Complex Roots of Polynomials (Cambridge International A-Level Mathematics)

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 z²+b=0 blank (a≠0) we can rearrange to solve
  • If in the form $az^{2}+bz+c=0(a\neq0)$ we can complete the square or use the quadratic formula

• We use the property $i=\sqrt{-1}$ 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 $z=m+ni(n\neq0)$ is a root of a quadratic with real coefficients then $z^{*}=m-ni$ 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 $z^{2}+bz+c=0$ if you are given a complex root in the form $m+ni$

  • We know that the complex conjugate $m-ni$ is another root,

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

  • Therefore z²+b z+c=left square bracket z−(m+n straight i) right square bracket left square bracket z−(m−n straight i) right square bracket

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

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

Worked Example

Exam Hint

• Once you have your final answers you can check your roots are correct by substituting your solutions back into the original equation.
• You should get 0 if correct! [Note: 0 is equivalent to $0+0i$]

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 $ax^{3}+bx^{2}+cx+d=0$ can have either:

    • 3 real roots
    • Or 1 real root and a complex conjugate pair

• A quartic equation of the form $ax^{4}+bx^{3}+cx^{2}+dx+e=0$ 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 $m+ni$ is a root, then we know $m-ni$ is also a root
  • This means that $( z−(m+n \mathrm{i} ) )$ and $(z-(m-ni))$ are factors of the cubic equation
  • Multiply the above factors together gives us a quadratic factor of the form $(Az^{2}+Bz+C)$
  • We need to find the third factor $(z-α)$
  • Multiply the factors and equate to our original equation to get

(A z²+B z+C)(z−α)=a x³+b x²+c x+d

• From there either

  • Expand and compare coefficients to find $α$
  • Or use polynomial division to find the factor (z−α)

• 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 $(z-α)$ and $(z-β)$

  • The unknown factors also form a quadratic factor $(z-α)(z-β)$

  • 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 $α$ and $β$

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 $f(z)=0$
  • 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

Worked Example

Exam Hint

• As with solving quadratic equations, we can substitute our solutions back into the original equation to check we get 0.