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Year 11 Specialist (Unit 1 & 2) Complex numbers

Multiplication and division of complex numbers

20 practice questions 0 video lessons Theory + worked examples

Learn multiplication and division of complex numbers for Year 11 Specialist Mathematics in Queensland (QCAA). Products expand like ordinary brackets with the single rule that i squared equals minus one, while division uses the complex conjugate to clear the denominator.

You will learn to multiply and square complex numbers, form the conjugate, divide and take reciprocals by multiplying by the conjugate, and find the modulus — the arithmetic that underpins polar form, Argand geometry and solving equations later in the course.

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Theory

Multiplying and dividing complex numbers extends ordinary algebra with the single rule \(i^2=-1\), in Year 11 Specialist Mathematics (QCAA, Queensland). You expand a product like any binomials and replace \(i^2\) with \(-1\); you divide by multiplying top and bottom by the conjugate of the denominator. This page shows both, with the conjugate, reciprocal and modulus, worked in full.

A complex number is written \(z=a+bi\), where \(a=\operatorname{Re}(z)\) is the real part, \(b=\operatorname{Im}(z)\) is the imaginary part, and \(i\) satisfies \(i^2=-1\).

To multiply, expand the brackets exactly as for real binomials, then replace every \(i^2\) with \(-1\) and collect the real and imaginary parts. For example \((a+bi)(c+di)\) becomes \((ac-bd)+(ad+bc)i\).

The complex conjugate of \(z=a+bi\) is \(\bar z=a-bi\): keep the real part, change the sign of the imaginary part. Its key property is that \(z\bar z=a^2+b^2\), which is always a real, non-negative number.

To divide, multiply the numerator and denominator by the conjugate of the denominator. This makes the denominator real \((a^2+b^2)\), so the quotient can be written back in the form \(a+bi\). The modulus \(\lvert z\rvert=\sqrt{a^2+b^2}\) is the distance of \(z\) from the origin, and \(\lvert z\rvert^2=z\bar z\).

A complex number and its conjugate on an Argand diagram The point z = 5 + 3i and its conjugate 5 - 3i are mirror images in the real (horizontal) axis: same real part, opposite imaginary part. Re Im z = 5 + 3i 5 − 3i 5 3 −3
The conjugate \(\bar z\) is the reflection of \(z\) in the real axis: same real part, opposite sign on the imaginary part.
Modulus of a complex number as a length The complex number z = 4 + 3i drawn from the origin. Its modulus is the length of the arrow, the hypotenuse of a right triangle with legs 4 and 3, so the modulus is 5. Re Im z = 4 + 3i 4 3 |z| = 5
The modulus \(\lvert z\rvert\) is the length of the arrow from the origin: for \(4+3i\), \(\sqrt{4^2+3^2}=5\).

Product of two complex numbers (expand, then use \(i^2=-1\)):

\[ (a+bi)(c+di) = (ac-bd) + (ad+bc)\,i \]
(a+bi)(c+di)=(ac−bd)+(ad+bc)i

Conjugate, and the real product with the conjugate:

\[ \bar z = a-bi, \qquad z\bar z = a^2+b^2 \]
z¯=a−bi,zz¯=a2+b2

Division and reciprocal (multiply by the conjugate of the denominator):

\[ \dfrac{z_1}{z_2} = \dfrac{z_1\,\bar z_2}{z_2\bar z_2}, \qquad \dfrac{1}{a+bi} = \dfrac{a-bi}{a^2+b^2} \]
1a+bi=a−bia2+b2

Modulus (distance from the origin):

\[ \lvert z\rvert = \sqrt{a^2+b^2}, \qquad \lvert z\rvert^2 = z\bar z \]
|z|=a2+b2
The dividing trick. To turn a quotient into \(a+bi\) form, multiply the numerator and denominator by the conjugate of the denominator. The denominator becomes the real number \(a^2+b^2\), so you can split into real and imaginary parts.

How to divide two complex numbers

  1. Write the conjugate of the denominator: for \(c+di\) it is \(c-di\).
  2. Multiply the numerator and denominator by that conjugate (this is multiplying by \(1\), so the value is unchanged).
  3. Expand and simplify: the denominator becomes the real number \(c^2+d^2\); in the numerator replace every \(i^2\) with \(-1\).
  4. Split into real and imaginary parts to give the answer as \(a+bi\), keeping fractions and surds exact.
Example 1 — Multiply two complex numbers
Expand and simplify \((3+2i)(4-i)\), giving your answer in the form \(a+bi\).
Solution

Expand the brackets, then replace \(i^2\) with \(-1\):

\((3+2i)(4-i)\)\(=\)\(12 - 3i + 8i - 2i^2\)
\(=\)\(12 + 5i - 2(-1)\)
\(=\)\(12 + 5i + 2\)
\(=\)\(14 + 5i\)

\((3+2i)(4-i) = 14 + 5i\).

