This is FAO anyone who has a clue about complex numbers…

Got a maths problem:

Find the two roots of 2+2(root 3)i
Did it the usual way, using polar form [mod, arg]
got an answer of (±)(root 3)(±)i

The question then went on to say
‘now use the alternative method started below…marks will NOT be awarded for substituting into an equation’

Let a+bi be one of the roots
then (a+bi)^2 = 2+2(root 3)i

Here’s how far I managed to get:
a^2 + (2ab)i + b^2 i^2 = 2 +2(root3)i
therefore, equating real and imaginary parts:
a^2 -b^2 = 2
and
ab = root 3

after that, I’m at sea…thought about substituting into the quadratic but thought that that would be in violation of the question…

Any mathdopers out there in the teeming millions able to ease my pain?

Solve for b in the second equation.
Substitute into the second equation.
Use the quadratic formula to solve for a[sup]2[/sup], and thus a.
Substitute back into either to find b.
Determine which roots are valid.

I suppose you knew that, but were thrown by: “marks will NOT be awarded for substituting into an equation.” I would interpret that as, “if you use a standard formula that directly produces the roots of a complex number, you will get no points.” Since using the quadratic formula is an accepted step in the solution procedure, I see no reason to forbid its use.

I suppose you could go the long way around and factor the quadratic into (a + k[sub]1[/sub])(a + k[sub]2[/sub]), and work through the product and the sum of the ks, but that seems like more work than necessary.