Modulus - z - and argument arg(z): Worked examples

You already met the two numbers that name any complex point: its length $∣ z ∣$ and its direction $ar g (z)$ (see the parent note). Knowing the formulas is not the same as never getting caught out. This page walks through every kind of input the topic can hand you, one example per trap, so that no exam scenario is new.

Two words we will use constantly, defined plainly first:

Figure — Modulus - z - and argument arg(z)

Figure s01 draws one arrow in each quadrant. The red arc marks the acute reference angle $α$ for the quadrant-II arrow (measured to the nearest axis), while the dashed black arc shows the full principal argument $π - α$ measured from the positive real axis. Keep this picture in mind: every quadrant fix below is just " $α$ plus the right multiple of $π$ ."

The scenario matrix

Every problem this topic throws is one of these cells. The right column names the example that clears it.

#	Case class	Signs $(x, y)$ / input	What's tricky	Covered by
1	Quadrant I	$(+, +)$	baseline, use $α$ directly	Ex A
2	Quadrant II	$(-, +)$	shift $π - α$	Ex B
3	Quadrant III	$(-, -)$	go negative: $α - π$	Ex C
4	Quadrant IV	$(+, -)$	go negative: $- α$	Ex D
5	On an axis	one of $x, y$ is $0$	$α$ undefined/degenerate	Ex E
6	The $- π$ vs $π$ trap	negative real axis	boundary of the range	Ex E
7	Zero input	$z = 0$	argument undefined	Ex E
8	Product (scale + rotate)	$z_{1} z_{2}$	radii $\times$ , angles $+$	Ex F
9	Limiting behaviour	$y \to 0^{\pm}$ near negative axis	argument jumps by $2 π$	Ex G
10	Real-world word problem	phasor / rotation	translate words to $(r, θ)$	Ex H
11	Exam-style twist	$ar g$ of a quotient / power	combine the rules	Ex I

Worked examples

Ex A — Cell 1: Quadrant I baseline

Ex B — Cell 2: Quadrant II, the $π - α$ shift

Worked example Find

$Arg (z)$ for $z = - 3 + i$ . Forecast: $x$ negative, $y$ positive — the arrow leans up and to the left. So the angle must be obtuse (between $\frac{π}{2}$ and $π$ ). Guess a value before computing.

Modulus. $∣ z ∣ = (- 3)^{2} + 1^{2} = 3 + 1 = 2$ . Why this step? Squaring kills the sign, so the leftward lean does not shorten the arrow.
Reference angle. $α = arctan \frac{1}{- 3} = arctan \frac{1}{3} = \frac{π}{6}$ . Why this step? $α$ only cares about steepness, so we drop the minus sign inside the $arctan$ .
Quadrant shift. $x < 0, y > 0 \Rightarrow$ QII $\Rightarrow Arg (z) = π - α = π - \frac{π}{6} = \frac{5 π}{6}$ . Why this step? Look at figure s01: from the positive real axis you must swing almost all the way round to the up-left arrow; that is $18 0^{\circ}$ minus the small acute gap $α$ to the negative real axis (exactly the red arc in that figure).

Verify: $r cos θ = 2 cos \frac{5 π}{6} = 2 (- \frac{3}{2}) = - 3 = x$ ✓, $r sin θ = 2 sin \frac{5 π}{6} = 2 \cdot \frac{1}{2} = 1 = y$ ✓.

Ex C — Cell 3: Quadrant III, going negative

Worked example Find

$Arg (z)$ for $z = - 1 - 3 i$ . Forecast: both parts negative — arrow points down-left. Below the real axis, so by our convention the principal argument is negative. Guess whether it is closer to $- π$ or to $- \frac{π}{2}$ .

