Argand plane — geometric representation: Visual walkthrough

Step 1 — A complex number is a pair of numbers

WHAT we just did: noticed that one $z$ = two ordinary numbers.

WHY it matters: a single number lives on a line (one coordinate). Two numbers need two directions to hold them — that forces a plane, not a line. This is the seed of everything.

PICTURE: on the left, a plain number line can only place a single value. On the right, we add a second axis at a right angle so the two numbers of $z$ each get their own direction.

Figure — Argand plane — geometric representation

Step 2 — Turn the pair $(x, y)$ into a point

WHAT we did: converted the algebra symbol $z$ into a location $P$ , and named the starting corner $O$ .

WHY this rule and not another: it is the only rule where the two independent numbers never interfere — moving right doesn't change your height, moving up doesn't change your rightness. Independent numbers deserve perpendicular axes.

PICTURE: the origin $O$ marked at the crossing of the axes; the orange dot is $z = 3 + 4 i$ . The blue arrow walks $x = 3$ right; the green arrow then walks $y = 4$ up. The dashed box shows the two walks are at right angles.

Step 3 — Draw the arrow $O P$ and find its length

Draw $O P$ now: connect the origin $O$ straight to the point $P$ .

WHAT the symbols do: $∣ z ∣$ is the arrow's length; $x^{2}$ and $y^{2}$ are the squared legs; the square root undoes the squaring to hand us back a plain length.

WHY Pythagoras and not, say, $x + y$ : we want the straight-line distance through the plane, not the distance walked along the two legs. Only Pythagoras gives the diagonal. Because it is a length, $∣ z ∣ \geq 0$ always, and $∣ z ∣ = 0$ only when $x = 0$ and $y = 0$ — i.e. when $z = 0$ sits exactly on the origin $O$ . That is the first degenerate case, handled.

PICTURE: the right triangle with legs labelled $x$ and $y$ , hypotenuse labelled $∣ z ∣$ , and the tiny right-angle square at the corner where the legs meet.

More on this quantity: Complex Numbers — Modulus and Argument.

Step 4 — The tilt of the arrow: introducing the angle

WHAT we did: admitted the arrow needs a direction number too, and named that number $θ = ar g z$ .

WHY anticlockwise from the positive real axis: we must fix one reference direction and one spin-sense so everyone gets the same angle for the same point. Convention picks the positive real axis as "angle zero" and anticlockwise as "positive".

PICTURE: three arrows of equal length pointing different ways — each has its own angle $θ = ar g z$ , shown as a coloured wedge swept anticlockwise from the positive real axis.

Step 5 — Which tool measures a tilt? The tangent

WHY tangent and not sine or cosine? Sine needs the hypotenuse ( $y /∣ z ∣$ ); cosine needs it too ( $x /∣ z ∣$ ). But tangent uses only the two legs $x$ and $y$ we already walked — no length required. Since the tilt of the arrow is exactly "how much up per unit across", the leg-ratio $y / x$ is the steepness. That is the cleanest possible encoding of the angle.

PICTURE: the same triangle, but now with the wedge $θ$ at the origin highlighted, the opposite leg $y$ in green, the adjacent leg $x$ in blue, and the ratio $y / x$ written on the arrow as "rise ÷ run".

Step 6 — Recovering $θ$ safely: the quadrant fix

To go back from the ratio to the angle we use $tan^{- 1}$ ("arctan") — it asks "which angle has this tangent?". But a calculator's $tan^{- 1}$ only ever answers with an angle between $- \frac{π}{2}$ and $\frac{π}{2}$ (that is, only the right half of the plane, quadrants I and IV).

Formula The full sign-aware rule

Quadrant	signs $(x, y)$	$ar g z$	reading
I	$+, +$	$α$	up-right, small positive angle
II	$-, +$	$π - α$	up-left, mirror across vertical
III	$-, -$	$- (π - α)$	down-left, go clockwise past $- \frac{π}{2}$
IV	$+, -$	$- α$	down-right, small negative angle

Degenerate directions (arrow lies on an axis):

on $+ x$ axis ( $y = 0, x > 0$ ): $θ = 0$ .
on $+ y$ axis ( $x = 0, y > 0$ ): $θ = \frac{π}{2}$ (here $y / x$ is undefined — division by zero — so tangent is useless and we read the angle straight off the axis).
on $- x$ axis ( $y = 0, x < 0$ ): $θ = π$ .
on $- y$ axis ( $x = 0, y < 0$ ): $θ = - \frac{π}{2}$ .
at the origin ( $x = 0, y = 0$ , i.e. $z = 0$ ): $ar g z$ is undefined. A zero-length arrow points nowhere, so there is no tilt to measure. (Its modulus is $0$ , but its argument simply does not exist.)

WHAT we did: split the angle recovery into size ( $α$ ) + quadrant (signs).

WHY: the tangent throws away sign information; the raw signs of $x, y$ put it back.

PICTURE: all four quadrants at once, one sample point in each ( $1 + i$ , $- 1 + i$ , $- 1 - i$ , $1 - i$ ), each with its reference angle $α$ drawn against the nearest horizontal and its true argument written in.

