Complex numbers: the two forms, conjugate, modulus and argument

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Further Pure Mathematics 1 (FP1) is the compulsory AS Level unit of IAL Further Mathematics, and its first topic is the complex number.

Complex numbers carried 43 of the 225 marks across the June 2024, January 2025 and June 2025 FP1 papers, about a fifth, and they appeared on all three.

This section sets up the language: what a complex number is, its two standard forms, and the words conjugate, modulus, argument, real part and imaginary part.

FP1 assumes P1 and P2, so you will use surds, exact trigonometric values, radians and the quadratic formula from those units without being reminded.

The idea: a number whose square is negative

No real number squares to $-1$ : every real square is zero or positive.

That is why $x^2 + 1 = 0$ has no real solutions.

Mathematicians in the sixteenth century, solving cubic equations, found their formulae passing through square roots of negative numbers on the way to perfectly real answers.

The fix was to allow a new number, written $\mathrm{i}$ , and agree on one rule:

\mathrm{i}^2 = -1.

Everything else follows from ordinary algebra with that one extra fact.

Then $\sqrt{-9} = \sqrt{9}\sqrt{-1} = 3\mathrm{i}$ , and $x^2 + 1 = 0$ has the two solutions $\mathrm{i}$ and $-\mathrm{i}$ .

A complex number combines a real part and an imaginary part.

You need to know complex numbers in two forms: $a + \mathrm{i}b$ and $r\cos\theta + \mathrm{i}\,r\sin\theta$ .

The papers write the first form as $a + b\mathrm{i}$ , and so will these notes: $a + \mathrm{i}b$ and $a + b\mathrm{i}$ are the same number.

A complex number is usually called $z$ , and $z = x + \mathrm{i}y$ is also common.

Powers of $\mathrm{i}$ cycle with period 4: $\mathrm{i}^1 = \mathrm{i}$ , $\mathrm{i}^2 = -1$ , $\mathrm{i}^3 = -\mathrm{i}$ , $\mathrm{i}^4 = 1$ , and then the pattern repeats.

So $\mathrm{i}^{10} = \mathrm{i}^8 \times \mathrm{i}^2 = -1$ .

Real part, imaginary part and equality

For $z = a + b\mathrm{i}$ with $a$ and $b$ real:

the real part is $\operatorname{Re} z = a$ ;

the imaginary part is $\operatorname{Im} z = b$ . Notice that $\operatorname{Im} z$ is the real number $b$ , not $b\mathrm{i}$ .

Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal.

One equation between complex numbers is therefore two equations between real numbers.

This is the most useful idea in the whole unit: it is how you find unknown real constants, and later how you compare coefficients.

Why must both parts match?

Suppose $a + b\mathrm{i} = c + d\mathrm{i}$ with $b \neq d$ .

Then $\mathrm{i} = \dfrac{a - c}{d - b}$ , a real number.

But no real number squares to $-1$ , so this is impossible; hence $b = d$ , and then $a = c$ .

The conjugate

The complex conjugate of $z = a + b\mathrm{i}$ is $z^* = a - b\mathrm{i}$ : the same real part with the sign of the imaginary part changed.

Two facts make it useful:

z + z^* = 2a = 2\operatorname{Re} z, \qquad zz^* = (a + b\mathrm{i})(a - b\mathrm{i}) = a^2 - b^2\mathrm{i}^2 = a^2 + b^2.

Both results are real.

The second one is how you divide by a complex number (next section) and why non-real roots of real equations come in pairs (later sections).

Modulus and argument

Picture $z = a + b\mathrm{i}$ as the point $(a, b)$ on a plane with a real axis across and an imaginary axis up.

This picture is the Argand diagram, the subject of a later section; here it gives two more numbers.

The modulus $|z|$ is the distance from the origin to the point: $|z| = \sqrt{a^2 + b^2}$ . It is never negative. Since $zz^* = a^2 + b^2$ , you also have $zz^* = |z|^2$ .

The argument $\arg z$ is the angle from the positive real axis to the line joining the origin to $z$ , measured anticlockwise as positive and clockwise as negative. The papers use the principal argument, in the interval $-\pi < \theta \leq \pi$ , in radians.

The diagram below shows $z = -3 + 2\mathrm{i}$ .

Its modulus is $r = \sqrt{9 + 4} = \sqrt{13}$ .

