B5 - Modulus and argument | AQA A-Level Further Mathematics

Can you locate a complex number using a distance and a direction instead of two coordinates? You will calculate its modulus and argument, handle the axes and zero carefully, and convert between Cartesian and modulus-argument forms. These descriptions prepare you to interpret multiplication geometrically.

Magnitude from Pythagoras

How far from the origin is $3+4\mathrm{i}$ ? Its horizontal and vertical displacements form the perpendicular sides of a right triangle, so its distance is $\sqrt{3^2+4^2}=5$ .

Modulus

The modulus of $z=x+\mathrm{i}y$ , written $|z|$ , is its distance from the origin on an equal-scale Argand diagram:

|z|=\sqrt{x^2+y^2}

It is nonnegative, and it is zero exactly when $z=0$ .

Coordinates can be negative; distances cannot. A point 3 units left and 4 up is also 5 units from the origin. Use the nonnegative square root, rather than $\pm5$ .

Worked example 1: a familiar triangle

Question. Find $|3+4\mathrm{i}|$ .

Plan. Use Pythagoras on the real and imaginary coefficients.

|3+4\mathrm{i}|=\sqrt{3^2+4^2}=\sqrt{25}=5

Answer. $5$ .

Check. The conjugate product is $(3+4\mathrm{i})(3-4\mathrm{i})=25=5^2$ .

Change only the sign of the real part. Squaring that coefficient removes its sign, so the distance should stay the same.

Worked example 2: a point to the left

Question. Find $|-3+4\mathrm{i}|$ .

Plan. Square the signed coefficients, then take the nonnegative root.

|-3+4\mathrm{i}|=\sqrt{(-3)^2+4^2}=\sqrt{25}=5

Answer. $5$ .

Check. Reflection in the imaginary axis takes $(3,4)$ to $(-3,4)$ and preserves distance from the origin, confirming Worked example 1's modulus.

The earlier conjugate identity $zz^*=x^2+y^2$ therefore has a geometric meaning:

Modulus and conjugation

For every complex $z$ ,

|z|^2=zz^*,\qquad |z^*|=|z|

The second equality follows because reflection in the real axis preserves distance. These identities are not separately supplied in the formula booklet: learn them.

For a real number, this agrees with ordinary absolute value: $|-7|=7$ . For a purely imaginary number, $|-7\mathrm{i}|=7$ too.

Quadrant-correct arguments in radians

A distance does not determine a point: many points are 5 units from the origin. You also need the direction of the arrow from the origin.

Argument

For $z\ne0$ , an argument is an angle from the positive real axis to the arrow representing $z$ . Anticlockwise angles are positive and clockwise angles negative. Angles differing by an integer multiple of $2\pi$ describe the same direction.

The principal argument, written $\arg z$ here, is the unique representative in

-\pi<\arg z\le\pi

unless a question specifies another interval. Use radians.

A half-turn is $\pi\,\mathrm{rad}=180^\circ$ . The abbreviation $\mathrm{rad}$ means radians; set your calculator to radians when evaluating these arguments.

For $x\ne0$ , $\tan\theta=y/x$ , but the inverse tangent alone cannot distinguish opposite directions. For example, $1+\mathrm{i}$ and $-1-\mathrm{i}$ both give $y/x=1$ . Locate the quadrant before choosing the angle.

For a point off the axes, calculate the positive reference angle $\alpha=\arctan|y/x|$ . The principal arguments are:

Quadrant	Signs of $(x,y)$	Principal argument
I	$(+,+)$	$\alpha$
II	$(-,+)$	$\pi-\alpha$
III	$(-,-)$	$-\pi+\alpha$
IV	$(+,-)$	$-\alpha$

[DIAGRAM: asset_name: Quadrants and principal arguments; asset_slug: aqa7367-principal-arguments; caption: The quadrant fixes the argument and the negative real axis uses positive pi; description: An equal-scale Argand diagram shows four rays through (1,square root of 3), (minus 1,square root of 3), (minus 1,minus square root of 3), and (1,minus square root of 3), labelled pi over 3, 2 pi over 3, minus 2 pi over 3, and minus pi over 3. Axis directions are labelled 0, pi over 2, pi, and minus pi over 2. A note says minus pi is excluded and pi is included.]
The quadrant fixes the argument and the negative real axis uses positive pi

Worked example 3: a quadrant I argument

Question. Find $\arg(1+\sqrt3\mathrm{i})$ exactly.

