A: Given $a_{n}=2a_{n-1}-2a_{n-2}$
The characteristic equation will be $r^{2}=2r-2$
$r^{2}-2r+2=0$
$r=\frac{2\pm \sqrt{(-2)^{2}-4*1*2}}{2}$
$r=\frac{2\pm \sqrt{-4}}{2}$
$r=\frac{2\pm 2\sqrt{-1}}{2}$
$r=\frac{2\pm 2i}{2}$
$r=1\pm i$
B: $a_{0}=1\, \, \, a_{1}=2$
The solution of recurrence relation is of form $a_{n}=\alpha r_{1}^{n} + \beta r_{2}^n$
$a_{n}=\alpha (1-i)^{n} + \beta (1+i)^n$
at n=0 and n=1 we have
$1=\alpha + \beta$
$2=(1-i)\alpha + (1+i)\beta$
Multiply first equation by $(1-i)$
$(1-i)=(1-i)\alpha + (1-i)\beta$
$2=(1-i)\alpha + (1+i)\beta$
subtract two equations
$1+i=2i\beta$
$\beta =\frac{i+1}{2i}$
$\beta =\frac{(i+1)i}{2i^2}$
$\beta =\frac{i-1}{-2}$
$\beta =\left ( \frac{1}{2}-\frac{i}{2} \right )$
$\alpha =1-\beta$
$\alpha =1-\left ( \frac{1}{2}-\frac{i}{2} \right )$
$\alpha =\left ( \frac{1}{2}+\frac{i}{2} \right )$
Hence the solution becomes:
$a_{n} =\left ( \frac{1}{2}+\frac{i}{2} \right )\cdot (1-i)^{n} + \left ( \frac{1}{2}-\frac{i}{2} \right )\cdot (1+i)^n$