Argand diagram of z = −3 + 2i in the second quadrant, joined to the origin O by a line of length r. The argument θ is the angle turned anticlockwise from the positive real axis to that line; the smaller angle α sits between the line and the negative real axis, so θ = π − α.

The calculator gives $\arctan\frac{2}{3} = 0.588$ for the small angle $\alpha$ inside the right-angled triangle, but that is the angle with the negative real axis.

The argument is measured from the positive real axis, so $\theta = \pi - 0.588 = 2.55$ (3 s.f.).

The rule for every quadrant is: find the acute angle $\alpha = \arctan\left|\frac{b}{a}\right|$ , then place it.

Quadrant of $z$

Signs of $a$ , $b$

$\arg z$

First

$a > 0$ , $b > 0$

$\alpha$

Second

$a < 0$ , $b > 0$

$\pi - \alpha$

Third

$a < 0$ , $b < 0$

$-(\pi - \alpha)$

Fourth

$a > 0$ , $b < 0$

$-\alpha$

Numbers on the axes are special cases: a positive real number has argument 0, a negative real number $\pi$ , a positive multiple of $\mathrm{i}$ has argument $\frac{\pi}{2}$ , and a negative multiple $-\frac{\pi}{2}$ .

The number 0 has modulus 0 and no argument.

Modulus-argument form

From the right-angled triangle in the diagram, a point at distance $r$ from the origin in direction $\theta$ has real part $r\cos\theta$ and imaginary part $r\sin\theta$ .

So every non-zero complex number can be written as

z = r(\cos\theta + \mathrm{i}\sin\theta), \qquad r = |z| > 0, \quad \theta = \arg z.

This is the second form, $r\cos\theta + \mathrm{i}\,r\sin\theta$ , with $r$ taken out as a factor.

Converting between the two forms is routine:

From modulus-argument to $a + b\mathrm{i}$ : evaluate $a = r\cos\theta$ and $b = r\sin\theta$ , exactly where the angle allows.

From $a + b\mathrm{i}$ to modulus-argument: $r = \sqrt{a^2 + b^2}$ and $\theta$ from the table above.

Learn to recognise the form.

A number printed as $4\left(\cos\frac{5\pi}{6} + \mathrm{i}\sin\frac{5\pi}{6}\right)$ tells you at a glance that its modulus is 4 and its argument is $\frac{5\pi}{6}$ ; you do not need to multiply anything out to know that.

The June 2025 FP1 report noted that some candidates did not recognise the modulus-argument form when a question used it.

Two points of caution.

The number in front must be positive: $-2(\cos\theta + \mathrm{i}\sin\theta)$ is not in modulus-argument form, because a modulus is never negative.

And the angle in both brackets must be the same angle, with a plus sign between the terms.

Why this matters, and where it leads

Complex numbers began as a trick for solving equations, and in FP1 that is still their main job: every quadratic, cubic and quartic with real coefficients can be solved completely once complex roots are allowed.

The modulus-argument form is the start of a longer story.

In FP2 you meet $r\mathrm{e}^{\mathrm{i}\theta}$ and De Moivre's theorem, which turn powers and roots of complex numbers into simple arithmetic on $r$ and $\theta$ .

Engineers describe alternating current and signals with complex numbers, because the argument records a phase and the modulus a size.

Two tempting wrong ideas

"The argument is just $\arctan\frac{b}{a}$ ." It is tempting because that is the button sequence.

But $\arctan$ only returns angles between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ , so it cannot tell $-3 + 2\mathrm{i}$ from $3 - 2\mathrm{i}$ : both have $\frac{b}{a} = -\frac{2}{3}$ .

They point in opposite directions.

Always sketch the point first and decide the quadrant.

"The modulus makes the number positive, like $|x|$ for real $x$ ." For a real number, $|-7| = 7$ is a distance from 0 on a line, and the two ideas agree.

For a complex number the distance is measured on a plane, so $|3 - 4\mathrm{i}| = \sqrt{9 + 16} = 5$ .

Changing the signs to get $3 + 4\mathrm{i}$ is not a modulus at all: a modulus is always one non-negative real number.

The June 2024 FP1 report noted this confusion between the modulus and the absolute value.

How the marks work

B1: an independent mark for a correct fact or value (a conjugate, a modulus, a plotted point). M1: a correct method applied with your own values; a slip later does not lose it.

