Pj6–Pj8 - Modulus-argument form | OCR A-Level Further Mathematics B (MEI)

Describe a complex number by its size and direction, then use those two pieces of information to multiply and divide. Pj6 supplies exact radian angles, Pj7 connects Cartesian and modulus-argument forms, and Pj8 turns a long expansion into a short calculation with lengths and angles.

Size, direction and the right quadrant

Can two complex numbers each have a negative real part, yet their product be positive and real? Try $(-1 + \sqrt3\,i)(-1 - \sqrt3\,i)$ : the product is $4$ . Size and direction will explain this without expanding the brackets.

For $1 + \sqrt3\,i$ , a right-angled triangle has horizontal side $1$ , vertical side $\sqrt3$ and hypotenuse $2$ . Its direction is $\frac{\pi}{3}$ , since $\cos\frac{\pi}{3} = \frac12$ and $\sin\frac{\pi}{3} = \frac{\sqrt3}{2}$ . The same length with a negative horizontal side points in a different direction.

You already know

For $z = x + yi$ , where $x$ and $y$ are real, $z^* = x - yi$ . Exact trigonometric values and the signs of sine and cosine identify an angle's quadrant. A full turn is $2\pi$ radians.

Here are the reference angles you will use repeatedly. Their values come from an equilateral triangle split in half and an isosceles right-angled triangle. Extend them to other quadrants using signs, rather than memorising a separate table for every angle.

$\alpha$	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$
$\cos\alpha$	$1$	$\frac{\sqrt3}{2}$	$\frac{\sqrt2}{2}$	$\frac12$	$0$
$\sin\alpha$	$0$	$\frac12$	$\frac{\sqrt2}{2}$	$\frac{\sqrt3}{2}$	$1$

These exact values are not listed in the formula booklet: learn them. In this topic, give arguments in radians.

[DIAGRAM: asset_name: Equal lengths in different quadrants; asset_slug: b003_007_modarg_quadrants; caption: The numbers 1 plus square root 3 i and minus 1 plus square root 3 i both have modulus 2 but different arguments; description: Two equal-scale Argand panels show the vectors from the origin to (1, square root 3) and (-1, square root 3). Right triangles label the signed horizontal components, the positive vertical component and length 2. Counterclockwise arcs label pi over 3 and 2 pi over 3 respectively.]
The numbers 1 plus square root 3 i and minus 1 plus square root 3 i both have modulus 2 but different arguments

Modulus and principal argument

The modulus $|z|$ is the non-negative distance of $z$ from the origin. For $z \ne 0$ , the principal argument $\arg z$ is its direction measured anticlockwise from the positive real axis, chosen in the interval

-\pi < \arg z \le \pi

For example, $\arg(-2) = \pi$ , not $-\pi$ . The number $0$ has modulus $0$ , but its argument is undefined.

Any angle differing by a whole number of turns gives the same direction. The principal interval selects one of them. Negative angles turn clockwise: the direction $\frac{5\pi}{4}$ has principal value $-\frac{3\pi}{4}$ .

Finding modulus and argument

For $z = x + yi \ne 0$ , put $r = |z| > 0$ and $\theta = \arg z$ . Then

\begin{aligned} r &= \sqrt{x^2 + y^2} \\ \cos\theta &= \frac{x}{r}, \qquad \sin\theta = \frac{y}{r} \\ zz^* &= (x + yi)(x - yi) = x^2 + y^2 = |z|^2 \end{aligned}

Use the signs of both coordinates to choose $\theta$ . The relation $\tan\theta = y/x$ , when $x \ne 0$ , does not identify the quadrant by itself.

Not in the formula booklet: learn it.

Find the length first, then choose the angle from its sine and cosine. This avoids a calculator's inverse tangent returning a direction on the opposite side of the origin.

Worked example 1: a first-quadrant number

Question. Find the modulus and principal argument of $z = 1 + \sqrt3\,i$ .

Plan. Calculate the distance, then identify the angle from both trigonometric ratios.

\begin{aligned} |z| &= \sqrt{1^2 + (\sqrt3)^2} = 2 && \text{Pythagoras} \\ \cos\theta &= \frac12, \qquad \sin\theta = \frac{\sqrt3}{2} && \text{divide coordinates by } 2 \\ \theta &= \frac{\pi}{3} && \text{both coordinates positive} \end{aligned}

Why this step? Both ratios are positive, so the direction must be in the first quadrant.

Answer. $|z| = 2$ and $\arg z = \frac{\pi}{3}$ .

