Answer:
- 0.5 + 2.985i
- 1 + 2.828i
- 1.5 + 2.598i
- 2 + 2.236i
Explanation:
Complex numbers have the general form a + bi, where a is the real part and b is the imaginary part.
Since, the numbers are neither purely imaginary nor purely real a ≠ 0 and b ≠ 0.
The absolute value of a complex number is its distance to the origin (0,0), so you use Pythagorean theorem to calculate the absolute value. Calling it |C|, that is:
Then, the work consists in finding pairs (a,b) for which:
You can do it by setting any arbitrary value less than 3 to a or b and solving for the other:

I will use b =0.5, b = 1, b = 1.5, b = 2

Then, four distinct complex numbers that have an absolute value of 3 are:
- 0.5 + 2.985i
- 1 + 2.828i
- 1.5 + 2.598i
- 2 + 2.236i
The answer would be A. When using Cramer's Rule to solve a system of equations, if the determinant of the coefficient matrix equals zero and neither numerator determinant is zero, then the system has infinite solutions. It would be hard finding this answer when we use the Cramer's Rule so instead we use the Gauss Elimination. Considering the equations:
x + y = 3 and <span>2x + 2y = 6
Determinant of the equations are </span>
<span>| 1 1 | </span>
<span>| 2 2 | = 0
</span>
the numerator determinants would be
<span>| 3 1 | . .| 1 3 | </span>
<span>| 6 2 | = | 2 6 | = 0.
Executing Gauss Elimination, any two numbers, whose sum is 3, would satisfy the given system. F</span>or instance (3, 0), <span>(2, 1) and (4, -1). Therefore, it would have infinitely many solutions. </span>