Answer
$a = -\frac{6}{7}$
$b = -\frac{2}{7}$
Work Step by Step
$(2-i)a + (1+3i)b + 2 = 0$
Expand the above expression:
$2a-ia+b+3ib+2 = 0$
Combine like terms:
$(2a+b+2) + (3b-a)i = 0$
In order for the complex number to equal zero, both the real and imaginary components must equal 0:
Equation 1: This is an expression from the real component of the complex number.
$2a + b + 2 = 0$
Equation 2: This is an expression from the imaginary component of the complex number.
$3b-a = 0$
Solve the above system of equations using substitution:
From Equation 2:
$a = 3b$
Substitute the above expression ($a = 3b$) into Equation 1, and solve for b:
$2(3b) + b + 2 = 0$
$6b + b + 2 = 0$
$7b + 2 = 0$
$7b = -2$
$b = -\frac{2}{7}$
Use b to solve for a using the above expression $a = 3b$:
$a = 3(-\frac{2}{7})$
$a = -\frac{6}{7}$
In summary:
$a = -\frac{6}{7}$
$b = -\frac{2}{7}$
