Correct explanation for the first part on why a is not a real number :


①

The complex conjugate root theorem states that if P is a polynomial in one variable with all coefficients being real , and a + bi is a root of P with a and b real numbers, then its complex conjugate a − bi is also a root of P.

This means that complex roots always occur in pairs.

②Following this, if the degree of a real polynomial is odd, it must have at least one real root.


Since the degree of f(z) is 3, if all the coefficients are real, it either has:

- 3 real roots or
-1 real root and a pair of complex conjugate roots.


③ But since it is given that there are no real solutions, all three solutions are either complex (or purely imaginary)

This means that not all of the coefficients are real. There is at least one non-real number coefficient.


④ But since the coefficients containing a are a, 2a - 4 and 2a - 8, all three are non-real since they all contain a and a real number.

(a = a + 0)


If any one of them is complex/non-real, it would imply the others are non - real as well.


Eg.

If 2a - 4 is non-real, neither would 2a - 8 since 2a - 8 = 2a - 4 - 4. Subtracting another 4 does not remove the imaginary portion of a.

vice versa for adding 4 to 2a - 8


So , by the same reasoning, 2a is non-real either and so a is also non-real.