Intuition Why a "scenario matrix" first?
When you convert z = x + i y into modulus-and-argument, the answer depends on WHERE the point sits . The formula tan θ = y / x is the same everywhere, but the correction you apply changes with the quadrant, and some inputs (like z on an axis, or z = 0 ) break the formula entirely. So before we grind examples, we list every situation that can occur . Then each example is tagged with the cell it kills — and together they leave no gap.
This page assumes only the parent note the parent . Every symbol used there — the point P ( x , y ) , the modulus r = ∣ z ∣ = x 2 + y 2 , the argument θ , the reference angle α = tan − 1 ∣ y / x ∣ — is used exactly as defined there.
Definition What can Complex Numbers throw at you here?
Every "find modulus and argument / plot / describe" problem falls into one of these cells . The columns say what makes the cell special; the last column names the example that handles it.
Cell
Situation
What's tricky
Example
A
Quadrant I ( + , + )
none — the "friendly" case
Ex 1
B
Quadrant II ( − , + )
naive tan − 1 gives wrong angle
Ex 2
C
Quadrant III ( − , − )
tan − 1 gives right ratio, wrong sign
Ex 3
D
Quadrant IV ( + , − )
angle is negative
Ex 4
E
On an axis (a leg is 0 )
tan − 1 ( y /0 ) undefined / y = 0
Ex 5
F
Zero z = 0
modulus 0 , argument undefined
Ex 6
G
Limiting behaviour (y → 0 , x → 0 )
angle sweeps toward 0 or ± π /2
Ex 7
H
Word problem (real-world)
translate a physical direction into z
Ex 8
I
Locus (geometry from algebra)
equation → shape
Ex 9
J
Exam twist (product / rotation)
combine argument-adds-under-multiply
Ex 10
Figure s01 (below) is the compass for this whole page: it shades each quadrant with the correction it needs and marks the axis cells. Keep glancing back at it — every example lands somewhere on this picture.
Caption s01. The four quadrants each carry their own sign/formula for arg z ; the axes (Cell E) and the origin (Cell F) are the borderlines where the formula must be replaced by a picture.
Worked example Example 1 · Cell A ·
z = 3 + 4 i
Find ∣ z ∣ , arg z , and the polar form.
Forecast: both parts are positive, so the point is up-and-right. Guess: angle is a positive acute number, no correction needed.
Step 1. Locate: ( x , y ) = ( 3 , 4 ) , Quadrant I.
Why this step? Signs decide which correction (if any) we need later — always place the point first.
Step 2. Modulus: ∣ z ∣ = 3 2 + 4 2 = 9 + 16 = 25 = 5 .
Why this step? Pythagoras on the right triangle with legs 3 , 4 (parent note's derivation).
Step 3. Reference angle: α = tan − 1 3 4 = tan − 1 ( 1.3333 ) ≈ 0.9273 rad ≈ 53.1 3 ∘ .
Why this step? α is the acute angle inside the triangle; in Q I the argument equals it.
Step 4. Argument: Q I ⇒ arg z = α ≈ 0.9273 rad .
Step 5. Polar form: z = 5 ( cos 0.9273 + i sin 0.9273 ) .
Verify: 5 cos 0.9273 = 5 ( 0.6 ) = 3 ✓ and 5 sin 0.9273 = 5 ( 0.8 ) = 4 ✓. Rebuilds 3 + 4 i .
Worked example Example 2 · Cell B ·
z = − 3 + 4 i
Forecast: point is up-and-left. A calculator's tan − 1 ( 4/ ( − 3 )) = tan − 1 ( − 1.333 ) returns a negative angle — but our point is clearly in the upper half. So the raw answer must be wrong. Guess: true angle is a bit less than 18 0 ∘ .
Step 1. Locate: ( − 3 , 4 ) , Quadrant II.
Why? x < 0 , y > 0 .
Step 2. Modulus: ∣ z ∣ = ( − 3 ) 2 + 4 2 = 25 = 5 .
Why? Squaring kills the sign — same length as Ex 1, mirror image.
Step 3. Reference angle: α = tan − 1 − 3 4 = tan − 1 ( 1.333 ) ≈ 53.1 3 ∘ .
Why the absolute value? α is the acute angle to the nearest part of the real axis (here the negative x -axis). We take ∣ y / x ∣ so α is always positive.
Step 4. Quadrant fix: Q II ⇒ arg z = π − α ≈ 18 0 ∘ − 53.1 3 ∘ = 126.8 7 ∘ ≈ 2.2143 rad .
Why π − α ? We measure from the positive real axis anticlockwise; the point is α past the negative axis, and the negative axis is at π , so we're α short of π .
Verify: 5 cos ( 2.2143 ) = − 3 ✓, 5 sin ( 2.2143 ) = 4 ✓.
