1. Stating the problem: We need to find the values of $a$ and $c$ such that the product of two complex numbers $(a + 4j)(c + 4j)$ equals $2 + 36j$. 2. Use the distributive property (FOIL) to expand the product: $$ (a + 4j)(c + 4j) = ac + 4aj + 4cj + 16j^2 $$ 3. Recall that $j^2 = -1$, so substitute: $$ ac + 4aj + 4cj + 16(-1) = ac + 4aj + 4cj - 16 $$ 4. Group real and imaginary parts: Real part: $ac - 16$ Imaginary part: $4a + 4c$ 5. Set the real and imaginary parts equal to those in $2 + 36j$: $$ ac - 16 = 2 $$ $$ 4a + 4c = 36 $$ 6. Simplify the equations: $$ ac = 18 $$ $$ a + c = 9 $$ 7. Solve the system: From $a + c = 9$, express $c = 9 - a$. Substitute into $ac = 18$: $$ a(9 - a) = 18 $$ $$ 9a - a^2 = 18 $$ $$ a^2 - 9a + 18 = 0 $$ 8. Solve the quadratic equation using the quadratic formula: $$ a = \frac{9 \pm \sqrt{81 - 72}}{2} = \frac{9 \pm \sqrt{9}}{2} = \frac{9 \pm 3}{2} $$ 9. Calculate the two possible values for $a$: $$ a = \frac{9 + 3}{2} = 6 $$ $$ a = \frac{9 - 3}{2} = 3 $$ 10. Find corresponding $c$ values: If $a = 6$, then $c = 9 - 6 = 3$. If $a = 3$, then $c = 9 - 3 = 6$. Final answer: The pairs $(a, c)$ are $(6, 3)$ or $(3, 6)$.