Plan. Locate the point, then use its reference triangle.

\begin{aligned} x&=1>0,\quad y=\sqrt3>0 && \text{quadrant I}\\ \alpha&=\arctan(\sqrt3)=\frac{\pi}{3} && \text{reference angle} \end{aligned}

Answer. $\arg(1+\sqrt3\mathrm{i})=\pi/3$ .

Check. Its modulus is 2, and $2\cos(\pi/3)=1$ , $2\sin(\pi/3)=\sqrt3$ . Both original coordinates are recovered.

Change only the real sign. The reference angle is unchanged, but the point moves into quadrant II.

Worked example 4: a quadrant II argument

Question. Find $\arg(-1+\sqrt3\mathrm{i})$ exactly.

Plan. Use $\pi-\alpha$ because the point is above and to the left.

\begin{aligned} \alpha&=\arctan\left|\frac{\sqrt3}{-1}\right|=\frac{\pi}{3}\\ \arg z&=\pi-\frac{\pi}{3}=\frac{2\pi}{3} && \text{quadrant II} \end{aligned}

Answer. $2\pi/3$ .

Check. With modulus 2, $2\cos(2\pi/3)=-1$ and $2\sin(2\pi/3)=\sqrt3$ . The calculator value $\arctan(-\sqrt3)=-\pi/3$ would put the point in the wrong quadrant.

On the axes, use the direction directly: positive real gives 0, positive imaginary gives $\pi/2$ , negative imaginary gives $-\pi/2$ , and negative real gives $\pi$ . The last uses the included endpoint: $-\pi$ represents the same direction but is outside this principal interval.

Conjugation reverses the direction angle modulo $2\pi$ . Usually $\arg(z^*)=-\arg z$ , but there is a boundary exception: for a negative real number, both $z$ and $z^*$ have principal argument $\pi$ , rather than $-\pi$ . Zero has no argument at all, as the next part explains.

Watch out

Wrong: $\arg(-1+\sqrt3\mathrm{i})=\arctan(-\sqrt3)=-\pi/3$ . The point is in quadrant II, so the principal argument is $2\pi/3$ . A ratio of coordinates loses quadrant information.

Try a point below the negative real axis.

Zero has no argument

The origin has a distance, zero, but no direction from itself. The arrow representing zero has zero length, so it cannot point at a particular angle.

For a nonzero number, dividing the coordinates by its positive modulus gives the direction components $\cos\theta=x/|z|$ and $\sin\theta=y/|z|$ . At zero these would require division by zero. The missing argument is a mathematical boundary, rather than a calculator setting you can fix.

You can write

0=0(\cos\theta+\mathrm{i}\sin\theta)

for every real $\theta$ . That equality does not select any direction. Consequently $\arg0$ is undefined, rather than zero or “all angles”. By contrast, $\arg5=0$ : a positive real number has a nonzero arrow pointing right.

Key idea

Always separate the two statements: $|0|=0$ is defined; $\arg0$ is undefined. Rules involving arguments require the relevant numbers to be nonzero.

Judge a conclusion that confuses a valid representation with a defined argument.

Cartesian and modulus-argument forms

A distance and direction locate a nonzero point by resolving its arrow horizontally and vertically. For instance, a length 4 at angle $\pi/3$ gives horizontal component 2 and vertical component $2\sqrt3$ .

