OK
The RSolve given above yields as answer
{{x[k] -> (1 - I)^k x0}}
Now define a Function fx
fx[k_, x0_] := FullSimplify[(1 - I)^k x0]
Look what happens to
fx[5, 2 + 3 I]
or
Table[fx[j, x0], {j, 1, 10}]
For a plot you must specify x0. Let us chose x0 = 1
tt = Table[{j, fx[j, 1]}, {j, 1, 15}];
pl1 = ListLinePlot[Re[#[[2]]] & /@ tt,PlotStyle->Red] (*real parts*)
pl2 = ListLinePlot[Im[#[[2]]] & /@ tt,PlotStyle->Blue]
Look at both plots in one
Show[pl1, pl2]