Argand plane — geometric representation: Question bank

Quick vocabulary refresh so nothing below uses a term you haven't pinned down:

Real axis = horizontal line, holds numbers $x + 0 i$ .
Imaginary axis = vertical line, holds numbers $0 + i y$ .
Modulus $∣ z ∣ = x^{2} + y^{2}$ = straight-line distance from the origin $O$ to the point $P (x, y)$ . We give this length its own letter $r$ , so $r$ always means $r = ∣ z ∣$ — the length of the arrow $O P$ .
Argument $ar g z = θ$ = anticlockwise angle from the positive real axis to the arrow $O P$ , taken in $(- π, π]$ .
Polar form $z = r (cos θ + i sin θ)$ = the same $z$ written using its length $r$ and direction $θ$ instead of its coordinates $(x, y)$ .
Quadrant = one of the four regions the two axes cut the plane into. Look at the picture below: Q I is $(+, +)$ , Q II is $(-, +)$ , Q III is $(-, -)$ , Q IV is $(+, -)$ — the signs of $(x, y)$ name the quadrant.
Locus / fixed point $z_{0}$ : in an equation like $∣ z - z_{0} ∣ = r$ , the number $z_{0}$ is a fixed, chosen point (a pin stuck in the map), while $z$ is the variable point that moves. A "locus" is just the set of all points $z$ obeying such a rule.

Figure — Argand plane — geometric representation

The figure above is your reference for every "quadrant" and "direction" item below. Notice how the sign pair $(x, y)$ alone tells you the region, and how the arrow's length is $r$ while its tilt is $θ$ .

True or false — justify

Every statement is either true or false. The reveal gives the reason, which is the whole point.

The modulus

∣ z ∣

can be negative if

y

is negative

False —

∣ z ∣ = x^{2} + y^{2}

is a length, and lengths are never negative; the sign of

y

is squared away before the root.

∣ z ∣ = 0

is possible for infinitely many complex numbers

False —

x^{2} + y^{2} = 0

forces both

x = 0

and

y = 0

, so only

z = 0

has zero modulus; the origin is the single point at distance

0

Two different complex numbers can occupy the same point on the Argand plane

False — the point

P (x, y)

carries both coordinates, so equal points mean equal

x

and equal

y

, hence the same

z

; the map from

z

to point is one-to-one.

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

then

z_{1} = z_{2}

False — same argument means same direction from the origin, but they can lie at different distances (different

r = ∣ z ∣

) along that ray, e.g.

1 + i

and

2 + 2 i

ar g z

and

ar g (2 z)

are always equal

True — multiplying by the positive real number

2

scales the arrow's length but leaves its direction unchanged, so the angle from the positive real axis is untouched.

The conjugate

\overset{z}{ˉ}

always has the same modulus as

z

True — reflecting

(x, y)

across the real axis gives

(x, - y)

, and

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

, so the distance from

O

is identical.

ar g \overset{z}{ˉ} = - ar g z

for every non-zero

z

Mostly true — reflection across the real axis flips the angle's sign; the one wrinkle is

z

on the negative real axis where

ar g z = π

and

\overset{z}{ˉ} = z

, so

ar g \overset{z}{ˉ} = π \neq = - π

(since

- π

is excluded from

(- π, π]

Adding a real number to

z

slides its point horizontally only

True — a real number is

(a, 0)

, so

z + a

shifts

x

a

and leaves

y

fixed; the point moves parallel to the real axis.

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

always

False — this is the triangle inequality's equality case, which holds only when the two arrows point the same direction; in general the diagonal of a parallelogram is shorter than the two sides summed.

The point for

- z

is the reflection of

z

across the imaginary axis

False —

- z = (- x, - y)

negates both coordinates, which is a

18 0^{\circ}

rotation about the origin (reflection through the origin), not a reflection across one axis.

Spot the error

Each line contains a flawed statement or step. The reveal names the flaw.

z = - 3 + 0 i

lies in Quadrant II because its real part is negative."

Error — a point with

y = 0

is on the real axis itself, not inside any quadrant; quadrants require both coordinates non-zero.

"For

z = - 1 - i

ar g z = tan^{- 1} (1) = π /4

Error — the calculator's

tan^{- 1}

only outputs

(- π /2, π /2)

, i.e. Q I/IV; the point is in Q III, so the true argument is

- (π - π /4) = - 3 π /4

"Since

tan θ = y / x

, the numbers

1 + i

and

- 1 - i

share the same argument."

Error — both give ratio

1

, but

tan

repeats every $π$ , so equal ratio does not mean equal angle; they sit on opposite rays (

π /4

- 3 π /4

ar g z = tan^{- 1} (y / x)

, so

ar g (0 + 2 i) = tan^{- 1} (2/0)

is undefined, hence

2 i

has no argument."

Error — division by

x = 0

breaks the formula, not the geometry; the arrow points straight up, so

ar g (2 i) = π /2

read directly off the picture.