Figure s02 (below) shows exactly this: the magenta arrow to ( − 3 , 4 ) , the orange arc that measures arg z from the positive real axis, and the small navy arc that is the reference angle α against the negative axis. Compare the two arcs — their sum is π .
Caption s02. Cell B — the argument (orange arc) is measured from + x ; the calculator's raw answer only ever finds the reference angle α (navy arc).
Worked example Example 3 · Cell C ·
z = − 1 − i
Forecast: down-and-left. tan θ = ( − 1 ) / ( − 1 ) = + 1 — positive! Same ratio as Q I. So the naive tan − 1 ( 1 ) = 4 5 ∘ would put us in Q I, totally wrong. Guess: true angle is negative and large in size.
Step 1. Locate: ( − 1 , − 1 ) , Quadrant III.
Why this step? Both parts negative fixes the point in the lower-left; this tells us the correction will use both a π and a minus sign.
Step 2. Modulus: ∣ z ∣ = ( − 1 ) 2 + ( − 1 ) 2 = 2 ≈ 1.4142 .
Why this step? Pythagoras on legs of length 1 and 1 ; squaring erases the signs, so the length is the same as for 1 + i .
Step 3. Reference angle: α = tan − 1 − 1 − 1 = tan − 1 ( 1 ) = 4 π = 4 5 ∘ .
Why this step? α is the acute angle inside the triangle, built from absolute values so it never carries the misleading positive ratio into the answer.
Step 4. Quadrant fix: Q III ⇒ arg z = − ( π − α ) = − ( π − 4 π ) = − 4 3 π ≈ − 2.3562 rad .
Why this step? Principal argument lives in ( − π , π ] . The point is below the real axis, so we rotate clockwise — a negative angle. It sits α past the negative real axis on the clockwise side, i.e. − ( π − α ) .
Verify: 2 cos ( − 4 3 π ) = 2 ( − 2 1 ) = − 1 ✓, 2 sin ( − 4 3 π ) = − 1 ✓.
Worked example Example 4 · Cell D ·
z = 1 − 3 i
Forecast: down-and-right. Here the naive tan − 1 actually works (Q IV is inside ( − π /2 , π /2 ) ). Guess: small negative angle.
Step 1. Locate: ( 1 , − 3 ) ≈ ( 1 , − 1.732 ) , Quadrant IV.
Why this step? x > 0 , y < 0 places the point below-and-right, telling us the angle will be a plain negative acute value.
Step 2. Modulus: ∣ z ∣ = 1 2 + ( 3 ) 2 = 1 + 3 = 4 = 2 .
Why this step? Pythagoras on legs 1 and 3 ; the squares 1 + 3 give a clean 4 = 2 .
Step 3. Reference angle: α = tan − 1 1 − 3 = tan − 1 ( 3 ) = 3 π = 6 0 ∘ .
Why this step? α is the acute angle inside the triangle; tan − 1 ( 3 ) is a standard exact value, so we get 6 0 ∘ without a calculator.
Step 4. Quadrant fix: Q IV ⇒ arg z = − α = − 3 π ≈ − 1.0472 rad .
Why just − α ? The point is just below the positive real axis, reached by a clockwise turn of α — no π to subtract.
Verify: 2 cos ( − 3 π ) = 2 ( 0.5 ) = 1 ✓, 2 sin ( − 3 π ) = 2 ( − 2 3 ) = − 3 ✓.
Worked example Example 5 · Cell E · all four axis directions
A point on an axis has one coordinate = 0 , so tan − 1 ( y / x ) either divides by zero or hides the axis behind a bland 0 . The cure is the same every time : read the direction straight off the picture. We do all four axis rays so no borderline is left out.
Case z a = 5 = 5 + 0 i — positive real axis, point ( 5 , 0 ) .
Modulus: ∣ z a ∣ = 5 2 + 0 2 = 5 .
Why? Pythagoras with one leg zero collapses to x 2 = ∣ x ∣ — the modulus is just the distance along the axis.
Argument: the arrow points straight right along + x ⇒ arg z a = 0 .
Why? The argument is the turn from the positive real axis; here we are already on it, so no turn — angle 0 .
Case z b = 2 i = 0 + 2 i — positive imaginary axis, point ( 0 , 2 ) .
Modulus: ∣ z b ∣ = 0 2 + 2 2 = 2 .
Why? One leg zero again, so length = ∣ y ∣ = 2 .
Argument: straight up ⇒ arg z b = 2 π .
Why the formula fails: tan θ = 2/0 is undefined — division by zero. Geometrically the angle is exactly 9 0 ∘ , where tan blows up; that "undefined tan" is precisely the signal for the vertical axis.
Case z c = − 5 = − 5 + 0 i — negative real axis, point ( − 5 , 0 ) .
Modulus: ∣ z c ∣ = ( − 5 ) 2 + 0 2 = 5 .