Modulus. $∣ z ∣ = (- 1)^{2} + (- 3)^{2} = 1 + 3 = 2$ . Why this step? Pythagoras on legs $∣ x ∣ = 1$ and $∣ y ∣ = 3$ ; the squares wipe out both minus signs, so a down-left arrow is exactly as long as its up-right mirror image.
Reference angle. $α = arctan \frac{- 3}{- 1} = arctan 3 = \frac{π}{3}$ . Why this step? The two minus signs cancel inside the ratio, but $α$ would use $∣ \cdot ∣$ anyway.
Quadrant shift. $x < 0, y < 0 \Rightarrow$ QIII $\Rightarrow Arg (z) = - (π - α) = α - π = \frac{π}{3} - π = - \frac{2 π}{3}$ . Why this step? We could measure $π + α$ going anticlockwise, but that leaves the range $(- π, π]$ . Turning clockwise the same amount gives the negative equivalent $α - π$ , which sits inside the allowed range.

Verify: $2 cos (- \frac{2 π}{3}) = 2 (- \frac{1}{2}) = - 1 = x$ ✓, $2 sin (- \frac{2 π}{3}) = 2 (- \frac{3}{2}) = - 3 = y$ ✓.

Ex D — Cell 4: Quadrant IV

Worked example Find

$Arg (z)$ for $z = 1 - i$ . Forecast: $x > 0, y < 0$ — arrow points down-right, just below the positive real axis. Small negative angle expected.

Modulus. $∣ z ∣ = 1^{2} + (- 1)^{2} = 2$ . Why this step? Pythagoras on legs $1$ and $1$ ; the minus sign on $y$ vanishes when squared, so the length matches $1 + i$ from Ex A's family.
Reference angle. $α = arctan \frac{- 1}{1} = arctan 1 = \frac{π}{4}$ . Why this step? Steepness is $∣ y ∣/∣ x ∣ = 1$ , and the angle whose tangent is $1$ is $\frac{π}{4}$ .
Quadrant shift. $x > 0, y < 0 \Rightarrow$ QIV $\Rightarrow Arg (z) = - α = - \frac{π}{4}$ . Why this step? In QIV the arrow is already close to the positive real axis, just below it — so the honest anticlockwise angle is the small acute $α$ measured clockwise, hence negative.

Verify: $2 cos (- \frac{π}{4}) = 2 \cdot \frac{1}{2} = 1 = x$ ✓, $2 sin (- \frac{π}{4}) = - 1 = y$ ✓.

Ex E — Cells 5, 6, 7: axes, the $π / - π$ trap, and zero

Worked example State

$∣ z ∣$ and $Arg (z)$ for $z = 7, z = - 7, z = 4 i, z = - 4 i, z = 0$ . Forecast: these sit on the axes, so no triangle exists and $α$ is degenerate. Predict each angle from the picture (figure s02) before reading.

$z = 7$ (positive real axis). $∣ z ∣ = 7$ , $Arg (z) = 0$ . Why? Arrow lies along the reference direction itself — zero turn.
$z = - 7$ (negative real axis). $∣ z ∣ = 7$ , $Arg (z) = π$ . Why not $- π$ ? $π$ and $- π$ point the same way, but the range is the half-open $(- π, π]$ — it includes $π$ and excludes $- π$ . So the negative real axis always gets $+ π$ . This is Cell 6, the classic trap. In figure s02 this is the single red arrow, flagged precisely because it is the one people get wrong.
$z = 4 i$ (positive imaginary axis). $∣ z ∣ = 4$ , $Arg (z) = \frac{π}{2}$ . Why? Straight up is a quarter turn anticlockwise.
$z = - 4 i$ (negative imaginary axis). $∣ z ∣ = 4$ , $Arg (z) = - \frac{π}{2}$ . Why negative? Straight down is a quarter turn clockwise.
$z = 0$ . $∣ z ∣ = 0$ , and $ar g (0)$ is undefined. Why? An arrow of length zero is just a dot — it has no direction to measure. Cell 7. (In figure s02 it is the lone black point at the origin.)

Verify: for $z = - 7$ : $r cos π = 7 \cdot (- 1) = - 7$ ✓, $r sin π = 0$ ✓. For $z = - 4 i$ : $4 cos (- \frac{π}{2}) = 0$ ✓, $4 sin (- \frac{π}{2}) = - 4$ ✓.

Ex F — Cell 8: product = scale and rotate

Worked example Let

$z_{1} = 1 + i$ and $z_{2} = 3 + i$ . Find $∣ z_{1} z_{2} ∣$ and $Arg (z_{1} z_{2})$ using the rules, then check by direct multiplication. Forecast: the parent note says radii multiply and angles add. Guess the final length and angle before grinding it out.

