OMG I just found out it *is* based on the \(\tan\). These are well known identities:
\[ \tanh(x)=\frac{e^{2x}-1}{e^{2x}+1},\quad \text{artanh}(x)={\frac {1}{2}}\log \left({\frac {1+x}{1-x}}\right),\quad i\tan(x) = \tanh(ix)\]
I am digging out long forgotten high school math, lol.
In our (new) formula ( remember \(d=\frac{1-s}{1+s}\) ):
\[ q_t(s) = \frac{1}{s}\frac{1-d^t}{1+d^t} = -\frac{1}{s}\frac{e^{t\log(d)}-1}{e^{t\log(d)}+1}
= -\frac{1}{s}\tanh\left(t\frac{\log(d)}{2}\right) = -\frac{1}{s}\tanh\left(-t\;\text{artanh}(s)\right) = \frac{1}{s}\tanh(t\;\text{artanh}(s))\]
And if now s is imaginary, \(s=ix\), then
\[ q_t(ix) = \frac{1}{ix}\tanh\left(t\;\text{artanh}(ix)\right) = \frac{1}{x}\tan\left(\frac{t}{i}i\arctan(x)\right) = \frac{1}{x}\tan\left(t\;\arctan(x)\right) \]
And from here its totally clear that \(t\mapsto q_t(ix)\) is a stretched \(\tan\) with period \(\frac{\pi}{\arctan(x)}\).
In the provided picture with \(c=-0.5\), \(s=i\sqrt{1/2}\), the period would be \(\frac{\pi}{\arctan\left(\sqrt{1/2}\right)} \approx \) 5.104 which matches the visual.
\[ \tanh(x)=\frac{e^{2x}-1}{e^{2x}+1},\quad \text{artanh}(x)={\frac {1}{2}}\log \left({\frac {1+x}{1-x}}\right),\quad i\tan(x) = \tanh(ix)\]
I am digging out long forgotten high school math, lol.
In our (new) formula ( remember \(d=\frac{1-s}{1+s}\) ):
\[ q_t(s) = \frac{1}{s}\frac{1-d^t}{1+d^t} = -\frac{1}{s}\frac{e^{t\log(d)}-1}{e^{t\log(d)}+1}
= -\frac{1}{s}\tanh\left(t\frac{\log(d)}{2}\right) = -\frac{1}{s}\tanh\left(-t\;\text{artanh}(s)\right) = \frac{1}{s}\tanh(t\;\text{artanh}(s))\]
And if now s is imaginary, \(s=ix\), then
\[ q_t(ix) = \frac{1}{ix}\tanh\left(t\;\text{artanh}(ix)\right) = \frac{1}{x}\tan\left(\frac{t}{i}i\arctan(x)\right) = \frac{1}{x}\tan\left(t\;\arctan(x)\right) \]
And from here its totally clear that \(t\mapsto q_t(ix)\) is a stretched \(\tan\) with period \(\frac{\pi}{\arctan(x)}\).
In the provided picture with \(c=-0.5\), \(s=i\sqrt{1/2}\), the period would be \(\frac{\pi}{\arctan\left(\sqrt{1/2}\right)} \approx \) 5.104 which matches the visual.