Visual walkthrough — Modulus - z - and argument arg(z)

Step 1 — Draw the plane and place the number

WHAT: we mark the point for a concrete number, say $z=3+2i$, so $x=3$ (right) and $y=2$ (up).

WHY: before we can talk about "how far" or "which way", we need a place. Every later idea is read straight off this dot.

PICTURE: the blue dot sits 3 units right, 2 units up. The two dashed guide-lines are $x$ and $y$ — remember them, they become the legs of a triangle in the next step.

• $x$ ::: the horizontal offset (real part), read on the horizontal axis
• $y$ ::: the vertical offset (imaginary part), read on the vertical axis

Step 2 — Grow the arrow, then measure its length

WHAT: we complete a right triangle. The bottom leg is $x$ (horizontal), the side leg is $y$ (vertical), and the arrow itself is the hypotenuse (the slanted longest side).

WHY Pythagoras and not something else? We want the straight-line distance from $0$ to the dot. Whenever a right angle sits between two known legs, the tool that turns those legs into the slanted distance is Pythagoras' theorem — that is exactly its job. No angle is needed, only the two legs.

$\underset{\text{arrow length}}{\underbrace{|z|}}=\sqrt{\underset{\text{bottom leg squared}}{\underbrace{x^2}}+\underset{\text{side leg squared}}{\underbrace{y^2}}}$

PICTURE: the yellow hypotenuse is the arrow; the little square marks the $90^{\circ }$ corner that licenses Pythagoras. For $z=3+2i$: $|z|=\sqrt{9+4}=\sqrt{13}\approx 3.61$.

Step 3 — Now ask "which direction?" — the angle appears

WHAT: we mark the angle $\theta$ between the positive real axis (our "east", the zero direction) and the arrow.

WHY a turn, and why anticlockwise? "How far" (Step 2) does not tell you where. Two arrows can both be length $\sqrt{13}$ yet point to totally different spots. The turn pins down the where. Anticlockwise-from-east is a fixed convention so everyone measures the same angle for the same arrow.

PICTURE: the green wedge is the angle $\theta$, opening from the horizontal axis up to the arrow. Notice the same right triangle from Step 2 is still here — we are about to read the angle off that triangle.

Step 4 — Read the angle off the triangle: why sin, cos, tan

WHAT: on the right triangle with angle $\theta$ at the origin, we name the sides relative to $\theta$:

• the leg next to $\theta$ (along the axis) has length $x$ — call it the adjacent;
• the leg across from $\theta$ (vertical) has length $y$ — the opposite;
• the arrow (hypotenuse) has length $r$.

WHY these three ratios? Each answers a different question:

$\underset{\text{"how much east?"}}{\underbrace{\cos \theta }}=\frac{x}{r},\underset{\text{"how much up?"}}{\underbrace{\sin \theta }}=\frac{y}{r},\underset{\text{"steepness"}}{\underbrace{\tan \theta }}=\frac{y}{x}$

• $\cos \theta =x/r$ measures how much of the arrow's length points east (adjacent ÷ hypotenuse).
• $\sin \theta =y/r$ measures how much points up (opposite ÷ hypotenuse).
• $\tan \theta =y/x$ measures pure steepness, up ÷ across — independent of length $r$.

PICTURE: each ratio is highlighted on the same triangle so you see which two sides it compares.

Step 5 — Why $\tan$ alone is NOT enough (the trap)

WHAT: we draw two arrows — $z=3+2i$ and $z=-3-2i$. They point opposite ways yet share $\tan \theta =\frac{2}{3}$.

WHY it fails: $\tan$ repeats every $180^{\circ }$ ($\pi$ radians). The button $\arctan$ can therefore only ever hand back an angle in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ — the right-hand half of the plane. Anything on the left ($x<0$) comes back wrong by exactly $\pi$.

PICTURE: two arrows, one blue (QI) one red (QIII), pointing dead opposite — but the naive formula returns the same angle for both. The signs of $x,y$ are the missing witnesses.

Step 6 — The full fix: every quadrant with its own picture

WHAT & WHY per quadrant (drawn all four, one panel each):

PICTURE: four arrows, one in each quadrant, each labelled with how $\alpha$ is bent into the true argument. Read them like a clock: QI small positive, QII large positive, QIII/QIV negative.

Step 7 — The degenerate & edge cases

WHAT (each read straight off an arrow):

• $z=5$ → arrow points east: $|z|=5$, $\mathrm{Arg}=0$.
• $z=-5$ → arrow points west: $\mathrm{Arg}=\pi$ (the endpoint of the allowed range, so $+\pi$).
• $z=3i$ → arrow points north: $\mathrm{Arg}=\frac{\pi }{2}$.
• $z=-3i$ → arrow points south: $\mathrm{Arg}=-\frac{\pi }{2}$.
• $z=0$ → no arrow at all: $|z|=0$ and $\arg (0)$ is undefined — a point has no direction.

WHY they matter: these are exactly where "compute $\arctan$" chokes (dividing by zero when $x=0$). The picture never chokes — you just look at where the arrow points.

PICTURE: the four axis arrows plus the lonely dot at the origin with no direction.

The one-picture summary

Everything collapses into a single arrow and its right triangle: length $r$ from Pythagoras, turn $\theta$ from the side ratios, quadrant fixed by the signs of $x$ and $y$.

$z=\underset{\sqrt{x^2+y^2}}{\underbrace{r}}(\cos \underset{\text{fixed by quadrant}}{\underbrace{\theta }}+i\sin \theta ),x=r\cos \theta ,y=r\sin \theta$

Recall