Polar data of each. $∣ z_{1} ∣ = 2, ar g z_{1} = \frac{π}{4}$ (Ex A style); $∣ z_{2} ∣ = 3 + 1 = 2, ar g z_{2} = arctan \frac{1}{3} = \frac{π}{6}$ (QI). Why this step? The multiplication rule only speaks the language of $(r, θ)$ , so we translate first — see Polar and Trigonometric form of complex numbers.
Combine. $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣ ∣ z_{2} ∣ = 2 \cdot 2 = 22$ ; $Arg (z_{1} z_{2}) = \frac{π}{4} + \frac{π}{6} = \frac{5 π}{12}$ . Why this step? Multiplying complex numbers scales by the second modulus and rotates by the second argument (this is exactly Euler's rule $e^{i θ_{1}} e^{i θ_{2}} = e^{i (θ_{1} + θ_{2})}$ ). Figure s03 shows the short black arrow $z_{1}$ being stretched to length $22$ and swung anticlockwise by $\frac{π}{6}$ onto the red product arrow.

Verify (direct): $z_{1} z_{2} = (1 + i) (3 + i) = 3 + i + 3 i + i^{2} = (3 - 1) + (3 + 1) i$ . Its modulus: $(3 - 1)^{2} + (3 + 1)^{2} = (4 - 23) + (4 + 23) = 8 = 22$ ✓. Its argument: $arctan \frac{3 + 1}{3 - 1} = \frac{5 π}{12} = 7 5^{\circ}$ ✓.

Ex G — Cell 9: limiting behaviour, the argument's jump

Worked example Track

$Arg (z)$ for $z = - 1 + ε i$ as $ε \to 0^{+}$ and as $ε \to 0^{-}$ . Forecast: both limits land on the point $- 1$ (the negative real axis). Do the arguments agree? Guess yes or no.

Approach from above ( $ε \to 0^{+}$ ). Then $y > 0$ , so we are in QII; $α = arctan \frac{ε}{- 1} \to 0$ , giving $Arg (z) = π - α \to π^{-}$ . Why this step? Just above the negative real axis the arrow is almost pointing straight left, i.e. angle just under $π$ .
Approach from below ( $ε \to 0^{-}$ ). Then $y < 0$ , so we are in QIII; $Arg (z) = α - π \to - π^{+}$ . Why this step? Just below the negative real axis the arrow's principal angle is just above $- π$ .
The jump. The two one-sided limits are $π$ and $- π$ — a discontinuity of size $2 π$ exactly along the negative real axis. Why this matters? This "cut" is why $Arg$ is not a smooth function everywhere; it is the price of squeezing every direction into one turn's worth of angle. Figure s04 draws both near-horizontal arrows (black just above, red just below the axis) and shades the seam where the value leaps.

Verify: at $ε = 0.01$ : QII value $= π - arctan (0.01) \approx 3.1316$ , close to $π$ ✓. At $ε = - 0.01$ : QIII value $= arctan (0.01) - π \approx - 3.1316$ , close to $- π$ ✓. The gap is $\approx 2 π$ .

Ex H — Cell 10: real-world word problem

Worked example A rotating machine part is modelled as a point on the plane. It starts

$5 cm$ from the axle along the positive real direction ( $z_{0} = 5$ ). It is then rotated $12 0^{\circ}$ anticlockwise and its distance is scaled by $\frac{1}{2}$ . Write the new position $z_{1}$ in Cartesian form. Forecast: rotating and scaling = multiplying by a complex number. Predict the new distance ( $2.5 cm$ ) and rough direction (up-left) before computing.

