Complex Roots of Polynomials (DP IB Analysis & Approaches (AA): HL): Revision Note

Complex roots of quadratics

When does a quadratic have complex roots?

• The quadratic equation $a{z}^2+bz+c=0$ where $a,b,c\in \mathrm{ℝ}$ has complex roots if

  • $b^2-4ac<0$

    • the discriminant is negative

How do I solve a quadratic with complex roots?

• To solve a quadratic equation with complex roots, $a{z}^2+bz+c=0$

  • either use the quadratic formula

    • e.g. $\pm \sqrt{-9}=\pm 3\mathrm{i}$

  • or complete the square

How are complex roots related to complex conjugates?

• The two complex roots to the quadratic equation $a{z}^2+bz+c=0$ where $a,b,c\in \mathrm{ℝ}$ are complex conjugate pairs

  • i.e. if $z=p+q\mathrm{i}$ is a root, then $z^{*}=p-q\mathrm{i}$ is the other root

• This is not true if any of the coefficients $a$, $b$ or $c$ are complex

How do I factorise a quadratic expression using complex roots?

• If a quadratic expression $a{z}^2+bz+c$ has a negative discriminant ($b^2-4ac<0$), then it can factorised as follows:

  • set the expression equal to zero

    • $a{z}^2+bz+c=0$

  • solve this equation to find the complex roots

    • $z_1=p+q\mathrm{i}$ and $z_2={z_1}^{*}=p-q\mathrm{i}$

  • rewrite $a{z}^2+bz+c$ in the factorised form $a\left(z-z_1\right)\left(z-z_2\right)$

    • You could expand inside each bracket

    • $a\left(z-p-q\mathrm{i}\right)\left(z-p+q\mathrm{i}\right)$

How do I find a quadratic equation when given its root?

• If you are given that $z_1=p+q\mathrm{i}$ is a root of a quadratic equation

  • then $z_2={z_1}^{*}=p-q\mathrm{i}$ is the other root

  • and $\left(z-z_1\right)\left(z-z_2\right)=0$ is the factorised equation

• To find the quadratic equation $z^2+bz+c=0$

  • expand $\left(z-z_1\right)\left(z-z_2\right)=0$

    • e.g. expanding $\left(z-5-6i\right)\left(z-5+6i\right)$

Complex roots of polynomials

How many roots does a polynomial equation have?

• The polynomial equation $a_n{z}^n+a_{n-1}{z}^{n-1}+...+a_2{z}^2+a_{1}z+a_0=0$

  • of degree $n$

  • where all coefficients are real

  • has $n$ roots

    • counting repeating roots individually

  • where not all roots have to be real

• Any complex roots must occur in conjugate pairs

  • e.g. a cubic equation can have

    • 3 real roots

    • or 1 real root and a complex conjugate pair

How do I solve a cubic equation given one real root?

• If given a real root, $z_1=k$, to the cubic equation $z^3+b{z}^2+cz+d=0$

  • Use the Factor theorem to find a linear factor

    • $\left(z-k\right)$

  • Then use polynomial division to divide $z^3+b{z}^2+cz+d$ by $\left(z-k\right)$ to find the quadratic factor

    • Solving this quadratic gives the other two roots

How do I solve a cubic equation given one complex root?

• If you are given that $z_1=p+q\mathrm{i}$ is a root of the cubic equation $z^3+b{z}^2+cz+d=0$

  • then the second root is $z_2={z_1}^{*}=p-q\mathrm{i}$

    • the complex conjugate

  • and the third root is real, which you can find using polynomial division

    • e.g. write $\left(z-z_1\right)\left(z-z_2\right)$ as the quadratic factor

    • Expand it into the form $z^2+Bz+C$

    • then divide $z^3+b{z}^2+cz+d$ by $z^2+Bz+C$ to get a linear factor which can be solved

How do I solve polynomial equations of higher degrees?

• The process of solving polynomial equations of degrees $n\ge 4$ follows the same method as cubics

  • If you are given one complex root, $z_1=p+q\mathrm{i}$

    • then $z_2={z_1}^{*}=p-q\mathrm{i}$ is the second root

    • so $\left(z-z_1\right)\left(z-z_2\right)$ is a quadratic factor

    • Expand this to $z^2+Bz+C$

    • then dividing the polynomial by $z^2+Bz+C$, etc

  • You may be given two complex roots

    • e.g. $1+\mathrm{i}$ and $4+5\mathrm{i}$

    • so $1-\mathrm{i}$ and $4-5\mathrm{i}$ are also roots

    • and $\left(z-1-\mathrm{i}\right)\left(z-1+\mathrm{i}\right)$ and $\left(z-4-5\mathrm{i}\right)\left(z-4+5\mathrm{i}\right)$ are both quadratic factors, etc