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If you have $xy = i $ and $x^2 + y^2 = 1$ then you get the solutions that have the golden ratio in them.

These are the solutions

Wolfram calculation: https://www.wolframalpha.com/input/?i=xy%3Di%2C+x%5E2%2By%5E2%3D1

What is exactly happening here? Why is that $-1$ in $ -1+\sqrt{5} $ instead of $+1$. Is there some spiral going on? Hopefully someone can show me what is happening here, I'm quite curious. I also don't really know how to visualise where $xy=i$ and $x^2 + y^2 =1 $ intersect.

It probably has to do with self-similarity, since that is basically the gist of the golden ratio.

The polynomial you get from those equations is $x^4 -x^2 -1 = y$, which is basically really similar to the golden ratio polynomial.

Also if you swap the constants: $x^2 + y^2 = i$ and $xy=1$, you seem to rotate the solutions wolfram calculation

Since $e$ is the hyperbolic constant and $\pi $ the circle constant, can you relate them all nicely in one equation?

Summing my questions and making them concrete:

  1. Why exactly is the golden ratio showing up here geometrically?
  2. How can you visualise what a 'complex' hyperbola and a 'real' circle is. They seem their 'conjugates' almost.
  3. Is there a way to better understand what is going on here? It is probably quite trivial but it seems very interesting.

$xy$ is basically a square that you squish and stretch while keeping the area constant. Maybe you are somehow constructing a golden square, but I don't clearly see it.

Thanks in advance!

EDIT: I still don't quite understand it, but I found this that maybe is related: Ideal triangle

bananenheld
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    Probably because with the change of variable $z=iy$ you get a system $xz=1,x^2−z^2=1$, i.e. $x^2−\frac{1}{x^2}=1$, which with another change of variable $w=x^2$ reduces to $w−\frac{1}{w}=1$ i.e. $w^2−w−1=0$ - having the golden ratio as a solution. –  Dec 31 '21 at 11:04
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    The Wolfram Alpha solutions could have been simplified to $x=\sqrt{\phi},y=i\sqrt{1/\phi}$ or $x=-\sqrt{\phi},y=-i\sqrt{1/\phi}$ or $x=i\sqrt{1/\phi},y=\sqrt{\phi}$ or $x=-i\sqrt{1/\phi},y=-\sqrt{\phi}$ where $\sqrt\phi=\sqrt{\frac12(\sqrt{5}+1)}$ and $\sqrt{1/\phi}=\sqrt{\frac12(\sqrt{5}-1)}$ – Henry Dec 31 '21 at 11:06
  • Why exactly is it the inverse of the golden ratio? I noticed it, seems kind of an unusual inverse. $ 1/ \phi = \frac{2}{1+\sqrt{5}} $ – bananenheld Dec 31 '21 at 12:58
  • Can be rationalised: $\frac{1}{\phi}=\frac{2}{1+\sqrt{5}}=\frac{2(\sqrt{5}-1)}{(1+\sqrt{5})(\sqrt{5}-1)}=\frac{2(\sqrt{5}-1)}{5-1}=\frac{\sqrt{5}-1}{2}$. –  Dec 31 '21 at 13:32
  • If you just want to check that something is a multiplicative inverse, just multiply: $\frac12(\sqrt{5}+1) \times \frac12(\sqrt{5}-1) = \frac14(5 - 1) = 1,$ proving that $\frac12(\sqrt{5}-1)$ is the multiplicative inverse of $\frac12(\sqrt{5}+1).$ – David K Jan 01 '22 at 19:20
  • Or use the defining property of the golden ratio: $\phi^2 - \phi - 1 = 0,$ so $(\phi - 1) - 1/\phi = 0,$ so $1/\phi = \phi - 1 = \frac12(\sqrt{5}+1) - 1 = \frac12(\sqrt{5}-1).$ – David K Jan 01 '22 at 19:22
  • Ah thanks, well that clears one question up. It indeed makes the answer even neater. Do you know the reason why it shows up there? Maybe it has to do with a logarithmic spiral? – bananenheld Jan 01 '22 at 21:14
  • I think a related math exchange post is here: https://math.stackexchange.com/a/2375897/985601 – bananenheld Jan 06 '23 at 15:50

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