"The polar form of

z = 0

0 (cos θ + i sin θ)

for the unique angle

θ = 0

Error — with

r = 0

the arrow has no direction, so any

θ

works; the argument of

0

is simply undefined, not

0

∣ z - z_{0} ∣ = r

is a circle, so

∣ z - z_{0} ∣ < r

is a smaller circle."

Error — the strict inequality is the set of points whose distance from the fixed point

z_{0}

is less than

r

, which is the filled disc's interior, not another circle.

"Because

∣ z ∣ = x^{2} + y^{2}

, the equation

∣ z ∣ = - 4

describes a circle of radius

- 4

Error — modulus is never negative, so

∣ z ∣ = - 4

has no solutions; it describes the empty set, not a circle.

Why questions

Explain the reasoning, not just the fact.

Why does a complex number need a whole plane, while a real number needs only a line?

A real number is one coordinate, which fits on a

1

-D line; a complex number carries two independent reals

(x, y)

, and two numbers demand two axes → a

2

-D plane.

Why is the argument measured from the positive real axis, and anticlockwise?

It is a shared convention so everyone assigns the same angle to the same point; anticlockwise makes increasing

θ

match the standard mathematical direction, and a fixed reference ray removes ambiguity.

Why does

tan θ = y / x

fail to pin down the quadrant by itself?

tan

has period

π

, so it gives identical values for a direction and its exact opposite (

18 0^{\circ}

apart); the single ratio cannot distinguish Q I from Q III, or Q II from Q IV — you must inspect the signs of

x

and

y

Why is

∣ z_{1} - z_{2} ∣

the distance between the two points, rather than

∣ z_{1} ∣ - ∣ z_{2} ∣

z_{1} - z_{2}

is the arrow that goes from

P_{2}

P_{1}

; its length is the straight-line gap between the points.

∣ z_{1} ∣ - ∣ z_{2} ∣

only compares their distances from the origin, which says nothing about how far apart they are.

Why does

∣ z - z_{0} ∣ = r

trace a circle and not, say, a square?

The equation says "distance from the fixed point

z_{0}

equals a constant

r

" — the set of all points at a fixed distance from a centre is, by definition, a circle of radius

r

centred at

z_{0}

Why does multiplying

z

- 1

rotate its point by

18 0^{\circ}

- 1

negates both

x

and

y

, sending

(x, y) \to (- x, - y)

, which is exactly the point diametrically opposite through the origin — a half-turn.

Why is polar form

r (cos θ + i sin θ)

preferred for multiplication over

x + i y

Because multiplication is really a stretch-and-turn: write

z_{1}

with length

r_{1}

, angle

θ_{1}

and

z_{2}

with length

r_{2}

, angle

θ_{2}

; the product has length

r_{1} r_{2}

and angle

θ_{1} + θ_{2}

, so you multiply the lengths and add the angles — two independent, easy moves. In

x + i y

form the same product produces four cross terms

(x_{1} x_{2} - y_{1} y_{2}) + i (x_{1} y_{2} + x_{2} y_{1})

that hide this simple geometry entirely.

Edge cases

The boundary inputs the naive formulas quietly ignore.

What is

ar g z

when

z

lies on the positive real axis, e.g.

z = 5

The arrow points along the reference ray itself, so

ar g z = 0

; the formula

tan^{- 1} (0/5) = 0

agrees here.

What is

ar g z

when

z

lies on the negative real axis, e.g.

z = - 5

The arrow points in the opposite direction, so

ar g z = π

(not

- π

, since the principal range is

(- π, π]

and includes

+ π

What is

ar g z

for a purely imaginary

z = + i y

with

y > 0

, like

z = 3 i

The arrow points straight up, so

ar g z = π /2

— read off the geometry, since

y / x = 3/0

makes the ratio formula blow up.

What is

ar g z

for

z = - 3 i

(purely imaginary,

y < 0

Straight down, so

ar g z = - π /2

; again the picture decides where

tan^{- 1}

cannot.

Does the modulus depend on which quadrant

z

sits in?

No —

x^{2} + y^{2}

squares away all signs, so the distance from the origin is quadrant-blind; only the argument carries quadrant information.

ar g z = π

the same point as

ar g z = - π

Geometrically yes (both point along the negative real axis), but the principal value convention keeps only

+ π

because the range

(- π, π]

is open at $- π$ and closed at $+ π$ to avoid naming one direction twice.

x = 0

and

y = 0

, what are the modulus and argument?

Modulus is

0

(the point is the origin), and argument is undefined because a zero-length arrow has no direction to measure.

Recall One-line summary of every trap

Modulus is a non-negative distance (never signed, zero only at the origin); argument is a direction that needs the signs of $x, y$ — not just their ratio — and is undefined at $z = 0$ ; equalities like $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ hold only in special aligned cases. When a formula divides by $0$ , read the angle off the picture.

Connections

3.5.03 Argand plane — geometric representation (Hinglish)
Complex Numbers — Modulus and Argument
Polar Form of Complex Numbers
Multiplication as Rotation and Scaling
De Moivre's Theorem
Vectors — Position Vectors and Addition
Loci in the Complex Plane