IB Analysis and Approaches HL · Complex numbers · Section 2 of 4

Modulus–argument and Euler form

3 worked examples · By Álvaro López

Key idea

A complex number can be given by its coordinates $(a, b)$ or by its distance from the origin and its direction. The modulus–argument (polar) form and the Euler form use the second description, and with it multiplying and dividing becomes very easy: you multiply (or divide) the distances and add (or subtract) the angles. It is the basis of De Moivre's theorem and of roots, and it appears in Paper 1 without a calculator.

Modulus and argument

Think of $z = a + bi$ as the point $(a, b)$ of the Argand diagram (if you do not remember the form $a + bi$ , review it in Cartesian form and operations). That point is fixed by two pieces of information: how far it is from the origin and the angle its segment makes with the positive real axis.

Argand diagram with the point z, its modulus r as the distance to the origin and its argument theta as the angle with the positive real axis

The right-angled triangle in the figure gives the relationships between $(a, b)$ and $(r, θ)$ :

$a = r cos θ, b = r sin θ, tan θ = \frac{b}{a} (a \neq = 0)$

The number $0$ has modulus $0$ and no argument (it does not point anywhere).

Two quick consequences of the diagram: the conjugate $z^{*}$ is its reflection in the real axis, so $∣ z^{*} ∣ = ∣ z ∣$ and $ar g z^{*} = - ar g z$ ; and $- z$ points in the opposite direction, with the same modulus and argument $ar g z \pm π$ .

Be careful with the quadrant

The equation $tan θ = \frac{b}{a}$ has two solutions in each full turn, separated by $π$ , and the calculator's $arctan$ key returns only the one in $] - \frac{π}{2}, \frac{π}{2} [$ , that is, the one in the first and fourth quadrants ( $a > 0$ ). If $a < 0$ you must correct it. The safest approach is to draw the point and work with the acute reference angle.

A useful check: the sign of $sin θ$ must match that of $b$ , and the sign of $cos θ$ must match that of $a$ .

Modulus–argument form and Euler form

Substituting $a = r cos θ$ and $b = r sin θ$ into $z = a + bi$ and taking out $r$ as a common factor:

There is a third way of writing it, even more compact, which relies on Euler's formula:

$e^{i θ} = cos θ + i sin θ$

For now, take it as a definition of what raising $e$ to an imaginary exponent means (in more advanced courses it is proved using series). The important point is that it obeys the rules of powers, and that explains why arguments add when multiplying, as we will see shortly.

All three forms describe the same number. For example, $z = 2 cis \frac{π}{3} = 2 e^{iπ /3} = 1 + 3 i$ . Two cases worth knowing by heart: $e^{iπ} = - 1$ and $e^{iπ /2} = i$ .

Product and quotient in modulus–argument form

In Cartesian form, multiplying requires expanding and grouping. In modulus–argument form it is almost immediate. Let $z = r_{1} cis θ_{1}$ and $w = r_{2} cis θ_{2}$ :

Why. When the brackets are expanded, the angle-addition identities appear:

$z w = r_{1} r_{2} (cos θ_{1} + i sin θ_{1}) (cos θ_{2} + i sin θ_{2}) = r_{1} r_{2} [(cos θ_{1} cos θ_{2} - sin θ_{1} sin θ_{2}) + i (sin θ_{1} cos θ_{2} + cos θ_{1} sin θ_{2})] = r_{1} r_{2} [cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2})]$

In Euler form the result is the familiar rule for powers: $r_{1} e^{i θ_{1}} \cdot r_{2} e^{i θ_{2}} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$ , and the quotient subtracts the exponents. For the quotient, reason in the same way, or use that dividing is multiplying by the inverse, which has modulus $\frac{1}{r}$ and argument $- θ$ .

After operating, the angle may fall outside $] - π, π]$ : add or subtract $2 π$ until it lies in the interval. It is the same number, just written with its principal argument.

Geometric interpretation

Multiplying by $w = r_{2} cis θ_{2}$ does two things at once to any point $z$ :

it stretches it (or shrinks it), multiplying its distance from the origin by $r_{2}$ ;
it rotates it through an angle $θ_{2}$ about the origin, anticlockwise if $θ_{2} > 0$ .

The points z and w and their product zw: the modulus is multiplied and the angles alpha and beta are added

Particular cases used very often:

Multiplying by	Effect
$i = cis \frac{π}{2}$	rotation of $9 0^{\circ}$ (does not change the modulus)
$- 1 = cis π$	rotation of $18 0^{\circ}$
$cis θ$ (modulus 1)	pure rotation through $θ$
$k > 0$ real	stretch by a factor of $k$ , no rotation

Dividing by $w$ does the opposite: it shrinks by $r_{2}$ and rotates through $- θ_{2}$ .

Worked examples

Example 1 Basic

From Cartesian to modulus–argument form and back

Write in modulus–argument form $r cis θ$ , with the principal argument in radians and exact values:

a) $z_{1} = - 3 + 3 i$ .

b) $z_{2} = - 23 - 2 i$ .

c) $z_{3} = - 7 i$ .

d) Convert $z_{4} = 6 cis \frac{2 π}{3}$ to Cartesian form.

