A-Level Complex numbers in cartesian form and the Argand diagram: worked solution

Question

Do not use a calculator. One root of ω⁴−2ω³+10ω²+pω+q=0 (p, q real) is 2+3i. Find p and q and the remaining roots.

Do not use a calculator. Step 1: Use the conjugate root theorem. The polynomial ω⁴ − 2ω³ + 10ω² + pω + q has REAL coefficients (p and q are real), so any non-real roots come in conjugate pairs. Since 2 + 3i is a root, its conjugate 2 − 3i is also a root. Step 2: Build a real quadratic factor from that pair. (ω − (2 + 3i))(ω − (2 − 3i)) Group the real parts so that it becomes a difference of two squares: = ((ω − 2) − 3i)((ω − 2) + 3i) = (ω − 2)² − (3i)² = ω² − 4ω + 4 − 9i² = ω² − 4ω + 4 + 9 [since i² = −1] = ω² − 4ω + 13 Step 3: The quartic must be this quadratic times another quadratic with real coefficients. The quartic is monic (leading coefficient 1) and so is ω² − 4ω + 13, so the other factor is monic too. Write ω⁴ − 2ω³ + 10ω² + pω + q = (ω² − 4ω + 13)(ω² + bω + c) with b, c real, and expand the right-hand side term by term: ω²(ω² + bω + c) = ω⁴ + bω³ + cω² −4ω(ω² + bω + c) = −4ω³ − 4bω² − 4cω 13(ω² + bω + c) = 13ω² + 13bω + 13c Adding and grouping by power of ω: ω⁴ + (b − 4)ω³ + (c − 4b + 13)ω² + (13b − 4c)ω + 13c Step 4: Compare coefficients with the left-hand side. ω³ : b − 4 = −2, so b = 2 ω² : c − 4b + 13 = 10, so c − 8 + 13 = 10, giving c = 5 ω¹ : p = 13b − 4c = 13(2) − 4(5) = 26 − 20 = 6 ω⁰ : q = 13c = 13(5) = 65 Step 5: Find the remaining roots from the second factor, ω² + 2ω + 5 = 0. It has no real factorisation, so complete the square: (ω + 1)² − 1 + 5 = 0 (ω + 1)² = −4 ω + 1 = ±√(−4) = ±2i ω = −1 ± 2i So p = 6 and q = 65, and the roots other than 2 + 3i are 2 − 3i, −1 + 2i and −1 − 2i.