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Cambridge International AS & A Level · Topic 3.9
A-Level Maths 9709 / Topic 3.9 / AS & A Level

Non-real roots come in pairs.

Paper 3 only. If a polynomial has real coefficients, its non-real roots always appear with their conjugates. Give one complex root of a cubic or quartic and the rest follow; and every complex number has exactly two square roots that you can find exactly.

Topic 3.9All complex numbers lessons →
The key idea

For a polynomial with REAL coefficients, if \(a + bi\) is a root then so is \(a - bi\), and \((z - (a + bi))(z - (a - bi)) = z^{2} - 2az + (a^{2} + b^{2})\) is a real quadratic factor. To find \(\sqrt{p + qi}\), write \((x + iy)^{2} = p + qi\) and equate parts: \(x^{2} - y^{2} = p\), \(2xy = q\).

Re Im 12−1 2 + i 2 − i 2i −2i z⁴ − 4z³ + 9z² − 16z + 20 = 0 real axis: mirror line
Fig. 1: the four roots of z⁴ − 4z³ + 9z² − 16z + 20 = 0 are 2 ± i and ±2i; because the coefficients are real, the roots are symmetric in the real axis
Section 01

Conjugate pairs.

Take the conjugate of a whole polynomial equation with real coefficients: the coefficients do not change, but every \(z\) becomes \(z^{*}\). So if \(z\) solves it, \(z^{*}\) does too, and the roots sit symmetrically about the real axis (Fig. 1).

Worked example A cubic from one complex root
Show that \(1 + 2i\) is a root of \(z^{3} - z^{2} + 3z + 5 = 0\), and find the other roots.
  1. \((1 + 2i)^{2} = 1 + 4i - 4 = -3 + 4i\) and \((1 + 2i)^{3} = (1 + 2i)(-3 + 4i) = -3 + 4i - 6i - 8 = -11 - 2i\).
  2. Substitute: \((-11 - 2i) - (-3 + 4i) + 3(1 + 2i) + 5 = (-11 + 3 + 3 + 5) + (-2 - 4 + 6)i = 0\) ✓.
  3. Real coefficients, so \(1 - 2i\) is also a root.
  4. Quadratic factor: sum \(= 2\), product \(= 1 + 4 = 5\), giving \(z^{2} - 2z + 5\).
  5. Divide (or compare coefficients): \(z^{3} - z^{2} + 3z + 5 = (z^{2} - 2z + 5)(z + 1)\).
Answer: The roots are \(1 + 2i\), \(1 - 2i\) and \(-1\).
Worked example Finding unknown coefficients
Given that \(1 + 2i\) is a root of \(z^{3} + pz^{2} + qz + 5 = 0\), where \(p\) and \(q\) are real, find \(p\) and \(q\).
  1. Use the powers found above: \((-11 - 2i) + p(-3 + 4i) + q(1 + 2i) + 5 = 0\).
  2. Real parts: \(-6 - 3p + q = 0\). Imaginary parts: \(-2 + 4p + 2q = 0\).
  3. From the first, \(q = 6 + 3p\); then \(-2 + 4p + 12 + 6p = 0\), so \(p = -1\) and \(q = 3\).
Answer: \(p = -1\), \(q = 3\) (the cubic above).
Examiner trap

When you substitute a complex root, ALL the real terms go into one equation and ALL the imaginary terms into the other. A single combined equation like \(-6 - 3p + q - 2 + 4p + 2q = 0\) throws away information and cannot give two unknowns.

Section 02

Quartics and long division.

For a quartic, one complex root gives one quadratic factor. Divide it out (or compare coefficients) to leave another quadratic, then solve that.

