AS & A Level · Practice questions · Complex numbers: arithmetic in Cartesian form

Minus one, every time.

Six original Cambridge-style questions on complex numbers in Cartesian form: arithmetic, division by the conjugate, equating real and imaginary parts, and complex roots of quadratics.

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

Algebra as usual, then \(i^{2} = -1\).

Add/subtract: real parts together, imaginary parts together.
Multiply: expand the brackets, then replace \(i^{2}\) by \(-1\).
Divide: multiply top and bottom by the conjugate of the bottom.
Unknown \(z\): write \(z = x + iy\) (and \(z^{*} = x - iy\)), then equate real and imaginary parts.
Quadratics: a negative discriminant gives a conjugate pair \(p \pm qi\).

Recall

[3 marks]

Given \(z = 2 - 5i\) and \(w = -1 + 3i\), find (a) \(z + w\) (b) \(z - w\) (c) \(zw\).

(a) \(1 - 2i\) ✓
(b) \(3 - 8i\) ✓
(c) \((2 - 5i)(-1 + 3i) = -2 + 6i + 5i - 15i^{2} = 13 + 11i\). ✓

Calculation

[3 marks]

Showing all your working, express \(\dfrac{4 - 3i}{1 + 2i}\) in the form \(x + iy\).

Multiply by \(\dfrac{1 - 2i}{1 - 2i}\). ✓
Top: \(4 - 8i - 3i + 6i^{2} = -2 - 11i\). Bottom: \(1 + 4 = 5\). ✓
\(-\dfrac{2}{5} - \dfrac{11}{5}i\), i.e. \(-0.4 - 2.2i\). ✓

Calculation

[3 marks]

Solve \(z^{2} + 4z + 29 = 0\), and verify that the sum of the roots is \(-4\) and their product is 29.

\(z = \dfrac{-4 \pm \sqrt{16 - 116}}{2} = \dfrac{-4 \pm 10i}{2} = -2 \pm 5i\). ✓
Sum: \((-2 + 5i) + (-2 - 5i) = -4\). ✓
Product: \((-2)^{2} + 5^{2} = 4 + 25 = 29\). ✓

Calculation

[4 marks]

Find the complex number \(z\) satisfying \(2z - iz^{*} = 4 + i\).

\(z = x + iy\): \(2x + 2iy - i(x - iy) = (2x - y) + (2y - x)i\). ✓
Equate: \(2x - y = 4\) and \(2y - x = 1\). ✓
\(y = 2x - 4\), so \(4x - 8 - x = 1\), \(x = 3\), \(y = 2\). ✓
\(z = 3 + 2i\). ✓

Proof

[4 marks]

Let \(u = 1 + i\). Find \(u^{2}\) and \(u^{4}\) in the form \(x + iy\), and hence show that \(u^{8}\) is a real number, stating its value.

\(u^{2} = 1 + 2i + i^{2} = 2i\). ✓
\(u^{4} = (2i)^{2} = 4i^{2} = -4\). ✓
\(u^{8} = (u^{4})^{2} = (-4)^{2} = 16\), which is real. ✓

Analysis

[6 marks]

The complex number \(w\) satisfies \((1 + i)w = 5 - i\).

(a) Find \(w\) in the form \(x + iy\). (b) Find \(ww^{*}\). (c) Find the real number \(k\) for which \(w + \dfrac{k}{w}\) is real.

(a) \(w = \dfrac{5 - i}{1 + i} \times \dfrac{1 - i}{1 - i} = \dfrac{5 - 5i - i + i^{2}}{2} = \dfrac{4 - 6i}{2} = 2 - 3i\). ✓
(b) \(ww^{*} = 2^{2} + 3^{2} = 13\). ✓
(c) \(\dfrac{1}{w} = \dfrac{w^{*}}{ww^{*}} = \dfrac{2 + 3i}{13}\), so \(w + \dfrac{k}{w} = \left(2 + \dfrac{2k}{13}\right) + \left(-3 + \dfrac{3k}{13}\right)i\). ✓
Real when \(-3 + \dfrac{3k}{13} = 0\), so \(k = 13\). ✓

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.