Solving a second order difference equation
You can indeed proceed with the complex roots $z_1$ and $z_2$ as well as with the real ones.

In order for a particular solution you may assume a solution of type:
$$y_p[n] = K ( \frac{1}{2})^n u[n]$$

And plug it into the LCCDE to find the undetermined coefficient $K=1/5$.

Then assuming a homogeneous solution of the form (for causal system) $$y_h[n] = C_1 z_1^n u[n] + C_2 z_2^n u[n]$$ 

You have the complete solution as:
$$y[n] = y_h[n] + y_p[n] = (C_1 z_1^n + C_2 z_2^n + \frac{1}{5} (\frac{1}{2})^n) u[n] $$

In order to find $C_1$ and $C_2$ you can get $2$ equations from $y[0]$ and $y[1]$ by explicitly solving them from the LCCDE via recursion, which yields
$y[0] = 1/3$ and $y[1] = 5/18$. Then you have the following:

$$ 
\begin{cases}
y[0] = \frac{1}{3} &amp;= C_1 + C_2 + \frac{1}{5} \\
y[1] = \frac{5}{18} &amp;= C_1 z_1 + C_2 z_2 + \frac{1}{10} \\
\end{cases}
$$

Solve this system for $C_1$ and $C_2$ to complete the procedure...

Alternateley you can also consider the following.
When have a LCCDE with real coefficients, its consequence is the fact that the roots associated with the homogeneous equation will be either real or in complex-conjugate pairs, as you have found. 

Furthermore you can deduce from the LCDDE with real coefficients that its solution $y[n]$ must be real valued, for a real input $x[n]$.

Hence you can deduce from these that the constants $C_1$ and $C_2$ of those two complex roots should be constrained like $C = C_1 = C_2$ or $C = C_1 = C_2^*$  then you can see that the complex roots will merge:

$$ C_1 z_1^n + C_2 z_2^n = C (z_1^n + (z_1^n)^* ) = 2C \mathcal{Re} \{z_1^n\}$$ 

or like

$$ C_1 z_1^n + C_2 z_2^n = C z_1^n + (C_1z_1^n)^*  = 2 \mathcal{Re} \{C z_1^n\}$$

Conversion to polar form helps like:
$$z_1 = |z_1| e^{j\phi_1} \Rightarrow 2C \mathcal{Re}\{z_1^n\} =2C |z_1|^n\cos(\phi_1 n) $$

or like
$$C = |C|e^{j\theta_C}, ~~z_1 = |z_1| e^{j\phi_1} \longrightarrow 2 \mathcal{Re}\{C z_1^n\} =2 |C| |z_1|^n\cos(\phi_1 n + \theta_C) $$