We all know the Golden Ratio.

\(\displaystyle \Phi=\frac{a}{b}=\frac{a+b}{a}\)

\(\displaystyle \Phi^2-\Phi-1=0 \Leftrightarrow \Phi = \frac{1 \pm \sqrt{5}}{2}\)

The Golden Ratio has all sorts of interesting properties like:

* subsequent powers of the Golden Ratios are summable (so-called "recurrence relation");

* there exists a direct relationship with the Fibonacci series;

* it can be expressed as a nested radical only using the number 1

* it is the "worst" real number for rational approximation because of its continued fraction representation (only using 1's)

But suppose we'd define an imaginary equivalent of this Golden Ratio (let's call it \(\displaystyle \Phi_{i}\)) as follows:

\(\displaystyle \Phi_{i}=\frac{a}{b}=\frac{a-b}{a}\)

\(\displaystyle \Phi_{i}^2-\Phi+1=0 \Leftrightarrow \Phi_{i} = \frac{1 \pm i\sqrt{3}}{2}\)

To my surprise, this Imaginary equivalent possesses all the same atrributes as the standard Golden Ratio:

There is a direct link to \(\displaystyle \pi\) and \(\displaystyle e\):

\(\displaystyle \Phi_{i} = e^{\pm \frac{\pi i}{3}}\)

There is a nice link to the Fibonacci series but then for the negative equivalent:

\(\displaystyle F_0 = 0\)

\(\displaystyle F_1 = -1\)

\(\displaystyle F_n = F_{n-1} - F_{n-2}\)

And this induces the series:

0,-1,-1, 0, 1, 1, 0,-1,-1, 0, 1, 1, 0...

and this series also follows a simple formula:

\(\displaystyle F_n = \frac{(1-i \sqrt(3))^n - (1+i \sqrt(3))^n}{2^ni\sqrt(3)}\)

or can be written as:

\(\displaystyle F_n = -\frac23 \sqrt{3} \sin(\frac{\pi n}{3})\)

Also, subsequent powers of \(\displaystyle \Phi_{i}\) can be subtracted (instead of summed):

\(\displaystyle \Phi_{i}^n =\Phi_{i}^{n-1} - \Phi_{i}^{n-2}\)

And for n = 1, the special relation occurs:

\(\displaystyle \Phi_{i} =1 - \frac{1}{\Phi_{i}}\)

And this can be used to establish the recursive fraction (by recursively substituting the full formula \(\displaystyle \Phi_{i}\) into the \(\displaystyle \Phi_{i}\) in the denoninator):

\(\displaystyle \Phi _{i}= 1- \frac{1}{1- \frac{1}{1- \frac{1}{1- \frac{1}{1- ...}}}}\) (yes, there seems to be another "worst" number for rational approximation).

And there also exists a representation in terms of a nested radical:

\(\displaystyle \Phi _{i}= \sqrt{-1+\sqrt{-1+\sqrt{-1+\sqrt{-1+\sqrt{-1+...}}}}}\)

Note:

This series should (I think) be interpreted as:

\(\displaystyle 1,\ \ \ 1-(\frac12+\frac12 i \sqrt3), \ \ \ 1-\frac{1}{1-(\frac12+\frac12 i \sqrt3)},\ \ \, \ 1-\frac{1}{1-\frac{1}{1-(\frac12+\frac12 i \sqrt3)}}\ \ \ \cdots\) and "converges" to \(\displaystyle \frac12-\frac12 i \sqrt3\).

I also tested whether there exists a similar connection to Pascal's triangle and it that also works out very nicely:

It's appears also true that the alternating sum of all n 'Imaginary' Fibonacci numbers equals the (n+2)-th Fibonacci number minus 1.

\(\displaystyle \displaystyle\sum_{k=0}^n (-1)^k F_{i(k)} = F_{i(n+2)}-1\)

And it is also true that the sum of n alternating squares of the 'Imaginary' Fibonacci series is \(\displaystyle F_{i(n)} * F_{i(n+1)}\):

\(\displaystyle \displaystyle\sum_{k=0}^n (-1)^k F_{i(k)}^2 = F_{i(n)}*F_{i(n+1)}\)

Another nice similarity is that by dividing each n-th (original) Fibonacci number by \(\displaystyle 10^n\) and then summing the results, the fraction \(\displaystyle \frac{1}{89}\) results. For the alternating series it is: \(\displaystyle \frac{1}{109}\).

When applied to the 'Imaginary' Fibonacci series: \(\displaystyle 0,-1,-1, 0, 1, 1, 0,-1,-1, 0, 1, 1, 0...\) the result is \(\displaystyle \frac{1}{91}\) and for the alternating version \(\displaystyle \frac{1}{111}\).

Just as for the normal Golden Ratio, the sine of certain complex numbers involving \(\displaystyle \Phi_{i}\) gives particularly simple answers, for example:

\(\displaystyle \sin(i \ln(\Phi_{i})) = -\frac12 \sqrt{3}\)

\(\displaystyle \sin(\frac12 \pi - i \ln(\Phi_{i})) = \frac12\)

This is as far as I've got, but guess there are many more advanced connections to be made.

