Exercises — Modulus - z - and argument arg(z)

WHAT we do: identify the real part and imaginary part, then apply Pythagoras.

• $x=3$ (horizontal offset), $y=4$ (vertical offset).
• $|z|=\sqrt{x^2+y^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5.$

WHY $\sqrt{x^2+y^2}$? The arrow to $(3,4)$ is the hypotenuse of a right triangle with legs $3$ and $4$; its length is Pythagoras. Answer: $|z|=5$.

Each point sits on an axis, so no triangle is needed — just point in the right direction.

• $z=7$: on the positive real axis ⇒ $\mathrm{Arg}=0$.
• $z=-7$: on the negative real axis ⇒ $\mathrm{Arg}=\pi$ (not $-\pi$; the range $(-\pi ,\pi ]$ includes $+\pi$ but excludes $-\pi$).
• $z=4i$: on the positive imaginary axis ⇒ straight up ⇒ $\mathrm{Arg}=\frac{\pi }{2}$.
• $z=-4i$: straight down ⇒ $\mathrm{Arg}=-\frac{\pi }{2}$.

Read the two signs $(x,y)$; that alone fixes the quadrant, and quadrant fixes the sign of the angle.

• $z_1=2+5i$: $(+,+)$ ⇒ QI ⇒ $\mathrm{Arg}\in (0,\frac{\pi }{2})$, positive.
• $z_2=-3+i$: $(-,+)$ ⇒ QII ⇒ $\mathrm{Arg}\in (\frac{\pi }{2},\pi )$, positive.
• $z_3=-2-6i$: $(-,-)$ ⇒ QIII ⇒ $\mathrm{Arg}\in (-\pi ,-\frac{\pi }{2})$, negative.
• $z_4=5-i$: $(+,-)$ ⇒ QIV ⇒ $\mathrm{Arg}\in (-\frac{\pi }{2},0)$, negative.

Modulus: $|z|=\sqrt{(-1{)}^2+(\sqrt{3}{)}^2}=\sqrt{1+3}=\sqrt{4}=2.$ Reference angle: $\alpha =\arctan \left|\frac{\sqrt{3}}{-1}\right|=\arctan (\sqrt{3})=\frac{\pi }{3}.$ Quadrant: $x=-1<0,y=\sqrt{3}>0$ ⇒ QII ⇒ $\mathrm{Arg}=\pi -\alpha =\pi -\frac{\pi }{3}=\frac{2\pi }{3}.$ WHY the shift? In QII the arrow leans up-and-left, so measured anticlockwise from the positive real axis it is obtuse — bigger than $\frac{\pi }{2}$. Subtracting $\alpha$ from $\pi$ gives that obtuse angle.

Modulus: $|z|=\sqrt{(-2{)}^2+(-2{)}^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}.$ Reference angle: $\alpha =\arctan \left|\frac{-2}{-2}\right|=\arctan (1)=\frac{\pi }{4}.$ Quadrant: $x<0,y<0$ ⇒ QIII ⇒ $\mathrm{Arg}=\alpha -\pi =\frac{\pi }{4}-\pi =-\frac{3\pi }{4}.$ WHY negative? The arrow points down-and-left. Going anticlockwise from the positive real axis you'd travel more than $\pi$, which leaves the range $(-\pi ,\pi ]$; the shorter, clockwise route is negative, landing at $-\frac{3\pi }{4}$.

Modulus: $r=\sqrt{1^2+(-1{)}^2}=\sqrt{2}.$ Reference angle: $\alpha =\arctan \left|\frac{-1}{1}\right|=\frac{\pi }{4}.$ Quadrant: $x>0,y<0$ ⇒ QIV ⇒ $\theta =-\alpha =-\frac{\pi }{4}.$ Polar form: $z=\sqrt{2}\left(\cos \left(-\frac{\pi }{4}\right)+i\sin \left(-\frac{\pi }{4}\right)\right).$ Check: $\sqrt{2}\cos (-\frac{\pi }{4})=\sqrt{2}\cdot \frac{1}{\sqrt{2}}=1$ ✓ and $\sqrt{2}\sin (-\frac{\pi }{4})=\sqrt{2}\cdot (-\frac{1}{\sqrt{2}})=-1$ ✓.

WHY not multiply? Because modulus is multiplicative: $|z_1{z}_2|=|z_1||z_2|$ (proved in the parent note via $z\bar{z}$). So we only need two easy Pythagoras sums.

• $|z_1|=\sqrt{3^2+4^2}=\sqrt{25}=5.$
• $|z_2|=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13.$
• $|z_1{z}_2|=5\times 13=65.$ (Sanity check: $z_1{z}_2=(3+4i)(5-12i)=15-36i+20i-48i^2=63-16i$, and $\sqrt{63^2+16^2}=\sqrt{3969+256}=\sqrt{4225}=65$ ✓.)

WHY add? Because $\arg (z_1{z}_2)=\arg z_1+\arg z_2(\mathrm{mod}2\pi )$ (angles add under multiplication — the "rotate" half of scale-and-rotate).