Example 2 — Square a complex number
Expand and simplify \((2+3i)^2\).
Solution

Square the binomial, then use \(i^2=-1\):

\((2+3i)^2\)\(=\)\((2+3i)(2+3i)\)
\(=\)\(4 + 6i + 6i + 9i^2\)
\(=\)\(4 + 12i + 9(-1)\)
\(=\)\(4 + 12i - 9\)
\(=\)\(-5 + 12i\)

\((2+3i)^2 = -5 + 12i\).

Example 3 — Divide by the conjugate
Express \(\dfrac{4+7i}{2+i}\) in the form \(a+bi\).
Solution

Multiply numerator and denominator by the conjugate of the denominator, \(2-i\):

\(\dfrac{4+7i}{2+i}\)\(=\)\(\dfrac{4+7i}{2+i}\times\dfrac{2-i}{2-i}\)
\(=\)\(\dfrac{(4+7i)(2-i)}{2^2+1^2}\)
\(=\)\(\dfrac{8 - 4i + 14i - 7i^2}{5}\)
\(=\)\(\dfrac{8 + 10i + 7}{5}\)
\(=\)\(\dfrac{15 + 10i}{5}\)
\(=\)\(3 + 2i\)

\(\dfrac{4+7i}{2+i} = 3 + 2i\).

Example 4 — Reciprocal, conjugate product and modulus
For \(z=2-3i\), find \(\dfrac{1}{z}\), \(z\bar z\) and \(\lvert z\rvert\).
Solution

Reciprocal — multiply top and bottom by the conjugate \(2+3i\):

\(\dfrac{1}{2-3i}\)\(=\)\(\dfrac{1}{2-3i}\times\dfrac{2+3i}{2+3i}\)
\(=\)\(\dfrac{2+3i}{2^2+3^2}\)
\(=\)\(\dfrac{2+3i}{13}\)
\(=\)\(\dfrac{2}{13}+\dfrac{3}{13}i\)

Conjugate product \(z\bar z=a^2+b^2\), and the modulus \(\lvert z\rvert=\sqrt{a^2+b^2}\):

\(z\bar z\)\(=\)\(2^2 + 3^2\)
\(=\)\(13\)
\(\lvert z\rvert\)\(=\)\(\sqrt{2^2+3^2}\)
\(=\)\(\sqrt{13}\)

\(\dfrac{1}{z}=\dfrac{2}{13}+\dfrac{3}{13}i\), \(\;z\bar z=13\), \(\;\lvert z\rvert=\sqrt{13}\).

Common pitfalls

Forgetting that \(i^2=-1\). After expanding, every \(i^2\) must be replaced by \(-1\). Leaving it as \(+i^2\), or treating it as \(+1\), changes the real part.
Conjugating the wrong part. The conjugate of \(a+bi\) is \(a-bi\): only the imaginary part changes sign. Do not also negate the real part.
Multiplying only the numerator. When dividing, multiply both the numerator and the denominator by the conjugate — otherwise you change the value.
Turning exact answers into decimals. Keep fractions and surds exact: write \(\tfrac{2}{13}+\tfrac{3}{13}i\) and \(\sqrt{13}\), not rounded decimals.

Frequently asked questions

How do you multiply two complex numbers?

Expand the brackets like ordinary binomials, then replace every \(i^2\) with \(-1\) and collect the real and imaginary parts. For example \((3+2i)(4-i)=14+5i\).

How do you divide complex numbers?

Multiply the numerator and denominator by the conjugate of the denominator. The denominator becomes the real number \(a^2+b^2\), so you can write the answer as \(a+bi\).

What is the complex conjugate?

The conjugate of \(z=a+bi\) is \(\bar z=a-bi\): keep the real part and change the sign of the imaginary part. It is the reflection of \(z\) in the real axis.

Why is \(z\bar z\) always real?

Because \((a+bi)(a-bi)=a^2-b^2i^2=a^2+b^2\). The imaginary terms cancel, leaving a real, non-negative number equal to \(\lvert z\rvert^2\).

How do you find the reciprocal of a complex number?

Write \(\dfrac{1}{a+bi}\) and multiply top and bottom by \(a-bi\). This gives \(\dfrac{a-bi}{a^2+b^2}\), which you can split into \(a+bi\) form.

What is the modulus of a complex number?

The modulus of \(z=a+bi\) is \(\lvert z\rvert=\sqrt{a^2+b^2}\), its distance from the origin on the Argand diagram. It satisfies \(\lvert z\rvert^2=z\bar z\).

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