Topic 1.14 · teacher page · AA Higher Level

Find one root, find them all

Equally spaced on a circle, so the rest are rotations.

The one thing to do with the animation

Sweep n and ask what stays the same.

The roots are always the same distance from the origin and always equally spaced. So one root plus an angle of 2pi/n gives every other, and nobody needs to solve n separate equations.

The conjugate pairing is a reflection, visible as the symmetry about the horizontal axis. That is a far better reason to believe "complex roots come in pairs" than being told it.

The sum-to-zero check

With no x^(n-1) term the roots sum to zero, because they are symmetrically placed about the origin. It is free and it catches a wrong angle immediately.

Pair it with the quadratic checks: roots sum to -b/a and multiply to c/a. For 2 +/- 3i that is 4 and 13, both of which can be read straight off the equation.

The answers

1. Imaginary part of the other root-3; the conjugate is 2 - 3i.
2. Modulus of each cube root of 82.
3. Fourth roots, degrees apart360/4 = 90.

Where the marks go

1 markOne root found properly, then 2pi/n added for the rest.

1 markArguments given in the required range.

1 markThe conjugate pair stated explicitly for a real polynomial.

What each wrong answer tells you

They giveWhat it means
+3 (Q1)Repeated the same root; the conjugate flips the sign.
2 (Q1)Gave the real part, which is unchanged.
8 (Q2)Gave the number rather than the root's modulus.
2.667 (Q2)Divided by 3 instead of taking a cube root.
120 (Q3)Used three roots instead of four.

Other things they will say

"Do I solve n equations?" No. One, then rotate. The figure is the argument.

"Why must complex roots pair up?" Because the coefficients are real, so the imaginary parts have to cancel. On the diagram it is a mirror in the horizontal axis.

"What if the coefficients are not real?" Then they need not pair, and the course does not ask. Worth saying so the rule is remembered with its condition.

A possible order

 What is happening
1Sweep n. What stays the same?
2De Moivre, forwards and then backwards for roots.
3The cube roots of 8 in full, with the sum-to-zero check.
4Conjugate pairs in quadratics and cubics.
5A root question given in Cartesian form, converted and solved.

Two things not to say

Do not find each root separately. It is slower, it is error-prone, and it hides the structure.

Do not state the conjugate rule without "with real coefficients". The condition is the whole of it.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

  1. Solve the quadraticSolve z² − 4z + 13 = 0.
    The discriminant is −36, so z = 2 ± 3i.
  2. Sum and productState the sum and product of those roots and check against the coefficients.
    Sum 4, product 13, matching −b and c.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

  1. Modulus and argumentFind the modulus and argument of 2 + 3i.
    |z| = √13 = 3.606 and arg z = arctan(3/2) = 0.983 radians.
  2. Cube rootsFind the three cube roots of 8.
    All have modulus 2 with arguments 0, 2π/3 and 4π/3: so 2, −1 + i√3 and −1 − i√3.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

  1. Why they come in pairsExplain why a polynomial with real coefficients cannot have 2 + 3i as a root without also having 2 − 3i.
    Conjugating the whole equation leaves the real coefficients unchanged and sends the root to its conjugate, so the conjugate satisfies it too. This is why complex roots of real polynomials always pair up.
  2. Build the polynomialFind a real quadratic with roots 2 ± 3i, and say why the pair is needed.
    z² − 4z + 13. Using only 2 + 3i gives z − 2 − 3i, which has complex coefficients. The pair multiplies out to something real because the imaginary parts cancel.

Practicalities

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