I am reading Nahin's popular book An Imaginary Tale. In chapter 4, section 4, Nahin gives an application where Leonardo's recurrence is solved by rewriting the recurrence relation as a complex-valued expression.
There's one part of the demonstration that I do not understand. We reduce the problem to its initial conditions, giving two equations in two unknowns:
$$k_{1} + k_{2} = 1$$
and
$$k_{1}2^{3/2}e^{i \pi / 4} + k_{2}2^{3/2}e^{-i \pi / 4} = 1$$
I've solved the equations getting
$$k_{1} = \frac{\frac{1}{c} - e^{-i \pi / 2}}{1 - e^{-i \pi / 2}}$$
and then, setting $c = 2^{3/2}e^{i \pi / 4}$,
$$k_{2} = 1 - \frac{\frac{1}{c} - e^{-i \pi / 2}}{1 - e^{-i \pi / 2}}$$
but I've struggled to simplify the equation from there. Have I made a mistake up to this point or have I've forgotten the relevant manipulations involving $e$ (or, likely, both)? The goal is to have $k_{1,2} = \frac{1}{2} \pm i \frac{1}{4} = \frac{1}{4}\sqrt{5}e^{\pm i \tan^{-1}{(1/2)}}$
I got the following:
$k_1 + k_2 = 1$
and
$k_1 - k_2 = \frac{i}{2}$
[I got this from your second equation. Put $2^{\frac{3}{2}} = 2\sqrt{2}$ and $e^{\pm i\frac{\pi}{4}} = \cos \frac{\pi}{4} \pm i \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \pm i \frac{1}{\sqrt{2}}$]
Solving,
$k_1 = \frac{1}{2} + \frac{i}{4}$
$k_2 = \frac{1}{2} - \frac{i}{4}$