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Module 12 · L08 of 16 ~40 min ⚡ +90 XP available

Division in Polar Form

In Cartesian form, dividing complex numbers requires multiplying by the conjugate and chasing a denominator $a^2 + b^2$. In polar form, division is the mirror image of multiplication: divide the moduli, subtract the arguments. Behind that one-line rule sits the elegant idea that dividing by $z$ is the same as multiplying by $\tfrac{1}{z}$ — and $\tfrac{1}{z}$ is just a reflection and rescaling in the Argand plane.

Today's hook — Let $z_1 = 12(\cos 80^\circ + i\sin 80^\circ)$ and $z_2 = 3(\cos 20^\circ + i\sin 20^\circ)$. Without expanding into Cartesian form, predict $\left|\tfrac{z_1}{z_2}\right|$ and $\arg\left(\tfrac{z_1}{z_2}\right)$. Then explain geometrically what dividing by $z_2$ does to the vector $z_1$. Check after card 05.

0/5QUESTS

Recall — your gut answer first

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Simplify $\dfrac{1 + i\sqrt 3}{1 + i}$ in Cartesian form using the conjugate trick. Then write each of $1 + i\sqrt 3$ and $1 + i$ in polar form and compute $\left|\tfrac{1 + i\sqrt 3}{1 + i}\right|$ and $\arg\left(\tfrac{1 + i\sqrt 3}{1 + i}\right)$. Do moduli divide or subtract? Do arguments divide or subtract?

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The two moves for dividing in polar form

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To divide two complex numbers given in polar form, two habits do all the work: divide the moduli and subtract the arguments. That is the entire rule — the mirror image of the multiplication rule from L07. Watch the order of subtraction: $\arg(z_1/z_2) = \arg z_1 - \arg z_2$ (numerator minus denominator).

The shrink-then-counter-rotate reading: dividing $z_1$ by $z_2$ (1) shrinks the length of $z_1$ by the factor $|z_2|$, and (2) rotates $z_1$ clockwise about $O$ by the angle $\arg z_2$.

$\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}\,[\cos(\alpha-\beta) + i\sin(\alpha-\beta)]$

$\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}, \;\; \arg\!\left(\dfrac{z_1}{z_2}\right) = \arg z_1 - \arg z_2$

Moduli divide

$\left|\tfrac{z_1}{z_2}\right| = \tfrac{|z_1|}{|z_2|}$. The length of the quotient is the ratio of the lengths — never the difference.

Arguments subtract

$\arg\!\left(\tfrac{z_1}{z_2}\right) = \arg z_1 - \arg z_2$ (mod $2\pi$). Numerator's argument minus denominator's argument — order matters.

Reduce to principal value

If $\alpha - \beta$ falls outside $(-\pi, \pi]$, add or subtract $2\pi$ to bring it back. The geometric direction is unchanged.

What you'll master

Know

Key facts

$\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}[\cos(\alpha-\beta) + i\sin(\alpha-\beta)]$
$\left|\tfrac{z_1}{z_2}\right| = \tfrac{|z_1|}{|z_2|}$ (moduli divide)
$\arg\!\left(\tfrac{z_1}{z_2}\right) = \arg z_1 - \arg z_2$ (arguments subtract, mod $2\pi$)
$\tfrac{1}{z} = \tfrac{1}{r}\,\text{cis}(-\theta)$ for $z = r\,\text{cis}\,\theta$

Understand

Concepts

Why dividing by $z$ is the same as multiplying by $\tfrac{1}{z}$, and why this gives the subtraction rule
Why $\tfrac{1}{z}$ has modulus $\tfrac{1}{|z|}$ and argument $-\arg z$
How polar division avoids the conjugate-trick mess of Cartesian division

Can do

Skills

Divide two complex numbers given in polar form quickly and reliably
Compute $\tfrac{1}{z}$ in polar form and confirm it equals $\bar z / |z|^2$ in Cartesian form
Use polar division to evaluate Cartesian quotients without the conjugate trick

Key terms

Reciprocal $\tfrac{1}{z}$For $z = r\,\text{cis}\,\theta \neq 0$, $\tfrac{1}{z} = \tfrac{1}{r}\,\text{cis}(-\theta)$. The modulus inverts and the argument negates.

