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I was messing around with the graphs of $\cos(\ln(x))$ and $\sin(\ln(x))$ in Desmos because I had noticed that for any $a^i$, where $i=\sqrt{-1}$, we can rewrite it as$$a^i=e^{i\ln(a)}=\cos(\ln(a))+i\sin(\ln(a))$$ and noticed some stuff. For example:

  1. It appears that $\displaystyle\lim_{x\to0^+}x^i=0$ (you can see this by rewriting the two equations in Desmos as $\cos(h\ln(x))$ and $\sin(h\ln(x))$ for very small $h$.
  2. $\displaystyle\lim_{x\to1}x^i=1$
  3. There appear to be other values for $x$ where $x^i=1+0i$, such as $x\approx4.81\text{ and }x\approx519$, however I decided to not look for any other values of $x$ where $x^i=1+0i$ since $\cos(\ln(x))$ and $\sin(\ln(x))$ starts to oscillate very slowly after around that.

However, my question is based off of (3). Is there a way I could find all of the values of $x$ on $x\in(0,\infty)$ where $x^i=1+0i$ without having to approximate solutions?

I have tried using the definition of $\cos(x)$, that being$$\dfrac12e^{-ix}+\dfrac12e^{ix}$$and then doing the transformation $x\to\ln(x)$ after setting it equal to $1$ and using the quadratic formula approach to get a general solution, but I just ended up getting$$x^i=1\implies x=\sqrt[\large e^{i\pi/2+2i\pi k}]1,k\in\mathbb Z$$which of course has to evaluate to just $1$.

I also tried using the definition of $\sin(x)$, that being$$\dfrac i2e^{-ix}-\dfrac i2e^{ix}$$but this time, after I do the transformation $x\to\ln(x)$, I set it equal to $0$ and attempted again to use the quadratic formula approach, but I ended up getting$$x^{2i}-1=0$$which can be factored as$$(x^i-1)(x^i+1)=0\implies x^i+1=0\text{ or }x^i-1=0$$which of course gets$$x=1\text{ or }e^{\pi+2\pi k}$$the only problem being$$\cos\left(\ln\left(e^{\pi+2\pi k}\right)\right)+i\sin\left(\ln\left(e^{\pi+2\pi k}\right)\right)=\color\red-1+0i$$instead of $1+0i$ like I want.

CrSb0001
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  • Exponents can be multivalued, but assuming you want the principal branch, our solutions are $x^i=1\iff i\ln(x)=2\pi i n, x=e^{2 π n}$ – Тyma Gaidash Jan 01 '24 at 03:14
  • @ТymaGaidash I wonder where I went wrong in trying to derive the solution then – CrSb0001 Jan 01 '24 at 03:15
  • How did you solve $x^i\pm1=0$ specifically? Based on the part where it says “which of course gets”, one assumes you solved $x^i+1=0\implies x=e^{(2k+1)\pi}$, the extraneous solutions. – Тyma Gaidash Jan 01 '24 at 03:38
  • @ТymaGaidash That's because I did$$x^i=e^{i\pi+2i\pi k}\implies x=e^{(i\pi+2i\pi k)/i}\gets x=e^{\pi+2\pi k}$$ – CrSb0001 Jan 01 '24 at 03:46
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    $e^{\pi i+2\pi i k}=-1$ meaning you solved $x^i=-1$ instead of $x^i=1=e^{2\pi i k}\implies x=e^{2\pi k}$ – Тyma Gaidash Jan 01 '24 at 03:50

1 Answers1

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You render

$x^i=1\implies x=1,$

but this is not true. You can easily compute

$[\exp(2\pi)]^i=\exp(2\pi i)=1$

but $\exp(2\pi)\not=1$.

Properly, start with

$x^i=1=\exp(2k\pi i)=[\exp(2k\pi)]^i;k\in\mathbb{Z}$

using the fact that $\exp(ab)=[\exp(a)]^b$. Continue from there.

Oscar Lanzi
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  • Your last formula doesn’t work for complex numbers, though? – Divide1918 Jan 01 '24 at 04:47
  • We do not generally have $e^{ab}=(e^a)^b$ in the complex domain. But the function $\exp(z)$ isactually the branch of $e^z$ where it does apply, by analytically continuing the real-variable Maclaurin series. – Oscar Lanzi Jan 01 '24 at 05:11
  • @OscarLanzi Not sure about what you mean by "it does apply". See https://math.stackexchange.com/questions/3911762/again-on-clausen-paradox – Jean-Claude Arbaut Jan 01 '24 at 05:57