Check. $zz^* = (1 + \sqrt3\,i)(1 - \sqrt3\,i) = 4 = |z|^2$ .

The second example changes only the sign of the real part. The distance and reference angle stay the same; the quadrant changes.

Worked example 2: the same length in the second quadrant

Question. Find the modulus and principal argument of $w = -1 + \sqrt3\,i$ .

Plan. Calculate the modulus, then use the negative cosine and positive sine to select the angle.

\begin{aligned} |w| &= \sqrt{(-1)^2 + (\sqrt3)^2} = 2 && \text{same squared coordinates} \\ \cos\phi &= -\frac12, \qquad \sin\phi = \frac{\sqrt3}{2} && \text{second quadrant} \\ \phi &= \pi - \frac{\pi}{3} = \frac{2\pi}{3} && \text{use the reference angle} \end{aligned}

Why this step? $\arctan(-\sqrt3) = -\frac{\pi}{3}$ points into the fourth quadrant and has the wrong signs for this number.

Answer. $|w| = 2$ and $\arg w = \frac{2\pi}{3}$ .

Check. $2\cos\frac{2\pi}{3} = -1$ and $2\sin\frac{2\pi}{3} = \sqrt3$ , recovering both parts.

The negative real axis is the one place where the choice of endpoint in the principal interval matters.

Watch out

Wrong: $\arg(-3) = 0$ , because its imaginary part is zero. Right: $\arg(-3) = \pi$ , because it lies on the negative real axis. Also, $|-3| = 3$ : a modulus is never negative.

Now change the sign of the imaginary part as well, and express the direction in the required interval.

Moving between the two forms

A point with length $r$ and direction $\theta$ has horizontal component $r\cos\theta$ and vertical component $r\sin\theta$ . This gives a second way to write exactly the same complex number.

Modulus-argument form

For $z \ne 0$ , with $r = |z| > 0$ ,

z = r(\cos\theta + i\sin\theta)

where $x = r\cos\theta$ and $y = r\sin\theta$ . Use $\theta = \arg z$ when a principal argument is required; angles differing by $2k\pi$ , for integer $k$ , represent the same number.

Not in the formula booklet: learn it.

The brackets matter: $r$ multiplies both components. For example, the previous part gives $-1 + \sqrt3\,i = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$ . To go in the other direction, evaluate the sine and cosine exactly.

Worked example 3: converting to Cartesian form

Question. Express $4\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$ in the form $x + yi$ .

Plan. Substitute the two exact trigonometric values, then multiply both terms by $4$ .

\begin{aligned} z &= 4\left(\frac{\sqrt3}{2} + \frac12 i\right) && \text{exact values at } \pi/6 \\ &= 2\sqrt3 + 2i && \text{multiply both parts} \end{aligned}

Why this step? The factor $4$ is the full length; each coordinate is the corresponding fraction of that length.

Answer. $z = 2\sqrt3 + 2i$ .

Check. $|z| = \sqrt{12 + 4} = 4$ , and both parts are positive, as the angle requires.

Change only the angle to $\frac{5\pi}{6}$ . The horizontal component changes sign, while the vertical component stays the same.

Worked example 4: converting a second-quadrant direction

Question. Express $4\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)$ in the form $x + yi$ .

Plan. Use the reference angle $\frac{\pi}{6}$ , applying the signs of the second quadrant.

\begin{aligned} z &= 4\left(-\frac{\sqrt3}{2} + \frac12 i\right) && \text{cosine negative, sine positive} \\ &= -2\sqrt3 + 2i && \text{multiply by } 4 \end{aligned}

Why this step? $\frac{5\pi}{6}$ is $\frac{\pi}{6}$ short of the negative real axis, rather than $\frac{\pi}{6}$ above the positive real axis.

Answer. $z = -2\sqrt3 + 2i$ .

Check. $|z| = 4$ , and $\operatorname{Re}(z) < 0$ , $\operatorname{Im}(z) > 0$ , matching its quadrant.

An exact-form answer should preserve exact lengths and angles, rather than replace them by rounded decimals.

Exam tip

“Express in exact modulus-argument form” needs the complete expression $r(\cos\theta + i\sin\theta)$ , with the same angle in both places. Giving only $r$ and $\theta$ , omitting $i$ , or using a different requested form loses credit.

The next number has a clockwise direction. Use the signs of sine and cosine before multiplying.

