If the equation, $x^2+bx+45 = 0 (b \in R)$ has conjugate complex roots and they satisfy $|z + 1| = 2\sqrt{10}$, then:
- $b^2 + b = 72$
- $b^2 - b = 30$
- $b^2 + b = 12$
- $b^2 - b = 42$
For x^2 + bx + 45 = 0, roots are conjugate complex, so b^2 - 4*45 < 0. Let roots be p + iq and p - iq. Product of roots = p^2 + q^2 = 45. Sum of roots = 2p = -b. |z + 1| = 2*sqrt(10) => (p+1)^2 + q^2 = 40. p^2 + 2p + 1 + q^2 = 40. Since p^2 + q^2 = 45, 45 + 2p + 1 = 40 => 2p = -6 => p = -3. Then -b = 2p = -6 => b = 6. Check: b^2 - b = 36 - 6 = 30.
If the equation has complex conjugate roots, its discriminant must be negative, which happens when b^2 is less than 180. We write the roots as z = u +/- vi and expand the given modulus equation |u + vi + 1| = 2 sqrt(10) to obtain (u + 1)^2 + v^2 = 40. From the equation x^2 + bx + 45 = 0, Vieta's formulas give u^2 + v^2 = 45, allowing us to substitute v^2 = 45 - u^2 into the previous equation. This yields (u + 1)^2 + 45 - u^2 = 40, which simplifies to 2u = -6, so u = -3. Since the sum of the roots is 2u = -6, we have -b = -6, giving b = 6. Substituting b = 6 into the expression b^2 - b gives 36 - 6 = 30.