Solving A Linear Recurrence Relation With Complex Roots Question:
For the given linear homogeneous difference equation, find the general solution:
$$y_{n+2} + y_{n+1} + y_n = 0$$ 
With the initial conditions of:
$$y(0)=\sqrt3, y(1) = 0$$ 
Attempted Answer:
I approached the problem normally as one would except by solving the auxiliary equation which yields:
$$m^2 + m + 1 = 0$$
$$\implies m = \frac{-1}{2} \pm \frac{i\sqrt3}{2}$$
Now the general solution is of form:
$$y_n = A \left(\frac{-1}{2} - \frac{i\sqrt3}{2}\right)^n
      + B \left(\frac{-1}{2} + \frac{i\sqrt3}{2}\right)^n $$
Here is the part where I get stuck when I substitute the initial conditions to form a system of equations:
$$A + B = \sqrt{3}$$
$$A \left(\frac{-1}{2} - \frac{i\sqrt3}{2}\right)
      + B \left(\frac{-1}{2} + \frac{i\sqrt3}{2}\right) = 0$$
Now to make my life easier (which it really didn't), I decided to take a 'shortcut' and rewrite the 2nd equation as the sum of an imaginary and real part:
$$\frac{-1}{2}(A + B) + i\frac{\sqrt3}{2}(-A + B) = 0 + 0i$$
Since the real part on the LHS must equal the real part on the RHS, and the same applies for the imaginary part, we obtain:
$$\frac{-1}{2}(A + B) = 0$$
$$\frac{\sqrt3}{2}(-A + B) = 0$$
So the above implies that $A = B = -B$ which can only be true if $A = B = 0$. However if that is the case, then the first equation in the system of equations ($A + B = \sqrt{3}$) implies that 0 = $\sqrt{3}$.
I know that there is a mistake somewhere in the logic of my reasoning (I think perhaps when I equated each real and imaginary part of the LHS to the RHS) but I don't know why and where exactly. Please point out my mistake. 

What happens
when the roots of the
characteristic polynomial are complex
is that the solutions
have a periodic component.
Your case is
$y_n 
= a \left(\frac{-1}{2} - \frac{i\sqrt3}{2}\right)^n
      + b \left(\frac{-1}{2} + \frac{i\sqrt3}{2}\right)^n
=a r^n + bs^n
$.
Note that
$r+s = -1$
and
$r-s = -i\sqrt{3}$.
If
$y_0 = u$ 
and
$y_1 = v$,
then
$a+b = u$
and
$ar+bs = v$.
Since
$b = u-a$,
$v 
= ar+(u-a)s
=a(r-s)+us
$,
so
$a
=\frac{v-us}{r-s}
$
and
$b
=u-a
=u-\frac{v-us}{r-s}
=\frac{u(r-s)-v+us}{r-s}
=\frac{ur-v}{r-s}
$.
In your case,
$u=\sqrt{3},
v = 0,
r=\frac{-1}{2} - \frac{i\sqrt3}{2},
s=\frac{-1}{2} + \frac{i\sqrt3}{2},
r-s=-i\sqrt{3}
$,
so
$a
=\frac{v-us}{r-s}
=\frac{-\sqrt{3}(\frac{-1}{2} + \frac{i\sqrt3}{2})}{-i\sqrt{3}}
=-i(\frac{-1}{2} + \frac{i\sqrt3}{2})
=\frac{i}{2} + \frac{\sqrt3}{2}
$
and
$b
=u-a
=\frac{\sqrt3}{2}-\frac{i}{2}
$.
(We are getting close.)
We have
$y_n
=ar^n+bs^n
$.
By the magic of
assigned problems,
$|r| = |s| = 1$.
$r = e^{-2i\pi/3}
$,
so
$r^n
=e^{-2ni\pi/3}
=\cos(-2n\pi/3)+i\sin(-2n\pi/3)
=\cos(2n\pi/3)-i\sin(2n\pi/3)
$.
Similarly,
$s = e^{2i\pi/3}
$,
so
$s^n
=e^{2in\pi/3}
=\cos(2n\pi/3)+i\sin(2n\pi/3)
$.
Also
$a
=e^{i\pi/6}
$
and
$b
=e^{-i\pi/6}
$.
Therefore
(finally!)
$\begin{align*}
y_n
&=ar^n+bs^n\\
&=e^{i\pi/6}e^{-2in\pi/3}+e^{-i\pi/6}e^{2in\pi/3}\\
&=e^{\pi i(1/6-2n/3)}+e^{\pi i(-1/6+2n/3)}\\
&=\cos(\pi (1/6-2n/3))+i\sin(\pi (1/6-2n/3))
+\cos(\pi (-1/6+2n/3))+i\sin(\pi (-1/6+2n/3))\\
&=\cos(\pi (1/6-2n/3))+\cos(\pi (-1/6+2n/3))
+i(\sin(\pi (1/6-2n/3))+\sin(\pi (-1/6+2n/3)))\\
&=2\cos(\pi (1/6-2n/3))
\qquad\text{since } 
\cos(-x)=\cos(x) \text{ and }\sin(x) = -\sin(-x)\\
\end{align*}
$
You can get explicit values for
values of $n$ mod 6,
but this shows how the result is periodic.
A check is that
the result is real,
with no imaginary part.
This has to hold,
since the initial values
and the recurrence coefficients
are all real.
