Complex Roots of Quadratics
What are complex roots?
- A quadratic equation can either have two real roots (zeros), a repeated real root or no real roots
- This depends on the location of the graph of the quadratic with respect to the x-axis
- If a quadratic equation has no real roots we would previously have stated that it has no real solutions
- The quadratic equation will have a negative discriminant
- This means taking the square root of a negative number
- Complex numbers provide solutions for quadratic equations that have no real roots
How do we solve a quadratic equation when it has complex roots?
- If a quadratic equation takes the form ax2 + bx + c = 0 it can be solved by either using the quadratic formula or completing the square
- If a quadratic equation takes the form ax2 + b = 0 it can be solved by rearranging
- The property i = √-1 is used
- If the coefficients of the quadratic are real then the complex roots will occur in complex conjugate pairs
- If z = m + ni (n ≠ 0) is a root of a quadratic with real coefficients then z* = m - ni is also a root
- The real part of the solutions will have the same value as the x coordinate of the turning point on the graph of the quadratic
- When the coefficients of the quadratic equation are non-real, the solutions will not be complex conjugates
- To solve these you can use the quadratic formula
How do we factorise a quadratic equation if it has complex roots?
- If we are given a quadratic equation in the form az2 + bz + c = 0, where a, b, and c ∈ ℝ, a ≠ 0 we can use its complex roots to write it in factorised form
- Use the quadratic formula to find the two roots, z = p + qi and z* = p - qi
- This means that z – (p + qi) and z – (p – qi) must both be factors of the quadratic equation
- Therefore we can write az2 + bz + c = (z – (p + qi))( z – (p - qi))
- This can be rearranged into the form (z – p – qi)(z – p + qi)
How do we find a quadratic equation when given a complex root?
- If we are given a complex root in the form z = p + qi we can find the quadratic equation in the form az2 + bz + c = 0, where a, b, and c ∈ ℝ, a ≠ 0
- We know that the second root must be z* = p - qi
- This means that z – (p + qi) and z – (p – qi) must both be factors of the quadratic equation
- Therefore we can write az2 + bz + c = (z – (p + qi))( z – (p - qi))
- Rewriting this as ((z – p) – qi))((z – p) + qi)) makes expanding easier
- Expanding this gives the quadratic equation z2 – 2pz + (p2 + q2)
- a = 1
- b = -2p
- c = p2 + q2
- This demonstrates the important property (x – z)(x – z*) = x2 - 2Re(z)x + |z|2
Exam Tip
- 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 ]
Worked Example
a)
Solve the quadratic equation z2 - 2z + 5 = 0 and hence, factorise z2 - 2z + 5.
b)
Given that one root of a quadratic equation is z = 2 – 3i, find the quadratic equation in the form az2 + bz + c = 0, where a, b, and c ∈ ℝ, a ≠ 0.
Complex Roots of Polynomials
How many roots should a polynomial have?
- We know that every quadratic equation has two roots (not necessarily distinct or real)
- 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 can have either:
- 3 real roots
- Or 1 real root and a complex conjugate pair
- A quartic equation of the form will have one of the following cases for roots:
- 4 real roots
- 2 real and 2 nonreal (a complex conjugate pair)
- 4 nonreal (two complex conjugate pairs)
- When a real polynomial of any degree has one complex root it will always also have the complex conjugate as a root
How do we solve a cubic equation with complex roots?
- Steps to solve a cubic equation with complex roots
- If we are told that p + qi is a root, then we know p - qi is also a root
- This means that z – (p + qi) and z – (p – qi) must both be factors of the cubic equation
- Multiplying the above factors together gives us a quadratic factor of the form (Az2 + Bz + C)
- We need to find the third factor (z - α)
- Multiply the factors and equate to our original equation to get
- From there either
- Expand and compare coefficients to find
- Or use polynomial division to find the factor (z - α)
- Finally, write your three roots clearly
How do we solve a polynomial of any degree with complex roots?
- When asked to find the roots of any polynomial when we are given one, we use almost the same method as 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 we solve polynomial equations with unknown coefficients?
- Steps to find unknown variables in a given equation when given a root:
- Substitute the given root p + qi 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 the polynomial
How do we factorise a polynomial when given a complex root?
- If we are given a root of a polynomial of any degree in the form z = p + qi
- We know that the complex conjugate, z* = p – qi is another root
- We can write (z – (p + qi)) and ( z – (p - qi)) as two linear factors
- Or rearrange into one quadratic factor
- This can be multiplied out with another factor to find further factors of the polynomial
- For higher order polynomials more than one root may be given
- If the further given root is complex then its complex conjugate will also be a root
- This will allow you to find further factors
Exam Tip
- You can speed up multiplying two complex conjugate factors together by
- rewrite (z – (p + qi))(z – (p - qi)) as ((z – p) - qi))( (z – p) + qi))
- Then ((z – p) - qi))( (z – p) + qi)) = (z - p)2 - (qi)2 = (z - p)2 + q2
- If you are working on a calculator paper read the question carefully to see how much of the working needs to be shown but always remember to use your GDC to check your working where you can
Worked Example
Given that one root of a polynomial p(x) = z3 + z2 – 7z + 65 is 2 – 3i, find the other roots.