Solving Recurrences via Characteristic Polynomial with Imaginary Roots
Yes, the solution is in fact $T(n) = \alpha(1+i)^n + \beta(1-i)^n$ for some constants $\alpha$ and $\beta$ determined by the base cases.  If the bases cases are real, then (by induction) all the complex terms in $T(n)$ will cancel, for all integer $n$.

For example, consider the recurrence $T(n) = 2T(n-1) - 2T(n-2)$, with base cases $T(0)=0$ and $T(1)=2$.  The characteristic polynomial of this recurrence is $x^2-2x+2$, so the solution is $T(n) = \alpha(1+i)^n + \beta(1-i)^n$ for some constants $\alpha$ and $\beta$.  Plugging in base cases gives us
$$
T(0) = \alpha(1+i)^0 + \beta(1-i)^0 = \alpha+\beta = 0\\
T(1) = \alpha(1+i)^1 + \beta(1-i)^1 = (\alpha+\beta) + (\alpha-\beta)i = 2
$$
which implies
$$
\alpha + \beta = 0 \\
\alpha - \beta = -2i
$$
which implies
$\alpha = -i$ and $\beta = i$.  So the solution is
$$
T(n) = i\cdot ((1-i)^n - (1+i)^n).
$$

This function oscillates between $\sqrt{2}^n$ and $-\sqrt{2}^n$ with a "period" of 4.  In particular, we have $T(4n) = 0$ for all $n$, because $(1-i)^4 = (1+i)^4 = -4$ (and because I chose the base case $T(0)$ carefully).