Conjugate $\bar z$For $z = a + ib$, $\bar z = a - ib$. In polar form, $\overline{r\,\text{cis}\,\theta} = r\,\text{cis}(-\theta)$. So $\tfrac{1}{z} = \tfrac{\bar z}{|z|^2}$.

Polar quotient rule$\tfrac{z_1}{z_2} = \tfrac{r_1}{r_2}\,\text{cis}(\alpha - \beta)$ — moduli divide, arguments subtract (numerator minus denominator).

Principal argumentThe unique value of $\arg z$ in $(-\pi, \pi]$. After a quotient, you may need to add or subtract $2\pi$ to bring the argument into this range.

Cartesian division$\tfrac{a + ib}{c + id} = \tfrac{(a+ib)(c-id)}{c^2 + d^2}$. Algebraically valid but rarely the easiest route — polar form is usually cleaner.

Rotation by $-\theta$Multiplying by $\text{cis}(-\theta)$ is a clockwise rotation by $\theta$. Dividing by $\text{cis}\,\theta$ is the same operation.

MEX-N1NESA outcome (Complex Numbers I): uses the polar form of complex numbers, including multiplication and division, and connects quotients to scaling and clockwise rotation in the Argand plane.

The polar quotient rule and its proof

core concept

Let $z_1 = r_1(\cos\alpha + i\sin\alpha)$ and $z_2 = r_2(\cos\beta + i\sin\beta)$ with $z_2 \neq 0$. Then

$$\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}\big[\cos(\alpha-\beta) + i\sin(\alpha-\beta)\big].$$

Proof (via the reciprocal). Use $\tfrac{z_1}{z_2} = z_1 \cdot \tfrac{1}{z_2}$ and find $\tfrac{1}{z_2}$ first:

$\dfrac{1}{z_2} = \dfrac{1}{r_2(\cos\beta + i\sin\beta)} \cdot \dfrac{\cos\beta - i\sin\beta}{\cos\beta - i\sin\beta} = \dfrac{\cos\beta - i\sin\beta}{r_2(\cos^2\beta + \sin^2\beta)} = \dfrac{1}{r_2}(\cos\beta - i\sin\beta)$.
Using $\cos(-\beta) = \cos\beta$ and $\sin(-\beta) = -\sin\beta$: $\dfrac{1}{z_2} = \dfrac{1}{r_2}\big[\cos(-\beta) + i\sin(-\beta)\big]$.
So $\tfrac{1}{z_2}$ has modulus $\tfrac{1}{r_2}$ and argument $-\beta$.
By the polar product rule from L07: $\dfrac{z_1}{z_2} = z_1 \cdot \tfrac{1}{z_2}$ has modulus $r_1 \cdot \tfrac{1}{r_2} = \tfrac{r_1}{r_2}$ and argument $\alpha + (-\beta) = \alpha - \beta$.
$\therefore \dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}\big[\cos(\alpha-\beta) + i\sin(\alpha-\beta)\big]$. $\blacksquare$

Reading off: $\left|\tfrac{z_1}{z_2}\right| = \tfrac{r_1}{r_2} = \tfrac{|z_1|}{|z_2|}$ and $\arg\!\left(\tfrac{z_1}{z_2}\right) = \alpha - \beta = \arg z_1 - \arg z_2$ (reduce mod $2\pi$ to the principal value if needed).

Connecting to the hook. $z_1 = 12\,\text{cis}\,80^\circ$ and $z_2 = 3\,\text{cis}\,20^\circ$. By the rule: $\left|\tfrac{z_1}{z_2}\right| = \tfrac{12}{3} = 4$ and $\arg\!\left(\tfrac{z_1}{z_2}\right) = 80^\circ - 20^\circ = 60^\circ$. So $\tfrac{z_1}{z_2} = 4\,\text{cis}\,60^\circ$. Geometrically, dividing by $z_2$ shrinks $z_1$ by a factor of 3 and rotates it $20^\circ$ clockwise — from a length-12 vector at $80^\circ$ to a length-4 vector at $60^\circ$.

