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Number and Algebra - Polynomials over the Complex field (HL)

Grade 12IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Fundamental Theorem of Algebra states that every non-constant polynomial of degree nn with complex coefficients has exactly nn complex roots, counting multiplicity.

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Conjugate Root Theorem: If a polynomial P(z)P(z) has real coefficients, then any complex roots must occur in conjugate pairs. That is, if z=a+biz = a + bi is a root, then z∗=a−biz^* = a - bi is also a root.

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Factorization over C\mathbb{C}: Any polynomial P(z)=anzn+an−1zn−1+⋯+a0P(z) = a_n z^n + a_{n-1} z^{n-1} + \dots + a_0 can be factored completely into nn linear factors of the form P(z)=an(z−z1)(z−z2)…(z−zn)P(z) = a_n(z - z_1)(z - z_2)\dots(z - z_n), where zi∈Cz_i \in \mathbb{C}.

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Factorization over R\mathbb{R}: A polynomial with real coefficients can be factored into a product of linear factors and irreducible quadratic factors (quadratics with a negative discriminant).

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Vieta's Formulas: For a polynomial anzn+an−1zn−1+⋯+a0=0a_n z^n + a_{n-1} z^{n-1} + \dots + a_0 = 0, the sum of the roots is −an−1an-\frac{a_{n-1}}{a_n} and the product of the roots is (−1)na0an(-1)^n \frac{a_0}{a_n}.

📐Formulae

P(z)=an(z−c1)(z−c2)…(z−cn)P(z) = a_n(z - c_1)(z - c_2)\dots(z - c_n)

For az2+bz+c=0:z1+z2=−ba,z1z2=ca\text{For } az^2 + bz + c = 0: z_1 + z_2 = -\frac{b}{a}, \quad z_1 z_2 = \frac{c}{a}

For az3+bz2+cz+d=0:∑zi=−ba,∑zizj=ca,z1z2z3=−da\text{For } az^3 + bz^2 + cz + d = 0: \sum z_i = -\frac{b}{a}, \quad \sum z_i z_j = \frac{c}{a}, \quad z_1 z_2 z_3 = -\frac{d}{a}

If P(z)∈R[z] and P(α+βi)=0, then P(α−βi)=0\text{If } P(z) \in \mathbb{R}[z] \text{ and } P(\alpha + \beta i) = 0, \text{ then } P(\alpha - \beta i) = 0

💡Examples

Problem 1:

Given that z=2+iz = 2 + i is a root of the polynomial P(z)=z3−6z2+13z−10P(z) = z^3 - 6z^2 + 13z - 10, find all other roots.

Solution:

  1. Since the coefficients of P(z)P(z) are all real (1,−6,13,−101, -6, 13, -10), the complex roots must occur in conjugate pairs. Therefore, z=2−iz = 2 - i is also a root.
  2. The product of the two known factors is: (z−(2+i))(z−(2−i))=((z−2)−i)((z−2)+i)=(z−2)2−i2=z2−4z+4+1=z2−4z+5(z - (2+i))(z - (2-i)) = ((z-2) - i)((z-2) + i) = (z-2)^2 - i^2 = z^2 - 4z + 4 + 1 = z^2 - 4z + 5
  3. Use polynomial division or Vieta's formulas to find the third root z3z_3. Using the product of roots: z1z2z3=−da=−−101=10z_1 z_2 z_3 = -\frac{d}{a} = -\frac{-10}{1} = 10 (2+i)(2−i)z3=10  ⟹  5z3=10  ⟹  z3=2(2+i)(2-i)z_3 = 10 \implies 5z_3 = 10 \implies z_3 = 2
  4. The roots are z=2+i,z=2−i,z=2z = 2+i, z = 2-i, z = 2.

Explanation:

We utilized the Conjugate Root Theorem because the coefficients are real, then used Vieta's product of roots property to find the final real root efficiently.

Problem 2:

Find a polynomial P(z)P(z) of degree 2 with complex coefficients such that its roots are ii and 1−i1-i. Write the answer in the form z2+bz+cz^2 + bz + c.

Solution:

  1. Using the factored form: P(z)=(z−i)(z−(1−i))P(z) = (z - i)(z - (1-i))
  2. Expand the brackets: P(z)=z2−z(1−i)−zi+i(1−i)P(z) = z^2 - z(1-i) - zi + i(1-i)
  3. Simplify the terms: P(z)=z2−z+zi−zi+i−i2P(z) = z^2 - z + zi - zi + i - i^2 P(z)=z2−z+i+1P(z) = z^2 - z + i + 1
  4. Result: P(z)=z2−z+(1+i)P(z) = z^2 - z + (1+i).

Explanation:

Since the coefficients are not restricted to being real, the roots do not need to be conjugates. We simply construct the polynomial from its linear factors (z−α)(z - \alpha) and (z−β)(z - \beta).