A1: a correct answer that needs its M mark. dM1 depends on the M mark before it (ddM1 on the two before). A1ft follows through your earlier value; A1cso needs a correct solution only.

A1*: the last mark of a "show that". The printed answer must be reached with no slip anywhere and at least one line before it.

awrt: any answer which rounds to the stated value. isw: ignore subsequent working. A misread that keeps the question's character costs two A or B marks in that part.

A quadratic earns its method mark by factorising, the formula with values in, or completing the square. Calculator roots count only once the quadratic itself is written down, and never under an all-working note.

Key terms

Complex number: a number $a + \mathrm{i}b$ , where $a$ and $b$ are real and $\mathrm{i}^2 = -1$ ; equivalently $r\cos\theta + \mathrm{i}\,r\sin\theta$ .

Real part, imaginary part: for $z = a + b\mathrm{i}$ , $\operatorname{Re} z = a$ and $\operatorname{Im} z = b$ (both real numbers).

Equality of complex numbers: $a + b\mathrm{i} = c + d\mathrm{i}$ if and only if $a = c$ and $b = d$ .

Conjugate: $z^* = a - b\mathrm{i}$ for $z = a + b\mathrm{i}$ ; $zz^* = a^2 + b^2 = |z|^2$ .

Modulus: $|z| = \sqrt{a^2 + b^2}$ , the distance of $z$ from the origin on an Argand diagram. Learn it (not in the booklet).

Argument: $\arg z$ , the angle from the positive real axis to the line from the origin to $z$ , taken in $-\pi < \arg z \leq \pi$ . Learn it (not in the booklet).

Worked example 1: real part, conjugate, modulus and argument

The complex number $z = -5 + 12\mathrm{i}$ .

(a) Write down $\operatorname{Im} z$ and $z^*$ . (2)

(b) Determine $|z|$ . (1)

1. (a) $\operatorname{Im} z = 12$ B1 (not $12\mathrm{i}$ )

2. $z^* = -5 - 12\mathrm{i}$ B1

3. (b) $|z| = \sqrt{(-5)^2 + 12^2} = \sqrt{169} = 13$ B1

4. (c) $z$ is in the second quadrant. $\alpha = \arctan\frac{12}{5} = 1.1760$ , so $\arg z = \pi - 1.1760$ M1 (correct quadrant used)

5. $\arg z = 1.97$ A1 (awrt 1.97)

Answer: (a) $12$ ; $-5 - 12\mathrm{i}$ (b) $13$ (c) $1.97$ .

A quick check: $1.97$ radians is about $113^\circ$ , just past a right angle, which matches a point a little to the left of the imaginary axis and well above the real axis.

Worked example 2: changing form both ways

(a) Express $w = 4\left(\cos\dfrac{2\pi}{3} + \mathrm{i}\sin\dfrac{2\pi}{3}\right)$ in the form $a + b\mathrm{i}$ , where $a$ and $b$ are exact. (2)

(b) Express $v = \sqrt{3} - \mathrm{i}$ in the form $r(\cos\theta + \mathrm{i}\sin\theta)$ , where $r > 0$ and $-\pi < \theta \leq \pi$ . (3)

1. (a) $\cos\frac{2\pi}{3} = -\frac{1}{2}$ and $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$ , so $w = 4\left(-\frac{1}{2}\right) + 4\left(\frac{\sqrt{3}}{2}\right)\mathrm{i}$ M1

2. $w = -2 + 2\sqrt{3}\,\mathrm{i}$ A1

3. (b) $r = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{4} = 2$ B1

4. Fourth quadrant: $\alpha = \arctan\frac{1}{\sqrt{3}} = \frac{\pi}{6}$ , so $\theta = -\frac{\pi}{6}$ M1

5. $v = 2\left(\cos\left(-\frac{\pi}{6}\right) + \mathrm{i}\sin\left(-\frac{\pi}{6}\right)\right)$ A1

Answer: (a) $w = -2 + 2\sqrt{3}\,\mathrm{i}$ (b) $v = 2\left(\cos\left(-\frac{\pi}{6}\right) + \mathrm{i}\sin\left(-\frac{\pi}{6}\right)\right)$ .