Polar quotient rule: $\tfrac{z_1}{z_2} = \tfrac{r_1}{r_2}\,\text{cis}(\alpha-\beta)$ · Moduli divide: $\left|\tfrac{z_1}{z_2}\right| = \tfrac{|z_1|}{|z_2|}$ · Arguments subtract: $\arg\!\left(\tfrac{z_1}{z_2}\right) = \arg z_1 - \arg z_2$ (mod $2\pi$) · Reciprocal: $\tfrac{1}{r\,\text{cis}\,\theta} = \tfrac{1}{r}\,\text{cis}(-\theta)$

Pause — copy the quotient rule $z_1/z_2 = (r_1/r_2)\,\text{cis}(\alpha-\beta)$, the law $|z_1/z_2| = |z_1|/|z_2|$, and the reciprocal $1/(r\,\text{cis}\theta) = (1/r)\,\text{cis}(-\theta)$ into your book.

Quick check: If $z_1 = 20\,\text{cis}\tfrac{5\pi}{6}$ and $z_2 = 4\,\text{cis}\tfrac{\pi}{3}$, what is $\dfrac{z_1}{z_2}$ in polar form?

Division = multiplication by the reciprocal

core concept

We just saw the polar quotient rule $z_1/z_2 = (r_1/r_2)\,\text{cis}(\alpha-\beta)$ — moduli divide, arguments subtract. That raises a question: how does the reciprocal $1/z$ fit into this framework, and what is its connection to the conjugate formula? This card answers it → $1/(r\,\text{cis}\theta) = (1/r)\,\text{cis}(-\theta)$, which is the same as $\bar{z}/|z|^2$ in Cartesian form.

The neatest way to think about polar division is to realise it is the product rule applied to $z_1 \cdot \tfrac{1}{z_2}$. The reciprocal $\tfrac{1}{z}$ has a clean polar form:

$$z = r\,\text{cis}\,\theta \;\Rightarrow\; \dfrac{1}{z} = \dfrac{1}{r}\,\text{cis}(-\theta).$$

Geometric reading of $\tfrac{1}{z}$: reflect $z$ in the real axis (giving $\bar z$ with argument $-\theta$), then scale by $\tfrac{1}{|z|^2}$. In one step: modulus inverts, argument negates.

Special cases worth knowing by heart:

$\tfrac{1}{i} = \text{cis}(-\tfrac{\pi}{2}) = -i$ (and indeed $\tfrac{1}{i} = \tfrac{-i}{i \cdot -i} = \tfrac{-i}{1} = -i$).
$\tfrac{1}{\text{cis}\,\theta} = \text{cis}(-\theta)$ — pure clockwise rotation when modulus is 1.
$\tfrac{1}{z} = \tfrac{\bar z}{|z|^2}$ — the polar identity rewritten in Cartesian terms.
Dividing by $\text{cis}\,\theta$ is the same as multiplying by $\text{cis}(-\theta)$ — i.e., rotating clockwise by $\theta$.

Common mistake. Students sometimes write $\arg(z_1/z_2) = \arg z_2 - \arg z_1$ (wrong order). The rule mirrors the algebra: numerator's argument minus denominator's argument. Reversing the order produces a quotient with the wrong sign on its argument — the reflection of the right answer in the real axis.

Division = multiplication by reciprocal: $\tfrac{z_1}{z_2} = z_1 \cdot \tfrac{1}{z_2}$ · $\tfrac{1}{r\,\text{cis}\,\theta} = \tfrac{1}{r}\,\text{cis}(-\theta)$ · $\tfrac{1}{i} = -i$ · $\tfrac{1}{z} = \tfrac{\bar z}{|z|^2}$ — same idea, Cartesian dress · Argument: numerator minus denominator (order matters)

Pause — copy $z_1/z_2 = z_1 \cdot (1/z_2)$, the Cartesian form $1/z = \bar{z}/|z|^2$, the special case $1/i = -i$, and the argument rule (numerator minus denominator) into your book.

Did you get this? True or false: for any non-zero complex number $z = r\,\text{cis}\,\theta$, the reciprocal $\tfrac{1}{z}$ satisfies $\left|\tfrac{1}{z}\right| = \tfrac{1}{r}$ and $\arg\!\left(\tfrac{1}{z}\right) = -\theta$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · STRAIGHT APPLICATION

Given $z_1 = 10\,\text{cis}\tfrac{3\pi}{4}$ and $z_2 = 2\,\text{cis}\tfrac{\pi}{6}$, find $\dfrac{z_1}{z_2}$ in polar form (principal argument) and in Cartesian form.