Why? x 2 = ∣ x ∣ ; the sign vanishes under squaring, so length is 5 .
Argument: the arrow points straight left ⇒ arg z c = π .
Why + π not − π ? The principal range is ( − π , π ] — it includes π but excludes − π , so left-pointing gets the value + π . (Naively tan θ = 0/ ( − 5 ) = 0 would suggest 0 ; the negative sign of x picks π .)
Case z d = − 3 i = 0 − 3 i — negative imaginary axis, point ( 0 , − 3 ) .
Modulus: ∣ z d ∣ = 0 2 + ( − 3 ) 2 = 3 .
Why? y 2 = ∣ y ∣ = 3 ; length is distance down the axis.
Argument: straight down ⇒ arg z d = − 2 π .
Why negative? Below the real axis means a clockwise quarter turn, and tan θ = − 3/0 is again undefined — the vertical-axis signal, now on the downward side.
Verify: 5 cos 0 = 5 , 5 sin 0 = 0 ✓ ; 2 cos 2 π = 0 , 2 sin 2 π = 2 ✓ ; 5 cos π = − 5 , 5 sin π = 0 ✓ ; 3 cos ( − 2 π ) = 0 , 3 sin ( − 2 π ) = − 3 ✓.
Worked example Example 6 · Cell F ·
z = 0
Forecast: the origin. It has no direction — you can't "point at yourself".
Step 1. Modulus: ∣0∣ = 0 2 + 0 2 = 0 .
Why this step? Pythagoras with both legs zero gives length zero; this is the one point whose distance from the origin is 0 , exactly matching the parent note's "∣ z ∣ = 0 only at z = 0 ".
Step 2. Argument: undefined .
Why undefined? Argument is the direction of the arrow O P . When P = O , there is no arrow — length zero, so no direction exists. Any angle would "work", so none is chosen. This is the one input the whole machinery genuinely cannot assign an angle to.
Verify: ∣ z ∣ = 0 ⟺ z = 0 , consistent with the parent note's statement that ∣ z ∣ = 0 only at the origin.
Common mistake Do not write
arg 0 = 0
A calculator forced to compute tan − 1 ( 0/0 ) may spit out 0 , but 0/0 is meaningless. The correct answer is "undefined ", full stop.
Worked example Example 7 · Cell G · watch
arg z as the point slides
Take z = 1 + i y and let y shrink toward 0 + , then grow toward + ∞ . What happens to the argument?
Forecast: as y → 0 + the point flattens onto the positive real axis (angle → 0 ); as y → ∞ it stands up vertical (angle → 2 π ).
Step 1. arg z = tan − 1 ( y /1 ) = tan − 1 ( y ) (Q I throughout, since x = 1 > 0 ).
Why no correction? x > 0 keeps us in Q I/IV where tan − 1 is exact.
Step 2. Limit y → 0 + : tan − 1 ( 0 ) = 0 . The arrow lies along the positive real axis.
Why this step? We are checking the boundary between Cell A and Cell E : as the vertical leg vanishes, the triangle flattens and the angle must collapse to 0 — matching z a 's answer in Ex 5. It confirms the limit and the axis case agree.
Step 3. Limit y → + ∞ : tan − 1 ( y ) → 2 π . The arrow tilts up toward vertical but never quite reaches 9 0 ∘ (it would need x = 0 ).
Why this step? We check the other boundary — toward Cell E's vertical axis. The steepness y / x grows without bound, yet the angle saturates because tan − 1 has a horizontal asymptote at 2 π ; the point approaches the imaginary axis but only reaches it when x is truly 0 .
Numeric checkpoints: tan − 1 ( 1 ) = 4 π ≈ 0.7854 , tan − 1 ( 10 ) ≈ 1.4711 (close to 2 π ≈ 1.5708 ).
Verify: tan − 1 ( 1 ) = 4 π exactly; tan − 1 ( 10 ) < 2 π ✓.
Figure s03 (below) plots arg z = tan − 1 ( y ) against y for Step 1–3: the orange dots mark the two numeric checkpoints and the violet dashed line is the 2 π ceiling the curve never crosses.
Caption s03. Cell G — as y grows the argument climbs but flattens toward the horizontal asymptote π /2 ; it only equals π /2 in the true axis case (Cell E).
Worked example Example 8 · Cell H · a drone's displacement
A drone flies 4 m east then 3 m north from base. Represent its position as a complex number and find its distance and bearing (angle east-of-north not needed — use angle from east).
Forecast: east = real axis, north = imaginary axis. Position = 4 + 3 i . Distance = 5 (it's a 3–4–5!).
Step 1. Model: east → + x , north → + y , so z = 4 + 3 i .
Why this mapping? The Argand plane is just a 2-D map; we align real axis with east.
Step 2. Distance from base: ∣ z ∣ = 4 2 + 3 2 = 5 m .