• $z_1=1+i$: QI, $\alpha =\arctan 1=\frac{\pi }{4}$ ⇒ $\mathrm{Arg}{z}_1=\frac{\pi }{4}.$
• $z_2=-1+i\sqrt{3}$: QII, $\mathrm{Arg}{z}_2=\frac{2\pi }{3}$ (from L2·Q1).
• Sum: $\frac{\pi }{4}+\frac{2\pi }{3}=\frac{3\pi }{12}+\frac{8\pi }{12}=\frac{11\pi }{12}.$

Is $\frac{11\pi }{12}$ in $(-\pi ,\pi ]$? Yes ($\frac{11\pi }{12}<\pi$), so no wrap-around needed. $\mathrm{Arg}(z_1{z}_2)=\frac{11\pi }{12}.$

• $z_2=-1-i\sqrt{3}$: QIII, $\alpha =\arctan \sqrt{3}=\frac{\pi }{3}$ ⇒ $\mathrm{Arg}{z}_2=\frac{\pi }{3}-\pi =-\frac{2\pi }{3}.$
• Sum of arguments: $\frac{2\pi }{3}+\left(-\frac{2\pi }{3}\right)=0.$ Answer: $\mathrm{Arg}(z_1{z}_2)=0.$ WHAT it means geometrically: $z_2=\overline{z_1}$ (conjugate; see Conjugate of a complex number). Multiplying a number by its conjugate gives $z\bar{z}=|z{|}^2$, a positive real number — which sits on the positive real axis, argument $0$. Directions cancelled exactly. ✓

WHY polar? Because a power multiplies the number by itself repeatedly, and under multiplication moduli multiply, arguments add. Raising to the $n$-th power therefore raises the modulus to the $n$-th and multiplies the argument by $n$ — this is De Moivre's Theorem: $(r(\cos \theta +i\sin \theta ){)}^n=r^n(\cos n\theta +i\sin n\theta ).$

• $1+i$: $r=\sqrt{2},\theta =\frac{\pi }{4}$ (QI).
• $r^8=(\sqrt{2}{)}^8=2^4=16.$
• $n\theta =8\cdot \frac{\pi }{4}=2\pi .$
• $(1+i{)}^8=16(\cos 2\pi +i\sin 2\pi )=16(1+0)=16.$ Answer: $(1+i{)}^8=16.$ (A real number — the eight quarter-turns of $\frac{\pi }{4}$ add up to two full turns, landing back on the positive real axis.)

WHY these rules? Division is the inverse of multiplication, so moduli divide and arguments subtract: $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|},\arg \left(\frac{z_1}{z_2}\right)=\arg z_1-\arg z_2(\mathrm{mod}2\pi ).$

• $|z_1|=2,|z_2|=\sqrt{2}$ ⇒ $\left|\frac{z_1}{z_2}\right|=\frac{2}{\sqrt{2}}=\sqrt{2}.$
• $\mathrm{Arg}{z}_1=\frac{2\pi }{3},\mathrm{Arg}{z}_2=\frac{\pi }{4}$ ⇒ difference $=\frac{2\pi }{3}-\frac{\pi }{4}=\frac{8\pi -3\pi }{12}=\frac{5\pi }{12}.$ $\frac{5\pi }{12}\in (-\pi ,\pi ]$, so $\mathrm{Arg}\left(\frac{z_1}{z_2}\right)=\frac{5\pi }{12}.$

WHAT the two conditions mean: $|z|=2$ pins the distance (a circle of radius $2$); $\mathrm{Arg}(z)=\frac{\pi }{6}$ pins the direction (a ray from the origin at $30^{\circ }$). Their intersection is a single point. Using $x=r\cos \theta ,y=r\sin \theta$:

• $x=2\cos \frac{\pi }{6}=2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}.$
• $y=2\sin \frac{\pi }{6}=2\cdot \frac{1}{2}=1.$ Answer: $z=\sqrt{3}+i.$ (Check: $|z|=\sqrt{3+1}=2$ ✓, QI with $\alpha =\arctan \frac{1}{\sqrt{3}}=\frac{\pi }{6}$ ✓.)

WHY polar? A cube root $u$ satisfies $u^3=w$; taking moduli and arguments turns this into two scalar equations, one for length and one for angle. Step 1 — put $w$ in polar.

• $|w|=\sqrt{(-4{)}^2+(4\sqrt{3}{)}^2}=\sqrt{16+48}=\sqrt{64}=8.$
• $w=-4+4\sqrt{3}i$ is in QII ($x<0,y>0$). Reference angle $\alpha =\arctan \left|\frac{4\sqrt{3}}{-4}\right|=\arctan \sqrt{3}=\frac{\pi }{3}$, so $\mathrm{Arg}(w)=\pi -\frac{\pi }{3}=\frac{2\pi }{3}.$
• So $w=8(\cos (\frac{2\pi }{3}+2\pi k)+i\sin (\frac{2\pi }{3}+2\pi k))$ for integers $k$.