Apply the rule: $\left|\tfrac{z_1}{z_2}\right| = \tfrac{10}{2} = 5$ and $\arg\!\left(\tfrac{z_1}{z_2}\right) = \tfrac{3\pi}{4} - \tfrac{\pi}{6} = \tfrac{9\pi}{12} - \tfrac{2\pi}{12} = \tfrac{7\pi}{12}$.

Moduli divide, arguments subtract. Use a common denominator when subtracting fractions of $\pi$.

PROBLEM 2 · CARTESIAN TO POLAR THEN DIVIDE

Compute $\dfrac{1 + i\sqrt 3}{1 + i}$ by first converting numerator and denominator to polar form. Give the answer in polar form, then in Cartesian form.

Convert: $1 + i\sqrt 3 = 2\,\text{cis}\tfrac{\pi}{3}$ (modulus $\sqrt{1+3} = 2$, argument $\tan^{-1}\sqrt 3 = \tfrac{\pi}{3}$). $1 + i = \sqrt 2\,\text{cis}\tfrac{\pi}{4}$ (modulus $\sqrt 2$, argument $\tfrac{\pi}{4}$).

Both points are in the first quadrant so $\arg z = \tan^{-1}(b/a)$ applies directly with no quadrant adjustment.

PROBLEM 3 · RECIPROCAL AND GEOMETRIC EFFECT

Let $z = 3 + 4i$. Find $\tfrac{1}{z}$ in (a) polar form and (b) Cartesian form. Describe the geometric transformation that takes $z$ to $\tfrac{1}{z}$.

$|z| = \sqrt{9 + 16} = 5$ and $\arg z = \tan^{-1}(4/3) \approx 0.927$ rad (first quadrant). So $z = 5\,\text{cis}(\tan^{-1}\tfrac{4}{3})$.

For Cartesian-form points, compute modulus and argument first, then take the reciprocal in polar form — it's quicker than the conjugate trick.

Fill the gap: If $z_1 = r_1\,\text{cis}\,\alpha$ and $z_2 = r_2\,\text{cis}\,\beta$ with $z_2 \neq 0$, then $\dfrac{z_1}{z_2} = $ $\,\text{cis}\,($ $)$. Moduli divide, arguments subtract.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Subtracting the arguments in the wrong order

$\arg(z_1/z_2) = \arg z_1 - \arg z_2$ — numerator minus denominator. Reversing the order gives the reflection of the correct answer in the real axis (a sign error on the argument). Mirror the algebra: top minus bottom.

Trap 02

Confusing division with subtraction

A common slip is $|z_1/z_2| = |z_1| - |z_2|$ or $\arg(z_1/z_2) = \arg z_1 / \arg z_2$. The correct pairing: moduli divide, arguments subtract — the opposite operations to what students sometimes write. Anchor it next to the L07 product rule (moduli multiply, arguments add).

Trap 03

Forgetting to reduce to principal argument

If $\alpha - \beta > \pi$ or $\leq -\pi$, the answer is geometrically correct but not in principal form. Adjust by $\pm 2\pi$. Example: $\arg z_1 = -\tfrac{3\pi}{4}$, $\arg z_2 = \tfrac{3\pi}{4}$ gives $-\tfrac{3\pi}{2}$, which should be written as $\tfrac{\pi}{2}$.

Did you get this? True or false: if $z_1 = 6\,\text{cis}(-\tfrac{3\pi}{4})$ and $z_2 = 2\,\text{cis}\tfrac{3\pi}{4}$, then $\dfrac{z_1}{z_2} = 3\,\text{cis}(-\tfrac{3\pi}{2})$, and this argument is in principal form.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Compute $\dfrac{15\,\text{cis}\tfrac{2\pi}{3}}{3\,\text{cis}\tfrac{\pi}{4}}$ in polar form with principal argument.

Find $\tfrac{1}{z}$ in both polar and Cartesian form for $z = 1 - i$. Verify using the identity $\tfrac{1}{z} = \tfrac{\bar z}{|z|^2}$.

Use polar form to simplify $\dfrac{\sqrt 3 - i}{1 + i}$.