Watch out

Wrong: $4(-\sqrt3/2+i/2)=-2\sqrt3+i/2$ . The outside factor multiplies both components. The correct result is $-2\sqrt3+2i$ .

Apply that same distribution to both coordinates below.

Multiply lengths and add directions

There is a surprisingly neat multiplication rule hiding inside the brackets. Put $z_1 = r_1(\cos\alpha + i\sin\alpha)$ and $z_2 = r_2(\cos\beta + i\sin\beta)$ . Expanding gives

\begin{aligned} z_1z_2 &= r_1r_2\bigl[(\cos\alpha\cos\beta - \sin\alpha\sin\beta) \\ &\qquad {}+ i(\sin\alpha\cos\beta + \cos\alpha\sin\beta)\bigr] \\ &= r_1r_2\bigl[\cos(\alpha+\beta) + i\sin(\alpha+\beta)\bigr] \end{aligned}

The angle-addition identities turn the expanded real and imaginary parts into one new direction. Those trigonometric identities are in the formula booklet; the complex multiplication rule itself is not.

Product in modulus-argument form

For non-zero $z_1$ and $z_2$ ,

z_1z_2 = r_1r_2\bigl[\cos(\alpha+\beta) + i\sin(\alpha+\beta)\bigr]

The moduli multiply and the directions add. If necessary, add or subtract a multiple of $2\pi$ to obtain the principal argument of the product.

Not in the formula booklet: learn it.

Adding directions can take you outside the principal interval. That does not make the product wrong; it means its direction needs an equivalent principal value.

Worked example 5: a product within the principal interval

Question. Multiply $u = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$ and $v = 3\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$ . Give the product in modulus-argument form and Cartesian form.

Plan. Multiply $2$ and $3$ , add the two angles, then evaluate the new sine and cosine.

\begin{aligned} |uv| &= 2 \times 3 = 6 && \text{multiply moduli} \\ \frac{\pi}{3} + \frac{\pi}{6} &= \frac{\pi}{2} && \text{add directions} \\ uv &= 6\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 6i && \text{evaluate exact values} \end{aligned}

Why this step? A product combines the two directions by addition, even though its coordinates do not multiply separately.

Answer. $uv = 6\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 6i$ .

Check. In Cartesian form, $u = 1 + \sqrt3\,i$ and $v = \frac{3\sqrt3}{2} + \frac32 i$ . Expanding gives real part $0$ and imaginary part $6$ .

Change only the direction of $v$ to $\frac{5\pi}{6}$ . The sum now passes $\pi$ , so the principal argument needs adjusting.

Worked example 6: reducing the product's argument

Question. For the same $u$ , let $w = 3\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)$ . Express $uw$ in modulus-argument form using its principal argument.

Plan. Multiply the moduli, add the directions and subtract one full turn.

\begin{aligned} |uw| &= 2 \times 3 = 6 && \text{multiply moduli} \\ \frac{\pi}{3} + \frac{5\pi}{6} &= \frac{7\pi}{6} && \text{add directions} \\ \arg(uw) &= \frac{7\pi}{6} - 2\pi = -\frac{5\pi}{6} && \text{principal value} \end{aligned}

Why this step? $\frac{7\pi}{6}$ and $-\frac{5\pi}{6}$ point in the same direction, but only the latter lies in $(-\pi, \pi]$ .

Answer. $uw = 6\left(\cos\left(-\frac{5\pi}{6}\right) + i\sin\left(-\frac{5\pi}{6}\right)\right)$ .

Check. This gives $-3\sqrt3 - 3i$ ; its modulus is $6$ and its direction is in the third quadrant.

The modulus rule and the argument rule do different jobs. Keep them on separate lines while learning the method.

Watch out

Wrong: $|uv| = |u| + |v|$ . Right: $|uv| = |u||v|$ . Arguments add, while moduli multiply. Also, $\arg(uv)$ equals the sum of principal arguments only after adjusting by a multiple of $2\pi$ when necessary.

For the next product, the new angle is negative already. Check that it still lies in the principal interval.

Divide lengths and subtract directions

Division reverses multiplication. If $q = z_1/z_2$ , then $qz_2 = z_1$ : the modulus of $q$ must multiply $r_2$ to give $r_1$ , and its direction must add to $\beta$ to give $\alpha$ .

Quotient in modulus-argument form

For $z_1 \ne 0$ , $z_2 \ne 0$ ,

\frac{z_1}{z_2} = \frac{r_1}{r_2}\bigl[\cos(\alpha-\beta) + i\sin(\alpha-\beta)\bigr]

Divide the moduli and subtract the denominator's direction from the numerator's direction. Adjust to the principal interval if required. Division by $0$ is undefined; a quotient with zero numerator is $0$ , whose argument is undefined.

