Solutions — Polynomial Equations with Complex Roots

Q1 — Solving Quadratics with Complex Roots

When the discriminant b²−4ac < 0, we use i = √(−1) to express the square root.

(a) x² + 4 = 0

x² = −4

x = ±√(−4) = ±√(4 × −1) = ±2i

x = 2i or x = −2i

(b) x² + 6x + 13 = 0

Discriminant: 6² − 4(1)(13) = 36 − 52 = −16

x = (−6 ± √(−16)) / 2 = (−6 ± 4i) / 2

x = −3 + 2i or x = −3 − 2i

Check: these are complex conjugates, as expected for a polynomial with real coefficients.

(c) x² − 4x + 8 = 0

Discriminant: (−4)² − 4(1)(8) = 16 − 32 = −16

x = (4 ± √(−16)) / 2 = (4 ± 4i) / 2

x = 2 + 2i or x = 2 − 2i

Q2 — Applying the Conjugate Root Theorem

The Conjugate Root Theorem states: if a polynomial has real coefficients and z = a + bi (b ≠ 0) is a root, then z̅ = a − bi must also be a root. We simply negate the imaginary part.

(a) Root: 3 + 4i → conjugate root: 3 − 4i

(b) Root: −1 + √2 i → conjugate root: −1 − √2 i

(c) Root: 5i = 0 + 5i → conjugate root: −5i

(Pure imaginary roots always come in pairs: ki and −ki.)

(d) Root: −3 − 7i → conjugate root: −3 + 7i

Q3 — Building a Quadratic from a Complex Root

For a complex root z = a + bi, the conjugate z̅ = a − bi is also a root. The corresponding monic quadratic factor is:

(x − z)(x − z̅) = x² − (z + z̅)x + z z̅ = x² − 2ax + (a² + b²)

(a) Root: 1 + 3i → a = 1, b = 3

Quadratic: x² − 2(1)x + (1² + 3²) = x² − 2x + 10

(b) Root: −2 + i → a = −2, b = 1

Quadratic: x² − 2(−2)x + ((−2)² + 1²) = x² + 4x + 5 → x² + 4x + 5

(c) Root: 4i → a = 0, b = 4

Quadratic: x² − 2(0)x + (0² + 4²) = x² + 16

Q4 — Factoring z⁴ − 1

Use the difference of two squares identity twice:

z⁴ − 1 = (z²)² − 1² = (z² − 1)(z² + 1)

= (z − 1)(z + 1)(z² + 1)

From (z − 1)(z + 1) = 0: z = 1 and z = −1 (both real roots).

From z² + 1 = 0: z² = −1 → z = ±i.

All four roots: z = 1, −1, i, −i

These four roots lie at the vertices of a square inscribed in the unit circle on the Argand diagram, equally spaced at 90° intervals — the fourth roots of unity.

Q5 — Cubic with One Complex Root Known

Given roots: 1 − 2i, −3, and (by conjugate root theorem) 1 + 2i.

Step 1: Form the quadratic factor from the complex conjugate pair z = 1 ± 2i (a = 1, b = 2):

(x − (1−2i))(x − (1+2i)) = x² − 2(1)x + (1² + 2²) = x² − 2x + 5

Step 2: Write the cubic using all three linear factors:

P(x) = (x + 3)(x² − 2x + 5)

Step 3: Expand:

= x(x² − 2x + 5) + 3(x² − 2x + 5)

= x³ − 2x² + 5x + 3x² − 6x + 15

= x³ + x² − x + 15

Verify by checking P(−3) = −27 + 9 + 3 + 15 = 0 ✓

Q6 — Solving a Quartic Using Factor Theorem

P(x) = x⁴ − 4x³ + 9x² − 20x + 20, with 2 + i given as a root.

Step 1: By the conjugate root theorem, 2 − i is also a root.

Step 2: Form the real quadratic factor from the pair 2 ± i (a = 2, b = 1):

(x − (2+i))(x − (2−i)) = x² − 4x + (4+1) = x² − 4x + 5

Step 3: Divide P(x) by (x² − 4x + 5) using polynomial long division.

Divide x⁴ − 4x³ + 9x² − 20x + 20 by x² − 4x + 5:

x⁴ ÷ x² = x². x²(x²−4x+5) = x⁴−4x³+5x². Subtract: 4x²−20x+20.

4x² ÷ x² = 4. 4(x²−4x+5) = 4x²−16x+20. Subtract: −4x+0 → 0. Wait: −20x − (−16x) = −4x; 20 − 20 = 0.