Step 7 — One angle or many? The principal value

WHAT we did: admitted the argument is multi-valued — infinitely many angles name the one direction.

WHY we must choose: for a formula to give a single answer, we pick one representative from that family. The agreed choice is the principal argument: the unique angle lying in $- π < θ \leq π .$

Worked meaning: for $z = - 1 - i$ the family includes $- \frac{3 π}{4}$ , but also $- \frac{3 π}{4} + 2 π = + \frac{5 π}{4}$ . Both point the same way, yet $+ \frac{5 π}{4}$ is bigger than $π$ , so it falls outside $(- π, π]$ — we reject it. The principal value is the in-range twin, $- \frac{3 π}{4}$ . That is exactly why the quadrant table in Step 6 keeps answers between $- π$ and $π$ .

PICTURE: one fixed arrow with several sweep-wedges stacked on it — $θ$ , $θ + 2 π$ , $θ - 2 π$ — all ending on the same arrow; the single wedge inside $(- π, π]$ is highlighted as the principal value.

Step 8 — Assembling polar form

We now have both numbers the arrow needs: its length $r = ∣ z ∣$ and its (principal) tilt $θ = ar g z$ . Read the legs back off the triangle using the length:

Substitute these into $z = x + i y$ : $z = x r cos θ + i y r sin θ = r (cos θ + i sin θ)$

WHAT we did: rewrote $z$ purely in terms of length and angle.

WHY this is the payoff: every symbol is now geometric — no more raw $x, y$ . This polar form is exactly what makes multiplication become rotation-and-scaling (see Polar Form of Complex Numbers, Multiplication as Rotation and Scaling, and De Moivre's Theorem).

PICTURE: the arrow with $r$ marked along it and $θ$ as its wedge; the horizontal shadow labelled $r cos θ$ , the vertical shadow labelled $r sin θ$ — showing $x$ and $y$ rebuilt from $r$ and $θ$ .

Worked check — $z = - 1 - i$ (the quadrant trap in action)

The one-picture summary

Everything collapses into a single diagram: the origin $O$ , the point $P (x, y)$ , its arrow $O P$ of length $r = ∣ z ∣ = x^{2} + y^{2}$ , its angle $θ = ar g z$ swept from the positive real axis, and the two shadows $x = r cos θ$ , $y = r sin θ$ that connect algebra to geometry.

Recall Feynman retelling — say it plainly

I start with a complex number, which is really just two ordinary numbers hiding together: one for right/left, one for up/down. Two numbers won't fit on a line, so I draw a flat map with two arrows-at-right-angles for axes. Where they cross is my starting corner, the origin $O$ . I drop a dot where the two numbers tell me to and call it $P$ , then draw a straight arrow $O P$ from $O$ to $P$ — the little over-arrow just means "arrow from $O$ to $P$ ". That arrow has a length — I get it with Pythagoras because the right/left and up/down walks are the two legs of a right triangle — and that length is the modulus. The arrow also has a direction — how tilted it is — and I measure tilt with tangent, "up-leg over across-leg", because that ratio uses only the two things I walked. To turn the ratio back into a real angle I must peek at the signs of my two numbers, otherwise I can't tell top-left from bottom-right. And since spinning a full turn lands back on the same arrow, infinitely many angles name it; I keep the one between $- π$ and $π$ , the principal value. If the arrow has zero length (that's $z = 0$ ), it points nowhere, so its argument simply doesn't exist. Finally I rewrite my number as "length times a direction dial": $r (cos θ + i sin θ)$ — pure geometry.

Recall Quick self-test

Why does a complex number force a plane, not a line?
Which theorem gives $∣ z ∣$ and why not just $x + y$ ?
Why tangent rather than sine/cosine for the tilt?
Where does the naive $tan^{- 1} (y / x)$ break, and how do we repair it?
Why do we pick a principal value, and what is its range?
What is $ar g 0$ ?

Why a plane and not a line?

A complex number carries two independent real numbers, so it needs two perpendicular directions to hold them.

What is the origin

O

The point where the real and imaginary axes intersect — the tail of every position vector

O P

What does

O P

mean?

The straight arrow (vector) starting at the origin

O

and ending at the point

P

Why Pythagoras for

∣ z ∣

The across-walk

x

and up-walk

y

are the legs of a right triangle; the modulus is the hypotenuse (straight-line distance), not the leg-sum.

Why tangent for the argument?

tan θ = y / x

uses only the two legs we already have — no hypotenuse/length needed.

Why does

tan^{- 1} (y / x)

fail?

tan

repeats every

18 0^{\circ}

, so opposite quadrants share a ratio; fix with reference angle

α

plus the signs of

x, y

Why is the argument multi-valued and what fixes it?

A full turn

2 π

returns the same direction, so

ar g z = θ + 2 π k

; the principal value in

(- π, π]

picks one.

What is

ar g 0

Undefined — a zero-length arrow has no direction.

Polar form of

z

z = r (cos θ + i sin θ)

with

r = ∣ z ∣

θ = ar g z

Connections

Argand plane — geometric representation
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