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When I first saw the Euler's identity $$e^{\theta i}=\cos(\theta)+i\sin(\theta)$$ I realized $$x=e^{\frac{πx}{2}}$$ must have a complex solution $x=i$. However, although I know the solution, I'm unable to find the way that leads to it. By some simple changes I've achieved $$x=-\frac{2}{π}W(-\frac{π}{2})$$. So I guess the equation has more solutions since the one I wrote is definitely not $i$. I searched a lot on the Internet but couldn't find anything that would show the approach to get exactly $x=i$. There might be something obvious I'm missing but I would be really happy if anyone could give me a hint how to mathematically get $x=i$ from that equation $x=e^{\frac{πx}{2}}$.

Blob Organic
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  • Numerical evidence suggests that the equation z = exp(πz/2) has infinitely many solutions for z ∈ ℂ. – Dan Asimov Jul 23 '23 at 23:20
  • So is there a mathematical way to achieve $x = i$ from the equation? – Blob Organic Jul 23 '23 at 23:22
  • What do you mean by achieve exactly? It clearly is a solution to the equation. Are you asking if it's unique? – Ennar Jul 23 '23 at 23:24
  • By achieve I mean get $x=i$ from the equation by multiplying, dividing and so on. I know it's clear it is a solution but it bothers me that I can't find it just by changing the equation – Blob Organic Jul 23 '23 at 23:27
  • Well, how do you get $i$ from the equation $x^2 = -1$ by dividing and multiplying? – Ennar Jul 23 '23 at 23:29
  • I edited my comment, I mean all operations including root. I just wrote those as an example – Blob Organic Jul 23 '23 at 23:30
  • Let me give you another example of what I mean by "mathematically achieve". Say we have equation $$sin(x)-x^2+lnx=sin(7)-49+ln(7)$$ It is obvious that $x=7$ is a solution but we see that using logic. I think this equation would be impossible to solve if there was $ln(6)$ at the end instead of $ln(7)$. So by " mathematically achieve" I mean to solve it without using logic – Blob Organic Jul 23 '23 at 23:35
  • The equation is transcendental equation, you can't simply solve it by radicals. So, what you did with Lambert W is the best that you can do (probably). Note that $W$ is multivalued, and inspecting some of its branches we get $W_0(-\pi/2) = i\pi/2$ and $W_{-1}(-\pi/2) = -i\pi/2$ which gives you solutions $\pm i$. – Ennar Jul 23 '23 at 23:40
  • Oh, I wrote my result into wolfram alpha and it automatically changed the $W$ function to $W_1$ and it returned some weird number so that confused me. It makes sense now, thank you – Blob Organic Jul 23 '23 at 23:46

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