Binet's Fibonacci Number Formula $\frac{φ^{n}-(-φ)^{-n}}{\sqrt 5}$ produces a sine-like curve for positive numbers and a spiral for negative numbers when plotted on a complex plane (see figs. 1., and 2.)
I'd like to know the parametric equations of this curve on the (real) Cartesian plane. In other words, the exact translation of this curve on to the Cartesian plane written in terms of $x(t),$ and $y(t)$.
Lets define differentiable paths $\gamma_k:\mathbb R \to \mathbb C$ so that they "match" formula
$$\gamma(t) = \frac{\varphi^t -(-\varphi)^{-t}}{\sqrt 5},$$ where $\varphi = \frac{1 + \sqrt 5}2$ is the golden ratio. To do so, use formula for exponentiation of complex numbers: $z^w = \exp(w\log(z))$ where $\log$ is the complex natural logarithm multi-function. We have $$\log\varphi = \ln \varphi + 2l\pi \mathrm i\,,\quad\log(-\varphi) = \ln\varphi + (2k + 1)\pi\mathrm i$$ where $\ln : (0, \infty) \to \mathbb R$ is the real natural logarithm function, and $k,l\in \mathbb Z$. Since $\varphi$ is positive real number, I will assume that $l = 0$, and complex logarithm is the same as real one for it.
Using Euler's formula we can write \begin{align} \gamma_k(t) &= \frac{\exp(t\ln\varphi) - \exp(-t\ln\varphi -t(2k+1)\pi\mathrm i)}{\sqrt 5} \\ &= \frac {\exp(t\ln\varphi) - \exp(-t\ln\varphi)\cos((2k+1)\pi t)} {\sqrt 5} + \mathrm i \frac{\exp(-t\ln\varphi)\sin((2k + 1)\pi t)}{\sqrt 5}. \end{align} Looking at real and imaginary parts we obtain family of curves that satisfy initial assumptions: $$ \begin{cases} x_k(t) = \frac {\varphi^t - \varphi^{-t}\cos((2k+1)\pi t)} {\sqrt 5}\\ y_k(t) = \frac{\varphi^{-t}\sin((2k + 1)\pi t)}{\sqrt 5} \end{cases}. $$ If you plot them you can see that your geometry is matched when $k=-1$ (it is a bit distorted, I think your first plot may have different scales on the axis).