A complex exponent

We are given $\displaystyle f(x) = \dfrac{1}{\left(ix + \sqrt{1 - x^2}\right)^i}$ . Notice first of all, that we can write $\ln{f(x)} = -i\ln{\left(ix + \sqrt{1 - x^2}\right)}$ . With this, we can see that we have the following: $-i\ln{\left(ix + \sqrt{1 - x^2}\right)} - i\ln{\left(-ix + \sqrt{1 - x^2}\right)} = 0.$

Thus, we must have $\left|ix + \sqrt{1 - x^2}\right| = 1$ , which implies that $ix + \sqrt{1 - x^2} \in \{e^{it} | \space t \in [0,\pi) \cup (\pi,2\pi)\}$ .

We know that $\left( e^{it} \right)^{-i} = e^t \in \mathbb{R}$ , and the only real numbers, $x$ , which permit these values for $f(x)$ are $x \in \left[-1,1\right]$ . Therefore, we see that $a = -1$ , and $b = 1$ .

So $f(a) = \left(-i\right)^{-i} = i^i = e^{-\pi/2}$ , and $f(b) = \left(i\right)^{-i} = \dfrac{1}{i^i} = e^{\pi/2}$ .

$\therefore \lfloor 1000 \left( f(a) + f(b) \right) \rfloor = \lfloor 1000 \left( e^{-\pi/2} + e^{\pi/2}\right) \rfloor = \lfloor 5018.3569 \rfloor = 5018.$