Modulus - z - and argument arg(z): Question bank

Recall The three facts every answer leans on

$∣ z ∣ = x^{2} + y^{2} \geq 0$ — a length, always non-negative. (See Complex Numbers - Cartesian form z = x + iy.)
$θ = ar g (z)$ satisfies $cos θ = x / r$ , $sin θ = y / r$ together — signs fix the quadrant.
Principal argument $== Arg (z) \in (- π, π] ==$ .

True or false — justify

State T/F, then give the one-sentence reason.

∣ z ∣

can be a negative number if

x

and

y

are both negative."

False.

∣ z ∣ = x^{2} + y^{2}

squares the coordinates first, so signs vanish; a length is never negative — e.g.

z = - 3 - 4 i

has

∣ z ∣ = 5

∣ z ∣ = ∣ - z ∣

for every complex number

z

True. Negating flips the arrow to the exact opposite direction but keeps its length:

∣ - z ∣ = (- x)^{2} + (- y)^{2} = x^{2} + y^{2} = ∣ z ∣

∣ z ∣ = ∣ \overset{z}{ˉ} ∣

where

\overset{z}{ˉ}

is the conjugate."

True. Conjugation reflects the arrow across the real axis (Conjugate of a complex number); reflection preserves distance from the origin, so the length is unchanged.

ar g (z) = ar g (\overset{z}{ˉ})

always."

False. Reflecting across the real axis flips the angle's sign:

ar g (\overset{z}{ˉ}) = - ar g (z)

(except on the real axis where both are

0

, and the negative real axis where both give

π

"If

∣ z_{1} ∣ = ∣ z_{2} ∣

then

z_{1} = z_{2}

False. Equal length only puts both points on the same circle around the origin; they can face any direction, e.g.

3

and

3 i

both have modulus

3

but differ.

Arg (z)

can equal exactly

- π

False. The principal range is

(- π, π]

, open at

- π

; the negative real axis is assigned

+ π

instead, so

- π

is never used.

"For any

z \neq = 0

, multiplying

z

by a positive real number leaves

ar g (z)

unchanged."

True. A positive real factor only stretches the arrow along its own direction; it does not rotate it, so the angle stays the same.

ar g (z_{1} z_{2}) = ar g (z_{1}) + ar g (z_{2})

holds exactly, with no adjustment."

False. The arguments add only modulo $2 π$ ; the raw sum can leave the principal range

(- π, π]

and must be shifted back by

\pm 2 π

"The modulus of a purely imaginary number

i y

equals

y

False. It equals

∣ y ∣

, the absolute value:

∣ i y ∣ = 0^{2} + y^{2} = ∣ y ∣

; for

y < 0

that is

- y

, not

y

Spot the error

Each line contains a plausible but flawed argument. Name the flaw.

z = - 1 - i

, so

Arg (z) = arctan \frac{- 1}{- 1} = arctan (1) = \frac{π}{4}

The signs

x < 0, y < 0

put

z

in quadrant III, but

arctan

only outputs

(- \frac{π}{2}, \frac{π}{2})

; the true argument is

\frac{π}{4} - π = - \frac{3 π}{4}

∣ z ∣^{2} = z^{2}

, since squaring a number gives its size squared."

z^{2} = (x + i y)^{2}

is generally a complex number, while

∣ z ∣^{2}

is real; the correct identity uses the conjugate,

∣ z ∣^{2} = z \overset{z}{ˉ} = x^{2} + y^{2}

∣ z ∣ = x + y

because the modulus adds up the parts."

The parts combine by Pythagoras, not by plain addition:

∣ z ∣ = x^{2} + y^{2}

; e.g.

3 + 4 i

gives

5

, not

7

z = 0

has

Arg (z) = 0

because it sits on the real axis."

A zero-length arrow has no direction, so

ar g (0)

is undefined;

0

does not "sit on" the positive real axis in any directional sense.

Arg (- 5) = - π

since

- 5

points left."

Left is angle

π

, and the convention

(- π, π]

assigns the negative real axis

+ π

;

- π

is excluded from the range.

∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣

for all

z_{1}, z_{2}

because moduli just add."

This is only the triangle inequality with equality; in general

∣ z_{1} + z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣

, with equality only when they point the same way (

ar g z_{1} = ar g z_{2}

"To get the polar form of

z = - 2

, write

z = 2 (cos 0 + i sin 0)

since

∣ z ∣ = 2

Modulus is right but the argument is wrong:

- 2

points left, so

Arg = π

, giving

z = 2 (cos π + i sin π)

Why questions

Answer with the underlying reason, not a formula.

"Why does

ar g (z)

need two equations (

cos θ = x / r

and

sin θ = y / r

) instead of just

tan θ = y / x

Because

tan

repeats every

π

: opposite arrows

z

and

- z

share the same ratio

y / x

, so tangent alone cannot tell them apart; the signs of

x

and

y

(encoded by

cos

and

sin

) pin the quadrant.

"Why is the modulus written as

∣ z ∣^{2} = z \overset{z}{ˉ}

so useful?"

It turns "size" into ordinary multiplication, which behaves nicely under products and lets us prove

∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣

cleanly (Conjugate of a complex number).

"Why do arguments add when complex numbers are multiplied?"

Because multiplication in polar form applies the angle-addition identities, so multiplying by

z

rotates the plane by

ar g (z)

— geometrically it is a rotation stacked on a scaling (Polar and Trigonometric form of complex numbers).

"Why is

ar g (z)

measured from the positive real axis and anticlockwise, not clockwise?"

It is a shared convention so everyone reads the same number; anticlockwise-positive matches the standard orientation of angles in Vectors and 2D coordinate geometry and makes

ar g

additive under multiplication.

"Why is the principal range chosen as

(- π, π]

rather than

[0, 2 π)

To keep the value symmetric about the real axis so that

ar g (\overset{z}{ˉ}) = - ar g (z)

and reflections read cleanly; either range works, but this one is the standard.

"Why does multiplying by

i

rotate a point by exactly

9 0^{\circ}

Because

i

has modulus

1

and argument

\frac{π}{2}

, so it adds no length but adds a quarter-turn to the argument (Euler's formula e^{i\θ} = cos\θ + i sin\θ).

"Why can we write

z = r e^{i θ}

instead of

r (cos θ + i sin θ)

Euler's formula e^{i\θ} = cos\θ + i sin\θ identifies

e^{i θ} = cos θ + i sin θ

, so the exponential is the unit arrow at angle

θ

— the same object in shorter notation.

Edge cases

The degenerate and boundary situations the topic hides.

"What is

ar g (z)

when

z = 0

Undefined — a point at the origin is an arrow of length

0

, which has no direction to measure.

"What is

Arg (z)

for a positive real number like

z = 7

Exactly

0

; the arrow lies along the positive real axis with no turn.

"What is

Arg (z)

for a negative real number like

z = - 7

π

(not

- π

), because the negative real axis is the one boundary point kept by the half-open range

(- π, π]

"What are

Arg (3 i)

and

Arg (- 3 i)

+ \frac{π}{2}

(straight up) and

- \frac{π}{2}

(straight down); the moduli are both

3

"If

∣ z ∣ = 1

, where does

z

live and what does that mean for powers

z^{n}

z

lies on the unit circle, so

∣ z^{n} ∣ = 1

for all

n

; by De Moivre's Theorem raising to

n

only multiplies the argument by

n

, spinning the point without stretching it.

"For which

z

does

ar g (z) = ar g (\overset{z}{ˉ})

Only on the real axis: at positive reals both are

0

, at negative reals both are

π

; everywhere else conjugation flips the sign of the argument.

"What happens to

Arg (z_{1} z_{2})

if the raw sum

ar g z_{1} + ar g z_{2}

exceeds

π

You subtract

2 π

to fall back into

(- π, π]

; e.g. two arrows near

\frac{3 π}{4}

each sum to

\frac{3 π}{2}

, which becomes

- \frac{π}{2}