How does one use the complex plane to solve this problem? Given:
$$a^2 + ab + b^2 = 1 + i$$
$$b^2 + bc + c^2 = -2$$
$$c^2 + ca + a^2 = 1$$
Find 
 $$(ab + bc + ca)^2.$$  
The solution says to use the complex plane. Can somebody explain to me (an average high school student) how I'd use the complex plane to solve this problem?
 A: Edit: I should have written the expansion of $a$ in terms of $a_r$ and $a_i$ differently. Sorry.
Think of this problem not as 3 problems with 3 variables, but 6 problems with 6 variables. Write $a = a_r + ia_i$, $b = b_r + ib_i$, and $c = c_r + ic_i$ (e.g. we explicitly write out the real and imaginary parts of a, b, and c. For example, if $a = 5+6i$, than $a_r=5$ and $a_i=6$.) The first equation would than read 
$(a_r + ia_i)^2+(a_r + ia_i)(b_r + ib_i)+(b_r + ib_i)^2=1+i$
$a_r^2+2ia_r a_i + (i^2)a_i^2 + a_rb_r+ia_rb_i+ia_ib_r+(i^2)a_ib_i+b_r^2+2ib_rb_i+(i^2)b_i^2=1+i$
$a_r^2+2ia_r a_i - a_i^2 + a_rb_r+ia_rb_i+ia_ib_r - a_ib_i+b_r^2+2ib_rb_i - b_i^2=1+i$
$(a_r^2 - a_i^2 + a_rb_r - a_ib_i+b_r^2 - b_i^2) + i(2a_r a_i +a_rb_i+a_ib_r +2b_rb_i )=1+i$
Equating the real and imaginary parts of this equation separately:
$a_r^2 - a_i^2 + a_rb_r - a_ib_i+b_r^2 - b_i^2=1$
$2a_r a_i +a_rb_i+a_ib_r +2b_rb_i =1$
This is now two equations in four variables. You can't solve it yet. Do the same for the other two equations, and you will have 6 equations in six variables ($a_r, a_i, b_r, b_i, c_r, c_i$). The real [or imaginary] parts on the left hand side of an equation must equal the real [or imaginary] part of the right hand side. (Recall that $-2$ may be though of as $-2 + 0i$ on the right hand side of the second equation.)
