Question

If ƒ(x) = x⁴ – 8x³ + 4x² + 4x + 39 and ƒ(3 + 2i) = a + ib, then a : b =

• A. $\frac{1}{8}$
• B. –$\frac{1}{4}$
• C. $\frac{1}{4}$
• D. –$\frac{1}{8}$

Answer 1

Answer: (D)

–$\frac{1}{8}$

Answer 2

Sol. Let x = 3 + 2i Ž (x – 3)² = 4i² = –4

Ž x² – 6x + 13 = 0

Thus, x⁴ – 8x³ + 4x² + 4x + 39

= (x² – 6x + 13) (x² – 2x – 21) – 96x + 312

\ ƒ(3 + 2i) = 0 (x² – 2x – 21) – 96 (3 + 2i) + 312

= 24 – 192i = a + ib

\ a = 24 and b = – 192

\ $\frac{a}{b} = \frac{24}{- 192}$ = – $\frac{1}{8}$.