Step 2 — take cube roots (De Moivre). A root $u=\rho (\cos \varphi +i\sin \varphi )$ needs:

• ${\rho }^3=8\Rightarrow \rho =2.$
• $3\varphi =\frac{2\pi }{3}+2\pi k\Rightarrow \varphi =\frac{2\pi }{9}+\frac{2\pi k}{3},k=0,1,2.$

Step 3 — list the three angles (and drag each into $(-\pi ,\pi ]$):

• $k=0:{\varphi }_0=\frac{2\pi }{9}(=40^{\circ })$ — QI.
• $k=1:{\varphi }_1=\frac{2\pi }{9}+\frac{2\pi }{3}=\frac{2\pi }{9}+\frac{6\pi }{9}=\frac{8\pi }{9}(=160^{\circ })$ — QII ✓.
• $k=2:{\varphi }_2=\frac{2\pi }{9}+\frac{4\pi }{3}=\frac{2\pi }{9}+\frac{12\pi }{9}=\frac{14\pi }{9}(=280^{\circ })$; subtract $2\pi$: $\frac{14\pi }{9}-\frac{18\pi }{9}=-\frac{4\pi }{9}(=-80^{\circ })$ — QIV.

Step 4 — the QII root. It is $u_1=2\left(\cos \frac{8\pi }{9}+i\sin \frac{8\pi }{9}\right)$, with modulus $2$ and argument $\frac{8\pi }{9}$. Cartesian form: $u_1=2\cos \frac{8\pi }{9}+2i\sin \frac{8\pi }{9}\approx 2(-0.9397)+2i(0.3420)\approx -1.879+0.684i.$ (Check the direction: $x<0,y>0$ — genuinely QII, exactly as asked.) Take-away: the three roots sit on a circle of radius $2$, spaced $\frac{2\pi }{3}=120^{\circ }$ apart — one per quadrant band, landing in QI, QII, and QIV here.

WHY use $z\bar{z}$? Because $|w{|}^2=w\bar{w}$ turns "length squared" into ordinary multiplication, and multiplication is easy to expand. Use $\overline{a+b}=\bar{a}+\bar{b}$. $|z_1+z_2{|}^2=(z_1+z_2)({\bar{z}}_1+{\bar{z}}_2)=z_1{\bar{z}}_1+z_1{\bar{z}}_2+z_2{\bar{z}}_1+z_2{\bar{z}}_2.$ $|z_1-z_2{|}^2=(z_1-z_2)({\bar{z}}_1-{\bar{z}}_2)=z_1{\bar{z}}_1-z_1{\bar{z}}_2-z_2{\bar{z}}_1+z_2{\bar{z}}_2.$ Add: the cross terms $\pm (z_1{\bar{z}}_2+z_2{\bar{z}}_1)$ cancel, leaving $2z_1{\bar{z}}_1+2z_2{\bar{z}}_2=2|z_1{|}^2+2|z_2{|}^2.■$ WHAT IT MEANS: in the parallelogram spanned by $z_1$ and $z_2$, the sum of the squares of the two diagonals equals the sum of the squares of all four sides. A geometry theorem, proven by algebra.

(i) $z=0$: $|z|=\sqrt{0^2+0^2}=0$. But $\mathrm{Arg}(0)$ is undefined — a zero-length arrow points nowhere, so "direction" has no meaning. This is the one genuinely degenerate input in the whole theory: modulus exists (it is $0$), argument does not. (ii) $z=-3$: $|z|=3$; on the negative real axis so $\mathrm{Arg}=\pi$ (the range endpoint, not $-\pi$). (iii) The number is already in polar form with $r=1$ and stated angle $\frac{7\pi }{6}$. But $\frac{7\pi }{6}>\pi$, so it is not a principal argument. Bring it into range by subtracting $2\pi$: $\frac{7\pi }{6}-2\pi =-\frac{5\pi }{6}.$ So $|z|=1$ and $\mathrm{Arg}(z)=-\frac{5\pi }{6}$ (the arrow points down-and-left, QIII). ✓

• L1 (Recognition): read the plane — modulus is a straight-line distance, argument is a direction, axis points are memorised.
• L2 (Application): reference angle then quadrant is the universal recipe; it never fails, and never needs a raw $\arctan$.
• L3 (Analysis): modulus is multiplicative and argument is additive — but arguments only add mod $2\pi$, so always land back in $(-\pi ,\pi ]$.
• L4 (Synthesis): powers and quotients are the same two rules pushed harder (De Moivre for powers, subtract-and-divide for quotients); a modulus + argument pair pins a unique point.
• L5 (Mastery): roots spread evenly around a circle; identities like the parallelogram law fall out of $|w{|}^2=w\bar{w}$; and the sole degenerate case is $z=0$, where the modulus is $0$ but the argument simply does not exist.

If you can reproduce every solution above with the collapsibles closed, you own this topic.