Units: metres, since both legs are metres.
Step 3. Direction (anticlockwise from east): arg z = tan − 1 ( 3/4 ) ≈ 0.6435 rad ≈ 36.8 7 ∘ .
Why Q I, no fix? Both east and north positive.
Verify: 5 cos 0.6435 = 4 m east ✓, 5 sin 0.6435 = 3 m north ✓. Total straight-line hop 5 m matches the 3–4–5 triangle.
Worked example Example 9 · Cell I · describe
∣ z − ( 1 − 2 i ) ∣ = 4
Forecast: ∣ z − z 0 ∣ is a distance. "Distance from a fixed point is constant" is the definition of a circle. Guess: circle, radius 4 .
Step 1. Match the template ∣ z − z 0 ∣ = r with z 0 = 1 − 2 i , r = 4 .
Why? From the parent note, ∣ z 1 − z 2 ∣ is the straight-line distance between the two points.
Step 2. Interpret: "z stays exactly 4 away from the point ( 1 , − 2 ) ."
Why this step? Turning the algebra into a spoken sentence exposes the geometric constraint hidden in the modulus bars.
Step 3. Conclusion: circle , centre ( 1 , − 2 ) , radius 4 .
Why this step? "The set of all points at a fixed distance r from a fixed centre" is the very definition of a circle — so a constant distance forces the locus to be exactly that circle, no other shape fits.
Cross-check with coordinates: write z = x + i y . Then
∣ ( x − 1 ) + i ( y + 2 ) ∣ = 4 ⟹ ( x − 1 ) 2 + ( y + 2 ) 2 = 16 ,
the standard circle equation with centre ( 1 , − 2 ) , radius 16 = 4 . ✓
Verify: the point ( 1 , 2 ) is on it: ( 1 − 1 ) 2 + ( 2 + 2 ) 2 = 0 + 16 = 16 ✓.
Figure s04 (below) draws that circle: violet centre at ( 1 , − 2 ) , an orange radius arrow of length 4 , and the navy test point ( 1 , 2 ) sitting on the rim (matching the Verify line).
Caption s04. Cell I — the equation ∣ z − ( 1 − 2 i ) ∣ = 4 is the circle of radius 4 centred at ( 1 , − 2 ) .
See Loci in the Complex Plane for perpendicular-bisector and ray loci.
Worked example Example 10 · Cell J ·
z 1 = 1 + i , z 2 = 3 + i ; find arg ( z 1 z 2 )
Forecast: multiplying complex numbers adds their arguments (and multiplies moduli). So don't expand the product first — add the two angles. Guess: 4 5 ∘ + 3 0 ∘ = 7 5 ∘ .
Step 1. arg z 1 : point ( 1 , 1 ) , Q I ⇒ tan − 1 ( 1 ) = 4 π = 4 5 ∘ .
Step 2. arg z 2 : point ( 3 , 1 ) , Q I ⇒ tan − 1 ( 1/ 3 ) = 6 π = 3 0 ∘ .
Step 3. Use the rule arg ( z 1 z 2 ) = arg z 1 + arg z 2 = 4 π + 6 π = 12 5 π = 7 5 ∘ .
Why this rule? See Multiplication as Rotation and Scaling : multiplying by z 2 rotates by arg z 2 . Angles compose by adding.
Step 4. Sanity by direct product: z 1 z 2 = ( 1 + i ) ( 3 + i ) = 3 + i + 3 i + i 2 = ( 3 − 1 ) + ( 3 + 1 ) i .
arg = tan − 1 3 − 1 3 + 1 , 3 − 1 3 + 1 ≈ 0.732 2.732 ≈ 3.732 = tan 7 5 ∘ .
Why this step? The "add the arguments" rule is only trustworthy if it agrees with brute-force multiplication. We expand the product independently and check its angle really is 7 5 ∘ — this closes the loop and shows the rotation rule is not a lucky coincidence.
Verify: tan − 1 ( 3.732 ) ≈ 1.3090 rad = 7 5 ∘ = 12 5 π ✓.
This is the seed of De Moivre's Theorem — repeatedly adding the same argument.
Recall Scenario self-test (hide answers)
For each, name the cell and the fix used.
z = − 3 + 4 i ::: Cell B, Q II, arg = π − α .
z = − 5 ::: Cell E, negative real axis, arg = π .
z = 0 ::: Cell F, argument undefined.
arg ( z 1 z 2 ) for two Q I numbers ::: Cell J, add the arguments.
∣ z − z 0 ∣ = r ::: Cell I, circle centre z 0 radius r .
Mnemonic The quadrant correction, in one breath
"I: as is · II: π minus · III: minus (π minus) · IV: just minus." The reference angle α is always positive; the quadrant only decides how you dress it .
build polar form r cos plus i sin