Two descriptions of one complex number

For $z\ne0$ , let $r=|z|>0$ and let $\theta$ be an argument. Then

z=x+\mathrm{i}y=r(\cos\theta+\mathrm{i}\sin\theta),\qquad x=r\cos\theta,\quad y=r\sin\theta

The expression on the right is modulus-argument form. These conversion rules are not separately supplied in the formula booklet: learn them.

The equations come from the right-triangle ratios, extended to all quadrants by the signed sine and cosine functions. Cartesian to modulus-argument means calculating $r$ and a quadrant-correct $\theta$ . The reverse means evaluating $r\cos\theta$ and $r\sin\theta$ .

Worked example 5: converting an upper point and back

Question. Write $z=-2+2\sqrt3\mathrm{i}$ in modulus-argument form with a principal argument, then recover its Cartesian coordinates from that form.

Plan. Calculate the distance and direction, then resolve the resulting arrow.

\begin{aligned} r&=\sqrt{(-2)^2+(2\sqrt3)^2}=4 && \text{Pythagoras}\\ \theta&=\pi-\arctan(\sqrt3)=\frac{2\pi}{3} && \text{quadrant II}\\ x&=4\cos\frac{2\pi}{3}=-2 && \text{horizontal component}\\ y&=4\sin\frac{2\pi}{3}=2\sqrt3 && \text{vertical component} \end{aligned}

Answer. $z=4(\cos(2\pi/3)+\mathrm{i}\sin(2\pi/3))$ ; converting back gives $(-2,2\sqrt3)$ .

Check. The conjugate product is $(-2)^2+(2\sqrt3)^2=16=r^2$ ; the recovered point has the required negative real and positive imaginary signs.

Change only the imaginary sign. Reflection preserves the modulus and reverses this non-boundary principal argument.

Worked example 6: converting the reflected point and back

Question. Write $w=-2-2\sqrt3\mathrm{i}$ in modulus-argument form with a principal argument, then recover its Cartesian coordinates.

Plan. Keep the distance calculation, choose quadrant III, then resolve.

\begin{aligned} r&=\sqrt{4+12}=4\\ \theta&=-\pi+\frac{\pi}{3}=-\frac{2\pi}{3} && \text{quadrant III}\\ x&=4\cos\left(-\frac{2\pi}{3}\right)=-2\\ y&=4\sin\left(-\frac{2\pi}{3}\right)=-2\sqrt3 \end{aligned}

Answer. $w=4(\cos(-2\pi/3)+\mathrm{i}\sin(-2\pi/3))$ ; its coordinates are $(-2,-2\sqrt3)$ .

Check. This is the conjugate of Worked example 5's number: cosine is even and sine is odd, so the resolved imaginary component alone changes sign.

An angle outside the principal interval still gives a valid representation. For example, $4\pi/3$ and $-2\pi/3$ have the same sine and cosine. Add or subtract $2\pi$ when a principal argument is requested. Keep the radius nonnegative; a negative leading factor must be absorbed into a half-turn before it can be called a modulus.

Watch out

In $r(\cos\theta+\mathrm{i}\sin\theta)$ , the factor $r$ multiplies both components. Writing $r\cos\theta+\mathrm{i}\sin\theta$ generally gives a different number.

Start from a distance and direction this time.

Quick Recap

$|x+\mathrm{i}y|=\sqrt{x^2+y^2}$ is a nonnegative distance.
Choose arguments from the quadrant and the requested interval, including its endpoint convention.
Zero has modulus zero and no argument.
Use $x=r\cos\theta$ and $y=r\sin\theta$ to convert back.

Yes: a positive distance and a direction locate any nonzero complex number. At zero, the distance remains meaningful but the direction disappears.

Last Accessed

Core Pure

Mechanics

Statistics

Discrete Mathematics

B5 - Modulus and argument

Magnitude from Pythagoras

Quadrant-correct arguments in radians

Zero has no argument

Cartesian and modulus-argument forms

Quick Recap

Learn faster with Nova