To solve for the possible value of b, we start with the complex number x = 3 + bi.
We need to compute x²:
x² = (3 + bi)² = 3² + 2(3)(bi) + (bi)² = 9 + 6bi – b²
Combining like terms, we have:
x² = (9 – b²) + (6b)i
Since we know that x² = 13, we compare the real and imaginary parts. The real part gives us:
9 – b² = 13
Simplifying this, we subtract 9 from both sides:
-b² = 4
This leads us to:
b² = -4
This indicates that either we made an error in our understanding or that the value of b must be a real number while yielding a complex solution. In the context of this problem, we can set 9 – b² = 0 to find valid values for b.
Rearranging gives:
b² = 9
Thus, b can be either 3 or -3, which isn’t in the provided list. Given the available options, the practical approach is ensuring that b gives us a real part consistent with x² = 13.
Testing the values:
- If b = 2: x² = (3 + 2i)² = 9 + 12i – 4 = 5 + 12i (not equal to 13)
- If b = 4: x² = (3 + 4i)² = 9 + 24i – 16 = -7 + 24i (not equal to 13)
- If b = 9: x² = (3 + 9i)² = 9 + 54i – 81 = -72 + 54i (not equal to 13)
- If b = 10: x² = (3 + 10i)² = 9 + 60i – 100 = -91 + 60i (not equal to 13)
Realistically, the most straightforward assessment is that none of the values yield valid calculations leading to a real and satisfactory result of 13.