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What are complex numbers, and how do you add, multiply, divide and represent them?

The imaginary unit, arithmetic of complex numbers, the complex conjugate, the Argand diagram, modulus and argument, and complex roots of real polynomial equations occurring in conjugate pairs.

A CCEA AS Further Maths answer on the imaginary unit, adding, multiplying and dividing complex numbers, the complex conjugate, the Argand diagram, modulus and argument, and why complex roots of real polynomials occur in conjugate pairs.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

CCEA wants you to work confidently with complex numbers: the imaginary unit ii, the four arithmetic operations, the complex conjugate, the Argand diagram, and modulus and argument. You must also know that a polynomial equation with real coefficients has any complex roots in conjugate pairs, and use this to build or solve quadratics and cubics.

The answer

The imaginary unit and the form of a complex number

This last fact, equating real and imaginary parts, is the single most useful tool for solving complex equations.

Arithmetic of complex numbers

Addition and subtraction act on the real and imaginary parts separately. Multiplication is ordinary expansion with i2=−1i^2 = -1:

The complex conjugate

The product zzˉz\bar{z} being real is exactly why multiplying by the conjugate clears ii from a denominator.

The Argand diagram, modulus and argument

A complex number z=a+biz = a + bi is plotted as the point (a,b)(a, b) on the Argand diagram, with the real axis horizontal and the imaginary axis vertical.

The modulus is the distance from the origin; the argument is the angle the line from the origin to zz makes with the positive real axis, measured anticlockwise.

Complex roots come in conjugate pairs

Examples in context

Example 1. Alternating-current circuits. Engineers represent voltages and currents as complex numbers, where the modulus gives the size of the signal and the argument gives the phase. Multiplying by a complex number with modulus 1 rotates the phasor without changing its size, which is exactly the Argand-diagram picture of an argument.

Example 2. Reconstructing a quadratic from one root. If a calculator reports a single complex root such as 1+2i1 + 2i for a quadratic with real coefficients, you immediately know 1−2i1 - 2i is the other root. The quadratic must be a multiple of z2−2z+5z^2 - 2z + 5, found from the sum 22 and product 55 of the conjugate pair.

Try this

Q1. Find (2+5i)−(4−3i)(2 + 5i) - (4 - 3i). [1 mark]

  • Cue. Subtract real and imaginary parts: (2−4)+(5−(−3))i=−2+8i(2 - 4) + (5 - (-3))i = -2 + 8i.

Q2. Find the modulus of z=−3+4iz = -3 + 4i. [2 marks]

  • Cue. ∣z∣=(−3)2+42=25=5|z| = \sqrt{(-3)^2 + 4^2} = \sqrt{25} = 5.

Q3. One root of a real quadratic is 5−i5 - i. Write down the quadratic. [3 marks]

  • Cue. Other root 5+i5 + i; sum 1010, product 2626; equation z2−10z+26=0z^2 - 10z + 26 = 0.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA AS 20216 marksThe complex numbers are z1=3+2iz_1 = 3 + 2i and z2=1−4iz_2 = 1 - 4i. Find z1z2z_1 z_2 and z1z2\dfrac{z_1}{z_2}, giving each in the form a+bia + bi.
Show worked answer →

For the product, expand and use i2=−1i^2 = -1:

z1z2=(3+2i)(1−4i)=3−12i+2i−8i2=3−10i+8=11−10i.z_1 z_2 = (3 + 2i)(1 - 4i) = 3 - 12i + 2i - 8i^2 = 3 - 10i + 8 = 11 - 10i.

For the quotient, multiply top and bottom by the conjugate of the denominator, z2‾=1+4i\overline{z_2} = 1 + 4i:

z1z2=(3+2i)(1+4i)(1−4i)(1+4i)=3+12i+2i+8i21+16=−5+14i17.\dfrac{z_1}{z_2} = \dfrac{(3 + 2i)(1 + 4i)}{(1 - 4i)(1 + 4i)} = \dfrac{3 + 12i + 2i + 8i^2}{1 + 16} = \dfrac{-5 + 14i}{17}.

So z1z2=−517+1417i.\dfrac{z_1}{z_2} = -\dfrac{5}{17} + \dfrac{14}{17}i.

Markers reward correct use of i2=−1i^2 = -1, multiplying by the conjugate to realise the denominator, and the final answers in a+bia + bi form.

CCEA AS 20195 marksGiven that 2−3i2 - 3i is a root of z2+pz+q=0z^2 + pz + q = 0 where pp and qq are real, find the values of pp and qq.
Show worked answer →

Because the coefficients are real, complex roots occur in conjugate pairs, so the other root is 2+3i2 + 3i.

The sum of the roots is (2−3i)+(2+3i)=4(2 - 3i) + (2 + 3i) = 4, and for z2+pz+q=0z^2 + pz + q = 0 the sum of the roots is −p-p, so −p=4-p = 4, giving p=−4p = -4.

The product of the roots is (2−3i)(2+3i)=4−9i2=4+9=13(2 - 3i)(2 + 3i) = 4 - 9i^2 = 4 + 9 = 13, and the product equals qq, so q=13q = 13.

Check: z2−4z+13=0z^2 - 4z + 13 = 0 has roots 4±16−522=2±3i\dfrac{4 \pm \sqrt{16 - 52}}{2} = 2 \pm 3i.

Markers reward stating the conjugate root, using the sum and product of roots, and the correct values.

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