Are a and b Always Real if c and Conjugate c are Roots of a Quadratic Equation?

[daniel_i_l]

Homework Statement

prove or disprove:
1) (z is complex and bar(c) is the conjugate of c)
If c and bar(c) are solutions to z^2 + az + b = 0 and c isn't real then a and b are real.
2) If S and T are subsets of the space V and intersection of Sp(S) and Sp(T) is {0} then the intersection of S and T is an empty set. (Sp is the span)

Homework Equations

The Attempt at a Solution

1) I wrote z as x+yi and got:
x^2 + 2xyi - y^2 + ax + ayi +b = x^2 -2xyi - y^2 +ax - ayi +b
and so:
4xyi + 2ayi = 0 => 2x + a = 0 => a=-2x which is real. and by putting that into the equation I get that b is also real. So the answer is True.

2) If S = T = {0} then the intersection of both Sp(S) and Sp(T) and S and T is {0}. So the answer is False.

Is that right? Am I missing anything here? especially the last one seemed too easy.
Thanks.

 

[AKG]

Both your answers are right. Note in your work for the first one, you have "4xyi + 2ayi = 0 => 2x + a = 0." Be aware that this is only valid because y is non-zero, which is because c isn't real. Another way to do 1) is this:

If c and \bar{c} are the solutions to z² + az + b = 0, then (z-c)(z-\bar{c}) = z^2 + az + b. Multiplying the left side out gives z^2 - (c+\bar{c})z + c\bar{c}, which is just z² - 2Re(c)z + |c|², and so a = - 2Re(c), and b = |c|², both of which are clearly real quantities.