Worked example A quartic
Given that \(2 + i\) is a root of \(z^{4} - 4z^{3} + 9z^{2} - 16z + 20 = 0\), find the other three roots.
  1. Conjugate pair: \(2 - i\) is also a root, giving the factor \(z^{2} - 4z + 5\) (sum 4, product \(4 + 1 = 5\)).
  2. Write \(z^{4} - 4z^{3} + 9z^{2} - 16z + 20 = (z^{2} - 4z + 5)(z^{2} + az + b)\).
  3. Constant terms: \(5b = 20\), so \(b = 4\). \(z^{3}\) terms: \(a - 4 = -4\), so \(a = 0\).
  4. Check the \(z\) terms: \(5a - 4b = -16\) ✓.
  5. \(z^{2} + 4 = 0\) gives \(z = \pm 2i\).
Answer: The other roots are \(2 - i\), \(2i\) and \(-2i\) (Fig. 1).
Your turn
Show that \(-1 + i\) is a root of \(z^{3} + 4z^{2} + 6z + 4 = 0\) and find the other roots.
Examiner trap

The quadratic factor is \(z^{2} - (\text{sum})z + \text{product}\). For roots \(2 \pm i\) that is \(z^{2} - 4z + 5\); writing \(z^{2} + 4z + 5\) (wrong sign) makes the division leave a remainder.

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\(3 - 2i\) is a root of a polynomial equation with real coefficients. Which number must also be a root?

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Which quadratic has roots \(1 \pm 3i\)?

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What is the least number of real roots a cubic equation with real coefficients can have?

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Which of these is a square root of \(2i\)?

Section 03

Square roots of a complex number.

Every non-zero complex number has two square roots, negatives of each other. Set \((x + iy)^{2}\) equal to the number, equate real and imaginary parts, and solve the resulting pair of equations; only real \(x\) are allowed, so one factor of the quartic in \(x\) is rejected.

Worked example Square roots in exact Cartesian form
Find the two square roots of \(-15 - 8i\).
  1. Let \((x + iy)^{2} = -15 - 8i\): \(x^{2} - y^{2} + 2xyi = -15 - 8i\).
  2. Real: \(x^{2} - y^{2} = -15\). Imaginary: \(2xy = -8\), so \(y = -\dfrac{4}{x}\).
  3. Substitute: \(x^{2} - \dfrac{16}{x^{2}} = -15\), so \(x^{4} + 15x^{2} - 16 = 0\).
  4. \((x^{2} + 16)(x^{2} - 1) = 0\). \(x\) is real, so \(x^{2} = 1\) and \(x = \pm 1\).
  5. \(x = 1\): \(y = -4\). \(x = -1\): \(y = 4\).
  6. Check: \((1 - 4i)^{2} = 1 - 8i + 16i^{2} = -15 - 8i\) ✓.
Answer: \(1 - 4i\) and \(-1 + 4i\), i.e. \(\pm(1 - 4i)\)
Worked example Using square roots in a quadratic
Solve \(z^{2} - 3z + (3 + i) = 0\).
  1. Discriminant: \(9 - 4(3 + i) = -3 - 4i\). (The coefficients are not all real, so expect no conjugate pair.)
  2. Square root of \(-3 - 4i\): \(x^{2} - y^{2} = -3\), \(xy = -2\) gives \(x^{4} + 3x^{2} - 4 = 0\), \(x^{2} = 1\), so \(\pm(1 - 2i)\).
  3. \(z = \dfrac{3 \pm (1 - 2i)}{2}\), so \(z = \dfrac{4 - 2i}{2}\) or \(z = \dfrac{2 + 2i}{2}\).
  4. Check \(1 + i\): \((1 + i)^{2} - 3(1 + i) + 3 + i = 2i - 3 - 3i + 3 + i = 0\) ✓.
Answer: \(z = 2 - i\) or \(z = 1 + i\)
Your turn
Find the square roots of \(7 - 24i\).
Examiner trap

Keep the signs linked: \(xy\) negative means \(x\) and \(y\) have OPPOSITE signs. Writing the roots as \(\pm 1 \pm 4i\) (four numbers) is wrong; there are exactly two.

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Given that \(2 - 3i\) is a root of \(z^{3} - 6z^{2} + 21z - 26 = 0\), find the other two roots.

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Given that \(1 + i\) is a root of \(z^{3} + pz + q = 0\), where \(p\) and \(q\) are real, find \(p\) and \(q\).

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Find the square roots of \(5 - 12i\).

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Given that \(1 - 2i\) is a root of \(z^{4} - 2z^{3} + 6z^{2} - 2z + 5 = 0\), find the other three roots.

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Go deeper · practice
Six original Cambridge-style questions
Building a quadratic from a complex root, exact square roots, cubics and quartics from one root, unknown coefficients, and roots plotted on an Argand diagram.