AS & A Level · Practice questions · Complex roots and square roots

Find one, get its partner.

Six original Cambridge-style questions on complex roots of polynomials with real coefficients and on square roots of complex numbers in exact Cartesian form.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Pair it, factor it, finish it.

Pair: real coefficients mean \(a + bi\) brings \(a - bi\).
Factor: \(z^{2} - 2az + (a^{2} + b^{2})\).
Finish: divide or compare coefficients for the remaining factor, then solve.
Unknown coefficients: substitute the root and equate real and imaginary parts.
Square roots: \(x^{2} - y^{2} = p\), \(2xy = q\); reject \(x^{2} < 0\); signs of \(x, y\) linked by \(q\).

Recall

[2 marks]

One root of a quadratic equation with real coefficients is \(4 - i\). Find the equation in the form \(z^{2} + bz + c = 0\).

The other root is \(4 + i\). Sum \(= 8\), product \(= 16 + 1 = 17\). ✓
\(z^{2} - 8z + 17 = 0\). ✓

Calculation

[4 marks]

Find the two square roots of \(21 + 20i\), giving your answers in the form \(x + iy\) with \(x\) and \(y\) exact.

\(x^{2} - y^{2} = 21\) and \(2xy = 20\), so \(y = \dfrac{10}{x}\). ✓
\(x^{2} - \dfrac{100}{x^{2}} = 21\), so \(x^{4} - 21x^{2} - 100 = 0\), \((x^{2} - 25)(x^{2} + 4) = 0\). ✓
\(x^{2} = 25\): \(x = 5\), \(y = 2\) or \(x = -5\), \(y = -2\). ✓
Square roots \(5 + 2i\) and \(-5 - 2i\). ✓

Calculation

[5 marks]

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

\((-1 + i)^{2} = 1 - 2i - 1 = -2i\); \((-1 + i)^{3} = -2i(-1 + i) = 2i + 2\). ✓
\((2 + 2i) + 4(-2i) + 6(-1 + i) + 4 = (2 - 6 + 4) + (2 - 8 + 6)i = 0\) ✓. ✓
\(-1 - i\) is also a root; factor \(z^{2} + 2z + 2\). ✓
\(z^{3} + 4z^{2} + 6z + 4 = (z^{2} + 2z + 2)(z + 2)\), so the roots are \(-1 - i\) and \(-2\). ✓

Analysis

[6 marks]

Given that \(2 + i\) is a root of \(z^{3} + pz^{2} + qz - 10 = 0\), where \(p\) and \(q\) are real, find \(p\) and \(q\) and the other two roots.

\((2 + i)^{2} = 3 + 4i\), \((2 + i)^{3} = (2 + i)(3 + 4i) = 2 + 11i\). ✓
\((2 + 11i) + p(3 + 4i) + q(2 + i) - 10 = 0\): real \(3p + 2q - 8 = 0\), imaginary \(4p + q + 11 = 0\). ✓
\(q = -11 - 4p\), so \(3p - 22 - 8p - 8 = 0\), \(p = -6\), \(q = 13\). ✓
\(z^{3} - 6z^{2} + 13z - 10 = (z^{2} - 4z + 5)(z - 2)\): the other roots are \(2 - i\) and \(2\). ✓

Calculation

[5 marks]

(a) Find the square roots of \(-3 - 4i\). (b) Hence solve \(z^{2} - 3z + 3 + i = 0\).

(a) \(x^{2} - y^{2} = -3\), \(xy = -2\): \(x^{4} + 3x^{2} - 4 = 0\), \(x^{2} = 1\). Roots \(\pm(1 - 2i)\). ✓
(b) Discriminant \(9 - 4(3 + i) = -3 - 4i\). ✓
\(z = \dfrac{3 \pm (1 - 2i)}{2}\): \(z = 2 - i\) or \(z = 1 + i\). ✓

Analysis

[7 marks]

(a) Given that \(2i\) is a root of \(z^{4} + 2z^{3} + 6z^{2} + 8z + 8 = 0\), find all four roots. (b) The roots are represented on an Argand diagram. Find the area of the quadrilateral with these four points as vertices.

(a) \(-2i\) is also a root, so \(z^{2} + 4\) is a factor. ✓
\(z^{4} + 2z^{3} + 6z^{2} + 8z + 8 = (z^{2} + 4)(z^{2} + 2z + 2)\) (constants \(4 \times 2 = 8\) ✓, \(z^{3}\): 2 ✓). ✓
\(z^{2} + 2z + 2 = 0\) gives \(z = -1 \pm i\). Roots: \(\pm 2i\), \(-1 \pm i\). ✓
(b) Vertices \((0, 2)\), \((-1, 1)\), \((-1, -1)\), \((0, -2)\): a trapezium with parallel vertical sides 4 and 2, a distance 1 apart. ✓
Area \(= \tfrac{1}{2}(4 + 2) \times 1 = 3\) square units. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.