Build the multiplier. "Rotate $12 0^{\circ}$ and scale by $\frac{1}{2}$ " is multiplication by $w = \frac{1}{2} (cos \frac{2 π}{3} + i sin \frac{2 π}{3})$ . Why this step? Cell 8's rule read backwards: to add an angle and multiply a length you multiply by a number with that argument and modulus.
Multiply. $z_{1} = z_{0} \cdot w = 5 \cdot \frac{1}{2} (cos \frac{2 π}{3} + i sin \frac{2 π}{3}) = \frac{5}{2} (- \frac{1}{2} + i \frac{3}{2}) = - \frac{5}{4} + \frac{5 3}{4} i$ . Why this step? $cos \frac{2 π}{3} = - \frac{1}{2}, sin \frac{2 π}{3} = \frac{3}{2}$ — the QII values from Ex B.
Interpret. $x = - 1.25 cm$ (left of axle), $y = \frac{5 3}{4} \approx 2.165 cm$ (above axle). New distance $= ∣ z_{1} ∣ = 2.5 cm$ . Why this step? Distance from origin is the modulus; the units ( $cm$ ) carry straight through since we only scaled and rotated.

Verify: $∣ z_{1} ∣ = (- 1.25)^{2} + (2.165)^{2} = 1.5625 + 4.6875 = 6.25 = 2.5 cm$ ✓, and $Arg (z_{1}) = \frac{2 π}{3} = 12 0^{\circ}$ ✓. Both forecasts confirmed.

Ex I — Cell 11: exam-style twist (quotient and power)

Worked example Given

$z = \frac{( 1 + i ) ^{4}}{- 1 + i 3}$ , find $∣ z ∣$ and $Arg (z)$ without expanding. Forecast: raising to a power multiplies the angle; dividing subtracts angles and divides moduli. Try to predict just the modulus first.

Numerator via De Moivre's Theorem. $1 + i$ has modulus $2$ and argument $\frac{π}{4}$ . So $(1 + i)^{4}$ has modulus $(2)^{4} = 4$ and argument $4 \cdot \frac{π}{4} = π$ . Why this step? De Moivre's theorem says a power of a complex number in polar form raises the modulus to that power and multiplies the argument by it: $[r (cos θ + i sin θ)]^{n} = r^{n} (cos n θ + i sin n θ)$ . That turns a fourth power into simple arithmetic on $(r, θ)$ — no binomial expansion.
Denominator (from Ex B). $- 1 + i 3$ has modulus $2$ and argument $\frac{2 π}{3}$ (QII). Why this step? We already found this exact number's polar data in Ex B, so we reuse it rather than recompute.
Divide: moduli divide, arguments subtract. $∣ z ∣ = \frac{4}{2} = 2$ ; $Arg (z) = π - \frac{2 π}{3} = \frac{π}{3}$ . Why this step? Division is the inverse of scale-and-rotate: you shrink the length by the divisor's modulus and unrotate by its argument. The result $\frac{π}{3}$ already lies in $(- π, π]$ , so no quadrant adjustment is needed.

Verify (direct): $(1 + i)^{4} = ((1 + i)^{2})^{2} = (2 i)^{2} = - 4$ . Then $z = \frac{- 4}{- 1 + i 3}$ . Multiply top and bottom by the conjugate $- 1 - i 3$ : the denominator becomes $(- 1)^{2} + (3)^{2} = 1 + 3 = 4$ , and the numerator becomes $- 4 (- 1 - i 3) = 4 + 43 i$ . So $z = \frac{4 + 4 3 i}{4} = 1 + 3 i$ . Its modulus is $1 + 3 = 2$ ✓ and its argument is $arctan 3 = \frac{π}{3}$ ✓ — matching the rule-based answer exactly.

Recall Quick self-test on the whole matrix

Which cell is $z = - 2 - 2 i$ , and what is $Arg (z)$ ? ::: Cell 3 (QIII); $Arg = - \frac{3 π}{4}$ . What is $Arg (- 6)$ and why not $- π$ ? ::: $π$ ; the range $(- π, π]$ excludes $- π$ and includes $π$ . Why does $Arg$ jump by $2 π$ across the negative real axis? ::: The one-sided limits are $π$ (from above) and $- π$ (from below); squeezing all directions into one turn forces a seam there. What is the difference between $ar g (z)$ and $Arg (z)$ ? ::: $ar g (z)$ is the whole family $θ + 2 π k$ ; $Arg (z)$ is the single member in $(- π, π]$ . For a product $z_{1} z_{2}$ , what happens to moduli and arguments? ::: Moduli multiply, arguments add (scale and rotate).