Not in the formula booklet: learn it.

When numbers are supplied in modulus-argument form, this is the natural route. If a detailed-reasoning question instead asks you to divide two Cartesian expressions, show the conjugate method you learnt earlier.

Worked example 7: a quotient

Question. Let $u = 6\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$ and $v = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$ . Find $u/v$ in modulus-argument form.

Plan. Divide $6$ by $2$ , then subtract $\frac{\pi}{6}$ from $\frac{2\pi}{3}$ .

\begin{aligned} \left|\frac{u}{v}\right| &= \frac62 = 3 && \text{divide moduli} \\ \arg\left(\frac{u}{v}\right) &= \frac{2\pi}{3} - \frac{\pi}{6} = \frac{\pi}{2} && \text{numerator minus denominator} \\ \frac{u}{v} &= 3\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) && \text{combine size and direction} \end{aligned}

Why this step? Multiplying this quotient by $v$ must restore $u$ , so the angle $\frac{\pi}{6}$ must be removed before it is added back.

Answer. $u/v = 3\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)$ .

Check. $3i(\sqrt3 + i) = -3 + 3\sqrt3\,i = u$ .

Change only the denominator's direction to $-\frac{5\pi}{6}$ . Subtracting a negative angle increases the result, and this time it exceeds $\pi$ .

Worked example 8: a negative denominator argument

Question. For the same $u$ , let $w = 2\left(\cos\left(-\frac{5\pi}{6}\right) + i\sin\left(-\frac{5\pi}{6}\right)\right)$ . Find $u/w$ using its principal argument.

Plan. Divide the moduli, subtract the whole negative angle, then reduce by $2\pi$ .

\begin{aligned} \left|\frac{u}{w}\right| &= 3 && \text{divide moduli} \\ \frac{2\pi}{3} - \left(-\frac{5\pi}{6}\right) &= \frac{3\pi}{2} && \text{subtract a negative} \\ \arg\left(\frac{u}{w}\right) &= \frac{3\pi}{2} - 2\pi = -\frac{\pi}{2} && \text{principal value} \end{aligned}

Why this step? The denominator's argument is one signed quantity. Brackets prevent its minus sign from being lost.

Answer. $u/w = 3\left(\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right)$ .

Check. The quotient is $-3i$ , and $(-3i)(-\sqrt3 - i) = -3 + 3\sqrt3\,i = u$ .

The opening puzzle now has a geometric explanation: $-1 + \sqrt3\,i$ and $-1 - \sqrt3\,i$ have modulus $2$ and arguments $\frac{2\pi}{3}$ , $-\frac{2\pi}{3}$ . Their product has modulus $4$ and direction $0$ , so it is the positive real number $4$ .

Problem-solving

Choose the form that makes the operation simple. Addition and subtraction suit $x + yi$ ; multiplication and division often suit modulus-argument form. Always check the final quadrant and the requested form.

The denominator below moves $\frac{\pi}{3}$ anticlockwise from the previous example. Track the change before reducing the answer.

Watch out

Wrong: $\frac{2\pi}{3}-(-\frac{5\pi}{6})=-\frac{\pi}{6}$ . Subtracting a negative adds it: the direction is $\frac{3\pi}{2}$ , whose principal value is $-\frac{\pi}{2}$ .

Bracket the denominator’s complete signed argument before subtracting it.

Quick Recap

For $z = x + yi$ , $|z| = \sqrt{x^2 + y^2}$ and $zz^* = |z|^2$ .
Use both coordinate signs to choose $-\pi < \arg z \le \pi$ ; $\arg 0$ is undefined.
Modulus-argument form is $r(\cos\theta + i\sin\theta)$ ; use exact trigonometric values for multiples of $\pi/6$ and $\pi/4$ .
Multiply moduli and add directions for a product; divide moduli and subtract directions for a quotient.
Adjust a final direction by a multiple of $2\pi$ when a principal argument is required.

Two negative real parts can produce a positive real product: the opposite arguments of $-1 \pm \sqrt3\,i$ cancel, while their moduli multiply to $4$ .

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Pj6–Pj8 - Modulus-argument form

Size, direction and the right quadrant

Moving between the two forms

Multiply lengths and add directions

Divide lengths and subtract directions

Quick Recap

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