Complex Roots of Polynomials (Cambridge (CIE) A Level Maths: Pure 3): Revision Note

Complex roots of quadratics

What are complex roots? 

• Complex numbers provide solutions for quadratic equations which have no real roots

• Complex roots occur when solving a quadratic with a negative discriminant

  • This leads to square rooting a negative number

How do I solve a quadratic equation with complex roots?

• We solve an equation with complex roots in the same way we solve any other quadratic equations

  • If in the form $a{z}^2+b=0 \left(a\ne 0\right)$ we can rearrange to solve

  • If in the form $a{z}^2+bz+c=0 (a\ne 0)$ we can complete the square or use the quadratic formula

• We use the property $\mathrm{i}=\sqrt{-1}$ along with a manipulation of surds

• $\sqrt{-a}=\sqrt{a\times -1}=\sqrt{a}\times \sqrt{-1}$

• When the coefficients of the quadratic equation are real, complex roots occur in complex conjugate pairs

  • If $z=m+n\mathrm{i}(\mathrm{n}\ne 0)$ is a root of a quadratic with real coefficients then $z^{*}=m-n\mathrm{i}$ is also a root

• When the coefficients of the quadratic equation are non-real, the solutions will not be complex conjugates

  • To solve these use the quadratic formula

How do I find a quadratic equation given one complex root?

• We can find the equation of the form ${\mathit{z}}^2+\mathit{b}\mathit{z}+\mathit{c}=0$ if you are given a complex root in the form $\mathit{m}+\mathit{n}\mathbf{i}$

  • We know that the complex conjugate $\mathit{m}\mathbf{-}\mathit{n}\mathbf{i}$ is another root, 

  • This means that $z-\left(m+n\mathrm{i}\right)$ and $z-\left(m-n\mathrm{i}\right)$are factors of the quadratic equation

  • Therefore $z^2+bz+c=[z-\left(m+n\mathrm{i}\right)][z-\left(m-n\mathrm{i}\right)]$

    • Writing this as the difference of two squares - $(\left(z-m\right)-n\mathrm{i})(\left(z-m\right)+n\mathrm{i})$ - will speed up expanding

  • Expanding and simplifying gives us a quadratic equation where $b$ and $c$ are real numbers

Complex Roots of cubics & quartics

How many roots should a polynomial have?

• We know from previous sections that every quadratic equation has two roots (not necessarily distinct)

• This is a particular case of a more general rule:

  • Every polynomial equation, with real coefficients, of degree n has n roots

  • The n roots are not necessarily all distinct and therefore we need to count any repeated roots that may occur individually

• From the above rule we can state the following:

  • A cubic equation of the form $a{x}^3+b{x}^2+cx+d=0$ can have either:

    • 3 real roots

    • Or 1 real root and a complex conjugate pair

• A quartic equation of the form $a{x}^4+b{x}^3+c{x}^2+dx+e=0$ will have the following cases for roots:

  • 4 real roots

  • 2 real and 2 non-real roots(a complex conjugate pair)

  • 4 non-real roots (two complex conjugate pairs)

Number of roots of a cubic function

Number of roots of a quartic function

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

• Steps to solve a cubic equation with complex roots

  • If we are told that $m+n\mathrm{i}$ is a root, then we know $m-n\mathrm{i}$ is also a root

  • This means that $(z-\left(m+n\mathrm{i}\right))$ and $(z-(m-ni\left)\right)$ are factors of the cubic equation

  • Multiply the above factors together gives us a quadratic factor of the form $(A{z}^2+Bz+C)$

  • We need to find the third factor $(z-\alpha )$

  • Multiply the factors and equate to our original equation to get

$\left(A{z}^2+Bz+C\right)\left(z-\alpha \right)=a{x}^3+b{x}^2+cx+d$

• From there either

  • Expand and compare coefficients to find $\alpha$

  • Or use polynomial division to find the factor $\left(z-\alpha \right)$

• Finally, write your three roots clearly

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

• When asked to find the roots of a quartic equation when we are given one, we use almost the same method as we did for a cubic equation

  • State the initial root and its conjugate and write their factors as a quadratic factor (as above) we will have two unknown roots to find, write these as factors $(z-\alpha )$ and $(z-\beta )$

  • The unknown factors also form a quadratic factor $(z-\alpha )(z-\beta )$

  • Then continue with the steps from above, either comparing coefficients or using polynomial division

    • If using polynomial division, then solve the quadratic factor you get to find the roots $\alpha$ and $\beta$

How do I find unknown coefficients of cubics and quartics?

• Steps to find unknown variables in a given equation when given a root:

  • Substitute the given root into the equation $\mathrm{f}\left(z\right)=0$

  • Expand and group together the real and imaginary parts (these expressions will contain our unknown values)

  • Solve as simultaneous equations to find the unknowns

  • Substitute the values into the original equation

  • From here continue using the previously described methods for finding other roots for either cubic/quartic equations