Solution

a) The point $(- 3, 3)$ lies in the 2nd quadrant. $r = 9 + 9 = 32$ . The reference angle is $α = arctan \frac{3}{- 3} = \frac{π}{4}$ , and in the 2nd quadrant $θ = π - \frac{π}{4} = \frac{3 π}{4}$ . So $z_{1} = 32 cis \frac{3 π}{4}$ .

b) The point $(- 23, - 2)$ lies in the 3rd quadrant. $r = 12 + 4 = 4$ . Reference angle: $α = arctan \frac{2}{2 3} = arctan \frac{1}{3} = \frac{π}{6}$ . In the 3rd quadrant, $θ = \frac{π}{6} - π = - \frac{5 π}{6}$ , which lies in $] - π, π]$ . So $z_{2} = 4 cis (- \frac{5 π}{6})$ .

c) It lies on the negative imaginary axis: $r = 7$ and $θ = - \frac{π}{2}$ . So $z_{3} = 7 cis (- \frac{π}{2})$ .

d) $z_{4} = 6 (cos \frac{2 π}{3} + i sin \frac{2 π}{3}) = 6 (- \frac{1}{2} + \frac{3}{2} i) = - 3 + 33 i$ .

Example 2 Medium

Product and quotient of two complex numbers

Let $z = 4 e^{3 π i /4}$ and $w = 2 e^{2 π i /3}$ .

a) Calculate $z w$ in Euler form with the principal argument.

b) Calculate $\frac{z}{w}$ in Euler form with the principal argument.

Solution

a) Multiply the moduli and add the arguments:

$z w = 4 \cdot 2 e^{i (\frac{3 π}{4} + \frac{2 π}{3})} = 8 e^{\frac{17 π}{12} i}$

As $\frac{17 π}{12} > π$ , subtract $2 π$ to bring it into $] - π, π]$ :

$\frac{17 π}{12} - 2 π = \frac{17 π - 24 π}{12} = - \frac{7 π}{12} ⟹ z w = 8 e^{- \frac{7 π}{12} i}$

b) Divide the moduli and subtract the arguments:

$\frac{z}{w} = \frac{4}{2} e^{i (\frac{3 π}{4} - \frac{2 π}{3})} = 2 e^{\frac{π}{12} i}$

The angle $\frac{π}{12}$ is already in $] - π, π]$ , so no adjustment is needed.

Example 3 Hard

A rotation with stretching

Let $u = 1 + 3 i$ and $v = 3 - i$ .

a) Write $u$ in Euler form.

b) Write $v$ in Euler form.

c) Calculate $\frac{u}{v}$ in Cartesian form, without working through fractions with the conjugate.

d) On a map of the complex plane, the point $P$ represents $u$ . $P$ is rotated through an angle of $\frac{π}{4}$ anticlockwise about the origin and, at the same time, its distance from the origin is multiplied by $2$ . Find, in Cartesian form, the number $p$ that represents the new point.

Solution

a) $u = 1 + 3 i$ lies in the 1st quadrant. $r = 1 + 3 = 2$ and $θ = arctan 3 = \frac{π}{3}$ . So $u = 2 e^{\frac{π}{3} i}$ .

b) $v = 3 - i$ lies in the 4th quadrant. $r = 3 + 1 = 2$ , $α = arctan \frac{1}{3} = \frac{π}{6}$ and in the 4th quadrant $θ = - \frac{π}{6}$ . So $v = 2 e^{- \frac{π}{6} i}$ .

c) Divide in Euler form: modulus $\frac{2}{2} = 1$ and argument $\frac{π}{3} - (- \frac{π}{6}) = \frac{π}{2}$ . Then

$\frac{u}{v} = e^{\frac{π}{2} i} = cos \frac{π}{2} + i sin \frac{π}{2} = i$

d) Stretching by $2$ and rotating through $\frac{π}{4}$ is equivalent to multiplying by $w = 2 cis \frac{π}{4} = 2 (\frac{2}{2} + \frac{2}{2} i) = 1 + i$ . Then

$p = u w = (1 + 3 i) (1 + i) = 1 + i + 3 i + 3 i^{2} = (1 - 3) + (1 + 3) i$

Check in polar form: $p$ must have modulus $22$ and argument $\frac{π}{3} + \frac{π}{4} = \frac{7 π}{12}$ , in the 2nd quadrant. Indeed, $1 - 3 < 0$ and $1 + 3 > 0$ , and $(1 - 3)^{2} + (1 + 3)^{2} = 8 = 22$ .

In the exam: when $\frac{u}{v}$ or a product is asked for and the numbers are "nice" in polar form, converting to Euler form saves a lot of calculation, but always write down the initial moduli and arguments (they earn marks) and reduce the final argument to $] - π, π]$ before giving the answer.

Common mistakes

Summary

Modulus $r = ∣ z ∣ = a^{2} + b^{2}$ . Argument $θ = ar g z$ : the angle with the positive real axis; the principal one lies in $] - π, π]$ . $0$ has no argument.
To find $θ$ : draw the point, calculate $α = arctan ∣ b / a ∣$ and adjust it to the quadrant ( $α$ , $π - α$ , $α - π$ , $- α$ ). $arctan (b / a)$ is only valid if $a > 0$ .
Polar form $z = r (cos θ + i sin θ) = r cis θ$ ; Euler form $z = r e^{i θ}$ (Euler: $e^{i θ} = cos θ + i sin θ$ ). From polar to Cartesian: $a = r cos θ$ , $b = r sin θ$ .
$z w = r_{1} r_{2} cis (θ_{1} + θ_{2})$ and $\frac{z}{w} = \frac{r _{1}}{r _{2}} cis (θ_{1} - θ_{2})$ . Reduce the final angle to $] - π, π]$ .
Geometrically, multiplying by $r cis θ$ stretches by $r$ and rotates by $θ$ ; multiplying by $i$ rotates by $9 0^{\circ}$ . Also $ar g z^{*} = - ar g z$ .

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