Check (b) by multiplying out: $2\cos\left(-\frac{\pi}{6}\right) = \sqrt{3}$ and $2\sin\left(-\frac{\pi}{6}\right) = -1$ .

Writing $\theta = \frac{11\pi}{6}$ gives the same point but lies outside the stated interval, so it would lose the A mark.

Worked example 3: equating real and imaginary parts

Given that $p$ and $q$ are real and $(2p + q) + (p - q)\mathrm{i} = 7 - \mathrm{i}$ , determine the value of $p$ and the value of $q$ .

(3)

1. Real parts: $2p + q = 7$ ; imaginary parts: $p - q = -1$ M1 (both equations formed)

2. Adding: $3p = 6$ , so $p = 2$ A1

3. $q = p + 1 = 3$ A1

Answer: $p = 2$ , $q = 3$ .

Check: $2p + q = 7$ and $p - q = -1$ , so the left side is $7 - \mathrm{i}$ .

One complex equation gave two real equations, which is exactly enough for two unknowns.

Worked example 4: a modulus-argument number with an unknown modulus (exam length)

The complex number $z = r\left(\cos\left(-\dfrac{3\pi}{4}\right) + \mathrm{i}\sin\left(-\dfrac{3\pi}{4}\right)\right)$ , where $r$ is a positive constant.

Given that $z + z^* = -6$ ,

(a) show that $r = 3\sqrt{2}$ . (3)

(b) Determine $z$ in the form $a + b\mathrm{i}$ , where $a$ and $b$ are integers. (1)

1. (a) $z = r\left(-\frac{\sqrt{2}}{2}\right) + r\left(-\frac{\sqrt{2}}{2}\right)\mathrm{i}$ , so $\operatorname{Re} z = -\frac{\sqrt{2}}{2}r$ M1 (exact values used)

2. $z + z^* = 2\operatorname{Re} z = -\sqrt{2}\,r$ , so $-\sqrt{2}\,r = -6$ M1

3. $r = \frac{6}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$ A1*

4. (b) $z = 3\sqrt{2}\left(-\frac{\sqrt{2}}{2}\right)(1 + \mathrm{i}) = -3 - 3\mathrm{i}$ B1

5. (c) $z^* = -3 + 3\mathrm{i}$ , the reflection of $z$ in the real axis, so $\arg z^* = \frac{3\pi}{4}$ B1

Answer: (a) shown (b) $z = -3 - 3\mathrm{i}$ (c) $\arg z^* = \frac{3\pi}{4}$ .

Conjugating changes the sign of the argument (here from $-\frac{3\pi}{4}$ to $\frac{3\pi}{4}$ ) and leaves the modulus alone, which is a useful check.

In (a), the line $\frac{6}{\sqrt{2}} = \frac{6\sqrt{2}}{2}$ is the step before the printed answer that the A1* needs.

Exam tip

Before you press $\arctan$ , write down which quadrant the number is in; the M mark for an argument goes to the correct adjustment ( $\pi - \alpha$ , $-\alpha$ , and so on), not to the calculator value.

Give the argument in radians unless the stem says otherwise: the June 2024 FP1 scheme gave M1 A0 to an answer in degrees.

Common mistakes

Writing $\operatorname{Im} z = b\mathrm{i}$ . The imaginary part is the real number $b$ . Fix: $\operatorname{Im}(4 - 7\mathrm{i}) = -7$ .

Using $\tan$ where $\arctan$ is meant, or giving degrees. The June 2024 report saw both. Fix: $\theta = \arctan\frac{b}{a}$ , adjusted for the quadrant, in radians.

Treating the modulus as "make it positive". Fix: $|a + b\mathrm{i}| = \sqrt{a^2 + b^2}$ , one non-negative real number; never change signs inside the number.

Missing the modulus-argument form. In $r(\cos\theta + \mathrm{i}\sin\theta)$ with $r > 0$ , the modulus is $r$ and the argument is $\theta$ ; read them off.

An argument outside $-\pi < \theta \leq \pi$ . Fix: subtract or add $2\pi$ ; $\frac{7\pi}{4}$ becomes $-\frac{\pi}{4}$ .

