Answer
$a = 3$
$b = -4$
Work Step by Step
$a(5+4i)+b(4+3i)+1 = 0$
Expand the above expression:
$5a+4ia+4b+3ib+1 = 0$
Combine like terms:
$(5a+4b+1) + (3b + 4a)i = 0$
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.
$5a+4b+ 1 = 0$
Equation 2: This is an expression from the imaginary component of the complex number.
$3b+4a = 0$
Solve the above system of equations using substitution:
From Equation 2:
$4a = -3b$
$a = -\frac{3}{4}b$
Substitute the above expression ($a = -\frac{3}{4}b$) into Equation 1, and solve for b:
$5(-\frac{3}{4}b)+4b+ 1 = 0$
$-\frac{15}{4}b+4b+ 1 = 0$
$-\frac{15}{4}b+\frac{16}{4}b+ 1 = 0$
$\frac{1}{4}b+ 1 = 0$
$\frac{1}{4}b = -1$
$b = -4$
Use b to solve for a using the above expression $a = -\frac{3}{4}b$:
$a = -\frac{3}{4}(-4)$
$a = 3$
In summary:
$a = 3$
$b = -4$