\(\displaystyle \Phi=\frac{a}{b}=\frac{a+b}{a}\)

\(\displaystyle \Phi^2-\Phi-1=0 \Leftrightarrow \Phi = \frac{1 \pm \sqrt{5}}{2}\)

The Golden Ratio has all sorts of interesting properties like:

* subsequent powers of the Golden Ratios are summable (so-called "recurrence relation");

* there exists a direct relationship with the Fibonacci series;

* it can be expressed as a nested radical only using the number 1

* it is the "worst" real number for rational approximation because of its continued fraction representation (only using 1's)

But suppose we'd define an imaginary equivalent of this Golden Ratio (let's call it \(\displaystyle \Phi_{i}\)) as follows:

\(\displaystyle \Phi_{i}=\frac{a}{b}=\frac{a-b}{a}\)

\(\displaystyle \Phi_{i}^2-\Phi+1=0 \Leftrightarrow \Phi_{i} = \frac{1 \pm i\sqrt{3}}{2}\)

To my surprise, this Imaginary equivalent possesses all the same atrributes as the standard Golden Ratio:

There is a direct link to \(\displaystyle \pi\) and \(\displaystyle e\):

\(\displaystyle \Phi_{i} = e^{\pm \frac{\pi i}{3}}\)

There is a nice link to the Fibonacci series but then for the negative equivalent:

\(\displaystyle F_0 = 0\)

\(\displaystyle F_1 = -1\)

\(\displaystyle F_n = F_{n-1} - F_{n-2}\)

And this induces the series:

0,-1,-1, 0, 1, 1, 0,-1,-1, 0, 1, 1, 0...

and this series also follows a simple formula:

\(\displaystyle F_n = \frac{(1-i \sqrt(3))^n - (1+i \sqrt(3))^n}{2^ni\sqrt(3)}\)

or can be written as:

\(\displaystyle F_n = -\frac23 \sqrt{3} \sin(\frac{\pi n}{3})\)

Also, subsequent powers of \(\displaystyle \Phi_{i}\) can be subtracted (instead of summed):

\(\displaystyle \Phi_{i}^n =\Phi_{i}^{n-1} - \Phi_{i}^{n-2}\)

And for n = 1, the special relation occurs:

\(\displaystyle \Phi_{i} =1 - \frac{1}{\Phi_{i}}\)

And this can be used to establish the recursive fraction (by recursively substituting the full formula \(\displaystyle \Phi_{i}\) into the \(\displaystyle \Phi_{i}\) in the denoninator):

\(\displaystyle \Phi _{i}= 1- \frac{1}{1- \frac{1}{1- \frac{1}{1- \frac{1}{1- ...}}}}\) (yes, there seems to be another "worst" number for rational approximation).

And there also exists a representation in terms of a nested radical:

\(\displaystyle \Phi _{i}= \sqrt{-1+\sqrt{-1+\sqrt{-1+\sqrt{-1+\sqrt{-1+...}}}}}\)

Note:

This series should (I think) be interpreted as:

\(\displaystyle 1,\ \ \ 1-(\frac12+\frac12 i \sqrt3), \ \ \ 1-\frac{1}{1-(\frac12+\frac12 i \sqrt3)},\ \ \, \ 1-\frac{1}{1-\frac{1}{1-(\frac12+\frac12 i \sqrt3)}}\ \ \ \cdots\) and "converges" to \(\displaystyle \frac12-\frac12 i \sqrt3\).

I also tested whether there exists a similar connection to Pascal's triangle and it that also works out very nicely:

It's appears also true that the alternating sum of all n 'Imaginary' Fibonacci numbers equals the (n+2)-th Fibonacci number minus 1.

\(\displaystyle \displaystyle\sum_{k=0}^n (-1)^k F_{i(k)} = F_{i(n+2)}-1\)

And it is also true that the sum of n alternating squares of the 'Imaginary' Fibonacci series is \(\displaystyle F_{i(n)} * F_{i(n+1)}\):

\(\displaystyle \displaystyle\sum_{k=0}^n (-1)^k F_{i(k)}^2 = F_{i(n)}*F_{i(n+1)}\)

Another nice similarity is that by dividing each n-th (original) Fibonacci number by \(\displaystyle 10^n\) and then summing the results, the fraction \(\displaystyle \frac{1}{89}\) results. For the alternating series it is: \(\displaystyle \frac{1}{109}\).

When applied to the 'Imaginary' Fibonacci series: \(\displaystyle 0,-1,-1, 0, 1, 1, 0,-1,-1, 0, 1, 1, 0...\) the result is \(\displaystyle \frac{1}{91}\) and for the alternating version \(\displaystyle \frac{1}{111}\).

Just as for the normal Golden Ratio, the sine of certain complex numbers involving \(\displaystyle \Phi_{i}\) gives particularly simple answers, for example:

\(\displaystyle \sin(i \ln(\Phi_{i})) = -\frac12 \sqrt{3}\)

\(\displaystyle \sin(\frac12 \pi - i \ln(\Phi_{i})) = \frac12\)

This is as far as I've got, but guess there are many more advanced connections to be made.

Last edited by a moderator: