides, where do we use the fact that gcd(m,n)=1?

     – Manatee May 07 '21 at 13:42
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    @Manatee 1. $\gcd(m,n)=1$ is just a symbolic way to mean that m and n are coprime.$\quad$ 2. Unfortunately, without relying on the definition of $z^{\frac mn}$ for the given $z,m,n,$ I can't justify/prove that it equals $\left(z^m\right)^{\frac1n}.$ Even the proof that $b^{\frac mn}$ for positive $b$ (no coprime condition necessary here) employs the definition $b^x:=e^{x\ln(b)}.$ – ryang May 07 '21 at 14:27
  • All right! I got it. I know that "gcd(m,n)=1 is just a symbolic way to mean that m and n are coprime", but I don't understand yet where we use this information. – Manatee May 07 '21 at 14:30
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    @Manatee 1. I've expanded my answer to clarify the use of the gcd condition in the proof.$\quad$ 2. A correction to my comment above: "Even the proof that $b^{\frac mn}=(b^m)^{\frac1n}$ for positive $b$ (no coprime..." – ryang May 07 '21 at 21:14
  • why is the last equality true? – Manatee Jul 17 '21 at 17:57
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    @Manatee My answer's point 2 has eight lines in total, incl. the first line of conditions; scratch out the coprime condition, then replace lines 5-8 with a simpler derivation leading to $“,\left(z^{\frac1n}\right)^m,”$ instead of $“,\left(z^m\right)^{\frac1n},”.$ – ryang Jul 18 '21 at 14:21