Complex Fibonacci numbers The Binet formula is a closed form expression for the \$n\$'th Fibonacci number:
$$F_n = \frac {\phi^n - (1-\phi)^n} {\sqrt 5}$$
where \$\phi = \frac {1 + \sqrt 5} 2\$ is the golden ratio. This formula works even when \$n\$ is negative or rational, and so can be a basis to calculating &quot;complex Fibonacci numbers&quot;.
For example, by setting \$n = \frac 1 2\$, we can calculate \$F_\frac 1 2\$ as:
$$F_\frac 1 2 = \frac 1 {\sqrt 5} \left( \sqrt \frac {1+\sqrt 5} 2 - \sqrt \frac {1-\sqrt 5} 2 \right) \\
\approx 0.56886-0.35158i$$
You are to take a floating point, \$-10 \le n \le 10\$, with up to 3 decimals after the point, and output \$F_n\$, accurate to at least 5 decimal places. You may either round or truncate, so long as it is consistent. You may also choose to input as a rational number if you wish, or as a (numerator, denominator) pair. You may also choose whether integers should be suffixed with .0 or not, so long as it is consistent across all 21 integer inputs.
As the output will be a complex number in all but 21 cases, you may output in any reasonable format for such a type, including outputting as a (real, imag) pair. For the integer inputs, the imaginary part will be \$0\$. You may choose whether to output the imaginary part in this case (and returning an integer  or float is perfectly fine).
This is code-golf, so the shortest code in bytes wins
Test cases
These all round their output, rather than truncate.

  n       Re(Fn)     Im(Fn)
-10      -55         0
  1        1         0
 -7       13         0
  3        2         0
  0.5      0.56886  -0.35158
  5.3      5.75045   0.02824
  7.5     16.51666   0.01211
 -1.5      0.21729  -0.92044
 -9.06    34.37587  -6.55646
  9.09    35.50413   0.00157
 -2.54     0.32202   1.50628
  5.991    7.96522   0.00071
 -6.033   -8.08507   0.84377
  8.472   26.36619  -0.00756

And a script to output all possible outputs in the same format (gets cut off on TIO due to the length).
Wolfram Language (Mathematica), 30 bytes
(h=2/--√5;h^#-(1-h)^#)√.2&amp;

Try it online!
Uses \$\frac{\sqrt 5+1}{2}=\frac{2}{\sqrt 5-1}\$. Mathematica's pre-increment/decrement operators still return the desired value when used on (non-variable) non-atoms.
The built-in Fibonacci is a continuous real function, the real part of the Binet formula: \$F_n=\frac{\phi^n-\cos(\pi n)\phi^{-n}}{\sqrt 5}\$.