Check your understanding

1. The complex number $z = 7 - 24\mathrm{i}$ .
(a) Write down $\operatorname{Re} z$ and $z^*$ . (2)
(b) Determine $|z|$ . (1)
(c) Determine $\arg z$ , giving your answer in radians to 3 significant figures. (2) Total [5]

2. (a) Express $6\left(\cos\dfrac{5\pi}{6} + \mathrm{i}\sin\dfrac{5\pi}{6}\right)$ in the form $a + b\mathrm{i}$ , where $a$ and $b$ are exact. (2)
(b) Show that $-1 - \sqrt{3}\,\mathrm{i} = 2\left(\cos\left(-\dfrac{2\pi}{3}\right) + \mathrm{i}\sin\left(-\dfrac{2\pi}{3}\right)\right)$ . (3) Total [5]

3. A student writes " $\arg(-4 + 3\mathrm{i}) = \arctan\left(\dfrac{3}{-4}\right) = -0.644$ ".
(a) Show that the student's value cannot be correct. (2)
(b) Determine $\arg(-4 + 3\mathrm{i})$ , giving your answer in radians to 3 significant figures. (2) Total [4]

4. Given that $x$ and $y$ are real and $(x + 2y) + (3x - y)\mathrm{i} = 5 - 6\mathrm{i}$ ,
(a) determine the value of $x$ and the value of $y$ . (3) For these values, $z = x + y\mathrm{i}$ .
(b) Determine the exact value of $|z|$ . (1) Total [4]

5. The complex number $z = k\left(\cos\dfrac{\pi}{3} + \mathrm{i}\sin\dfrac{\pi}{3}\right)$ , where $k$ is a positive constant. Given that $\operatorname{Im}(z) - \operatorname{Re}(z) = 2\sqrt{3} - 2$ ,
(a) show that $k = 4$ . (3)
(b) Determine $zz^*$ . (1) Total [4]

Answers

1. (a) $\operatorname{Re} z = 7$ B1; $z^* = 7 + 24\mathrm{i}$ B1
(b) $|z| = \sqrt{49 + 576} = \sqrt{625} = 25$ B1
(c) Fourth quadrant: $\arg z = -\arctan\frac{24}{7}$ M1; $= -1.29$ A1 (awrt $-1.29$ )

2. (a) $6\left(-\frac{\sqrt{3}}{2}\right) + 6\left(\frac{1}{2}\right)\mathrm{i}$ M1; $= -3\sqrt{3} + 3\mathrm{i}$ A1
(b) $|-1 - \sqrt{3}\,\mathrm{i}| = \sqrt{1 + 3} = 2$ B1; third quadrant, $\alpha = \arctan\frac{\sqrt{3}}{1} = \frac{\pi}{3}$ , so $\arg = -\left(\pi - \frac{\pi}{3}\right)$ M1; $= -\frac{2\pi}{3}$ , so $-1 - \sqrt{3}\,\mathrm{i} = 2\left(\cos\left(-\frac{2\pi}{3}\right) + \mathrm{i}\sin\left(-\frac{2\pi}{3}\right)\right)$ A1*. (Checking by multiplying out the right-hand side, $2\cos\left(-\frac{2\pi}{3}\right) = -1$ and $2\sin\left(-\frac{2\pi}{3}\right) = -\sqrt{3}$ , is also accepted, with a concluding line.)

3. (a) $-4 + 3\mathrm{i}$ has a negative real part and a positive imaginary part, so it lies in the second quadrant M1; its argument is therefore between $\frac{\pi}{2}$ and $\pi$ , but $-0.644$ is negative (a fourth-quadrant angle), so the student's value cannot be correct A1*
(b) $\arg(-4 + 3\mathrm{i}) = \pi - \arctan\frac{3}{4}$ M1; $= 2.50$ A1 (awrt 2.50)

4. (a) $x + 2y = 5$ and $3x - y = -6$ M1; $x = -1$ A1; $y = 3$ A1
(b) $|-1 + 3\mathrm{i}| = \sqrt{10}$ B1ft

5. (a) $z = \frac{k}{2} + \frac{\sqrt{3}}{2}k\,\mathrm{i}$ M1; $\frac{\sqrt{3}}{2}k - \frac{k}{2} = 2\sqrt{3} - 2$ , so $\frac{k}{2}(\sqrt{3} - 1) = 2(\sqrt{3} - 1)$ M1; $\frac{k}{2} = 2$ , so $k = 4$ A1*
(b) $zz^* = |z|^2 = 16$ B1

Complex numbers: the two forms, conjugate, modulus and argument

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