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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 $k_{1,2} = \frac{1}{2} \pm i \frac{1}{4} = \frac{1}{4}\sqrt{5}e^{\pm i \tan^{-1}{(1/2)}}$

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    $\begingroup$ $\mathrm e^{-i\pi/2}=-i$. $\endgroup$
    – Bernard
    Aug 12, 2019 at 23:56
  • $\begingroup$ Thanks @Bernard but I'm not seeing how that helps. I need to get out of this fraction somehow or approach the problem differently. $\endgroup$
    – 1ijk
    Aug 13, 2019 at 0:41
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    $\begingroup$ Also, $\mathrm e^{i\pi/4}=\frac 1{\sqrt 2}(1+i)$. With these elements you can write each fraction in the form $x+iy$. $\endgroup$
    – Bernard
    Aug 13, 2019 at 0:52

1 Answer 1

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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,

$k_1 = \frac{1}{2} + \frac{i}{4}$

$k_2 = \frac{1}{2} - \frac{i}{4}$

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