Remainder: −4x. This is non-zero; the quotient is x²+4 with remainder −4x+0. Let’s redo carefully.

After subtracting x²(x²−4x+5) from x⁴−4x³+9x²−20x+20:

Remainder: (9−5)x²−20x+20 = 4x²−20x+20.

4x² ÷ x² = 4. 4(x²−4x+5) = 4x²−16x+20.

Final remainder: ((−20)−(−16))x + (20−20) = −4x. Hmm, non-zero remainder means 2+i is not a root of this P(x). Let’s use a cleaner polynomial.

Using P(x) = x⁴ − 4x³ + 9x² − 16x + 20 (which does have 2+i as a root):

Dividing by x²−4x+5: first term x²(x²−4x+5) = x⁴−4x³+5x². Subtract: 4x²−16x+20.

4(x²−4x+5) = 4x²−16x+20. Subtract: 0. Quotient: x²+4.

Step 4: x² + 4 = 0 → x² = −4 → x = ±2i.

All roots: 2+i, 2−i, 2i, −2i

P(x) = (x²−4x+5)(x²+4)

Q7 — Finding a Polynomial from Its Roots

Given roots: x = 2 (real) and x = 1 + 3i (complex).

Step 1: By the conjugate root theorem, x = 1 − 3i is also a root.

Step 2: Form the quadratic factor from 1 ± 3i (a = 1, b = 3):

x² − 2(1)x + (1² + 3²) = x² − 2x + 10

Step 3: Multiply all linear factors together:

P(x) = (x − 2)(x² − 2x + 10)

Step 4: Expand:

= x(x²−2x+10) − 2(x²−2x+10)

= x³ − 2x² + 10x − 2x² + 4x − 20

= x³ − 4x² + 14x − 20

Verify: P(2) = 8 − 16 + 28 − 20 = 0 ✓

Q8 — Sum and Product of Complex Roots

For a monic quadratic x² + px + q with roots α and β:

α + β = −p and αβ = q

Given α + β = 4 and αβ = 13, the monic quadratic is:

x² − 4x + 13

Finding the roots using the quadratic formula:

Discriminant: 4² − 4(1)(13) = 16 − 52 = −36

x = (4 ± √(−36)) / 2 = (4 ± 6i) / 2 = 2 ± 3i

α = 2 + 3i and β = 2 − 3i (a complex conjugate pair, as required for real coefficients).

Verification:

α + β = (2+3i) + (2−3i) = 4 ✓

αβ = (2+3i)(2−3i) = 4 + 9 = 13 ✓

Q9 — Solving a Cubic Completely

P(x) = x³ − 3x² + 4x − 12, given that 2i is a root.

Step 1: By the conjugate root theorem, −2i is also a root.

Step 2: Form the quadratic factor from ±2i (a = 0, b = 2):

(x − 2i)(x + 2i) = x² + 4

Step 3: Perform polynomial long division of P(x) ÷ (x² + 4):

x³ ÷ x² = x. x(x²+4) = x³+4x. Subtract from P(x):

x³−3x²+4x−12 − (x³+4x) = −3x² + 0x − 12 = −3x² − 12

−3x² ÷ x² = −3. −3(x²+4) = −3x²−12. Subtract: 0.

Quotient: x − 3 (exact division, zero remainder ✓)

Step 4: P(x) = (x − 3)(x² + 4)

The third root comes from x − 3 = 0: x = 3.

All roots: x = 3, x = 2i, x = −2i

Over ℝ: P(x) = (x − 3)(x² + 4), where x²+4 is irreducible over ℝ.

Q10 — Quartic with Two Complex Conjugate Pairs

Given roots: 1 + i and 3i. Since coefficients are real, the conjugate root theorem gives us all four roots:

1 + i, 1 − i, 3i, −3i

Step 1: Form the quadratic factor from 1 ± i (a = 1, b = 1):

x² − 2(1)x + (1² + 1²) = x² − 2x + 2

Step 2: Form the quadratic factor from ±3i (a = 0, b = 3):

x² − 2(0)x + (0² + 3²) = x² + 9

Step 3: Multiply the two quadratic factors:

P(x) = (x² − 2x + 2)(x² + 9)

= x⁴ + 9x² − 2x³ − 18x + 2x² + 18

= x⁴ − 2x³ + 11x² − 18x + 18

Verification: P(1+i) should equal 0. The factored form (x²−2x+2)(x²+9) has 1±i as roots of the first factor, confirming P(1+i) = 0 ✓