Side comment first: yes, root of the polynomial is something that makes it to be equal to zero, so there is 0 in the right side of every equation. Square equations You're completely right up to the point where you're trying to find complex roots of a square equation. It is done using exactly the same way as for real roots: So if your equation is 1x² + 1x + 1 = 0, then a=1, b=1, c=1 and roots are Since √(-3) = i√3 (because i√3 x i√3 = (i x i) x (√3 x √3) = -1 x 3 = -3), the roots are -1/2 ± (√3/2)i It is possible to find complex roots your way You can find the roots the way you calculate things, but be more careful with imaginary part; let's represent x not as 'a + b' as you do but as 'a + bi', where both a and b are real numbers, and i is, well, square root of -1. Then we will get: a² + 2abi + (bi)² + a + bi + 1 = 0 => a² + 2abi - b² + a + bi + 1 = 0 => (a² - b² + a + 1) + (2ab + b)i = 0 Now look at the number at the left side: it has real part (a² - b² + a + 1) and imaginary part (2ab + b)i. For this number to be zero, both parts should be zero (you cannot make zero out of real number by adding imaginary part to it). So we got that I'm going to look at the imaginary part first. 2ab + b = 0 => (2a + 1)b = 0 Now we have two numbers (2a + 1 and b) such that their product is zero, so one of those numbers is zero. It cannot be b (try to substitute b = 0 -- the equation turns out to a² + a + 1 = 0, which has no real solutions, and a was a real number). Hence Now, when we know a, let's look at the real part. 1/4 - b² - 1/2 + 1 = 0 => b² = 3/4 => b = ±√3/2 We substituted x = a + bi, so we can get our two roots now: -1/2 ± (√3/2)i General situation There is a theorem that the number of (complex) roots a polynomial has is equal to its degree. Well, the whole thing is a bit less straightforward as some roots can repeat themselves, like in x² - 2x + 1 = 0 -- there is only one solution, x = 1, but since the same equation can be represented as (x - 1)(x - 1) = 0, the solution x = 1 is, well, counted twice -- once for each "x - 1". I'm not going deeper here, but what's important to know: if you take random polynomial of degree n, almost always it will have exactly n different complex roots. It is not possible to calculate exact roots for general polynomial of degree over 4 -- not that we don't know how, it's proven theorem that the formula cannot exist. About my picture So, I have equations which look like ..x^24 + ..x^23 + ...and so on... + ..x^2 + ..x + .. = 0, where every '..' is either 1 or -1. I lose nothing assuming that x^24 always has 1 (think about it), so there are 24 coefficients left. It gives me 2^24 = 16777216 equations. I find approximate solutions for each of those, and each has exactly 24 solutions. In this way I'm getting 2^24 x 24 = 402653184 complex numbers. For each such number, if it is a + bi, I put a point with coordinates (a, b). And that's how I've got my picture! x = (-b ± √(b² - 4ac)) / 2a
(-b ± √(b² - 4ac)) / 2a = (-1 ± √(1 - 4)) / 2 = (-1 ± √(-3)) / 2
1(a + bi)² + 1(a + bi) + 1 = 0 =>
a² - b² + a + 1 = 0 and (2ab + b)i = 0
(2ab + b)i = 0 =>
2a + 1 = 0 => a = -1/2
a² - b² + a + 1 = 0 =>