Interpolated Fibonacci numbers - real or complex? The common Binet-formula for the Fibonacci-numbers
$$ f_n = {\varphi^n- (1-\varphi)^n \over \sqrt 5 } \small {\qquad \qquad \text{ where }\varphi={1+\sqrt 5\over 2}}$$
allows interpolation to negative and fractional indexes, however for non-natural indexes we have in general complex values (and also we lose uniqueness due to fractional powers of a negative basis). This is the interpolation-scheme which I'm used to.      
In the mathworld-article on Fibonacci-numbers I found now the formula for the 
interpolation to noninteger indexes as
$$ f^*_x = {\varphi^x-\cos(\pi x) \varphi^{-x} \over \sqrt 5 } $$
which guarantees real values also for the interpolated fibonacci-numbers. 
While I like the idea to find some real-number-interpolation for the index which leads also to real numbers in the range, it still looks a bit ugly because of asymmetry and feels a bit like handwaving; compare the discussion about which is the best interpolation for the factorial and why we choose the Euler-gamma-function for this.... : one argument which made me confident that the Binet-interpolation is "the correct one" is similar to the same (possible) argument for the Euler's-gammafunction: it coincides also with the concept of indefinite summation, extended to fractional and even complex summation-indexes.
Now in the mathworld-article there is no further reasoning about this and so I thought I ask here:    
Q: is there any other reason for this type of interpolation other than the non-imaginary-ness of the resulting values for real indexes?
(an instructive reference were as well welcome...)

[update] Just for the intuition: here is a picture which shows the curve in the complex plane for fibonacci-numbers due to the Binet-formula for fractional (but strictly real) indexes. To avoid the overlap of $\text{fib}(1)=\text{fib}(2)=1$ I used a slightly shifted version for the fibonacci-numbers, by setting $\text{fib}^*(0)=0.1, \text{fib}^*(1)=1.1$ and let the curve begin at index $-4$ :
:

and the detail of a self-overlapping segment of the curve 
(The "real-only" version using the cos()-cofactor in the formula would give simply the straight line on the x-axis itself, but with back-an-forth-movements for the negative indexes.)
 A: I like the "talk" section of wikipedia, Talk:Generalizations_of_Fibonacci_numbers.  I am quoting from their writeup.
According to Binet's formula, 
$$F_n = \frac{\varphi^n-\psi^n}{\varphi-\psi} = \frac{\varphi^n-\psi^n}{\sqrt 5}$$ 
Since $\psi = -\frac{1}{\varphi}$, this formula can also be written as
$$F_n = \frac{\varphi^n-(-\varphi)^{-n}}{\sqrt 5}$$
Now if you factor the $-1$ out of the $-\varphi$, you get
$$F_n = \frac{\varphi^n-(-1)^{-n}\varphi^{-n}}{\sqrt 5}$$
...
And from Euler's identity, $-1 = e^{i\pi}$, so
$$F_n = \frac{\varphi^n-e^{i\pi n}\varphi^{-n}}{\sqrt 5}$$
Ok, now continuing on my own.... And from Euler's formula, $e^{i\pi n} = \cos (\pi n) + i \sin (\pi n)$, so Binet's solution can be equivalently expressed as $F_1$, to distinguish it from its complex conjugate. 
$$F_1(z) = \frac{\varphi^z-(\cos (\pi z) + i \sin (\pi z))\varphi^{-z}}{\sqrt 5}$$
There is an alternative definition, with $-1 = e^{-i\pi}$, which leads to the complex conjugate solution, which I will label $F_2$.
$$F_2(z) = \frac{\varphi^z-(\cos (\pi z) - i \sin (\pi z))\varphi^{-z}}{\sqrt 5}$$
Due to the fact that linear combinations of solutions to the Fibonacci recurrence relation are also solutions, we can average $F_1$ and $F_2$ together, and get the real valued solution, $F_{\text{real}}(z)$, where at the real axis, the imaginary term cancels with its conjugate.
$$F_\text{real}(z) = \frac{F_1(z)+F_2(z)}{2} = \frac{\varphi^z-\cos (\pi z) \varphi^{-z}}{\sqrt 5}$$
I think both solutions are equally valid, and perhaps sometimes it seems more natural to have a real valued solution, and this derivation shows where the "cos" term in that solution comes from.  As I noted in earlier comments, there is also some interesting behavior for the "ratio" function for these various solutions, $r(z)=\frac{F(z+1)}{F(z)}$, which I may post later, which connects the complex solution to the formal Schroeder function solution for the ratio function.