Let $z_1 = 4\,\text{cis}\tfrac{\pi}{6}$ and $z_2 = 2\,\text{cis}(-\tfrac{5\pi}{6})$. Find $\tfrac{z_1}{z_2}$ in polar form with principal argument.

Prove using the polar quotient rule and the product rule: $\left|\dfrac{z_1 z_2}{z_3}\right| = \dfrac{|z_1||z_2|}{|z_3|}$ and $\arg\!\left(\dfrac{z_1 z_2}{z_3}\right) = \arg z_1 + \arg z_2 - \arg z_3$ (mod $2\pi$), for $z_3 \neq 0$.

Odd one out: Three of the following statements about $\dfrac{z_1}{z_2}$ (with $z_1 = r_1\,\text{cis}\,\alpha$, $z_2 = r_2\,\text{cis}\,\beta$, $z_2 \neq 0$) are correct. Which one is NOT?

Revisit your thinking

Earlier you predicted $\left|\tfrac{z_1}{z_2}\right|$ and $\arg\!\left(\tfrac{z_1}{z_2}\right)$ for $z_1 = 12\,\text{cis}\,80^\circ$ and $z_2 = 3\,\text{cis}\,20^\circ$.

By the polar quotient rule: $\left|\tfrac{z_1}{z_2}\right| = \tfrac{12}{3} = 4$ and $\arg\!\left(\tfrac{z_1}{z_2}\right) = 80^\circ - 20^\circ = 60^\circ$, so $\tfrac{z_1}{z_2} = 4\,\text{cis}\,60^\circ$. Geometrically, dividing by $z_2$ shrinks the vector $\overrightarrow{Oz_1}$ by factor 3 and rotates it $20^\circ$ clockwise — the exact mirror of multiplying by $z_2$, which would scale by 3 and rotate $20^\circ$ anticlockwise. Reading every quotient as "shrink then counter-rotate" is the cleanest way to handle Cartesian division in Module 12.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Find the quotient $\dfrac{8\,\text{cis}\tfrac{5\pi}{12}}{2\,\text{cis}\tfrac{\pi}{4}}$ in polar form, with argument in $(-\pi, \pi]$. (2 marks)

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ApplyBand 43 marks

Q2. Express $z = -\sqrt 3 + i$ in polar form. Hence find $\tfrac{1}{z}$ in polar form and in Cartesian form. (3 marks)

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AnalyseBand 53 marks

Q3. Let $z_1 = 2\,\text{cis}\tfrac{\pi}{3}$ and $z_2 = 4\,\text{cis}(-\tfrac{2\pi}{3})$. Find $\dfrac{z_1}{z_2}$ in polar form with principal argument. Then describe geometrically what division by $z_2$ does to any non-zero complex number $z$, justifying with reference to $|z_2|$ and $\arg z_2$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Modulus $\tfrac{15}{3} = 5$; argument $\tfrac{2\pi}{3} - \tfrac{\pi}{4} = \tfrac{8\pi}{12} - \tfrac{3\pi}{12} = \tfrac{5\pi}{12}$. Answer: $5\,\text{cis}\tfrac{5\pi}{12}$.

2. $z = 1 - i$: $|z| = \sqrt 2$, $\arg z = -\tfrac{\pi}{4}$. Polar: $\tfrac{1}{z} = \tfrac{1}{\sqrt 2}\,\text{cis}\tfrac{\pi}{4} = \tfrac{1}{\sqrt 2}\!\left(\tfrac{1}{\sqrt 2} + i\tfrac{1}{\sqrt 2}\right) = \tfrac{1}{2} + \tfrac{1}{2}i$. Check $\tfrac{\bar z}{|z|^2} = \tfrac{1 + i}{2} = \tfrac{1}{2} + \tfrac{1}{2}i$. $\checkmark$

3. $\sqrt 3 - i = 2\,\text{cis}(-\tfrac{\pi}{6})$; $1 + i = \sqrt 2\,\text{cis}\tfrac{\pi}{4}$. Quotient: $\tfrac{2}{\sqrt 2}\,\text{cis}(-\tfrac{\pi}{6} - \tfrac{\pi}{4}) = \sqrt 2\,\text{cis}(-\tfrac{5\pi}{12})$.

4. Modulus $\tfrac{4}{2} = 2$; argument $\tfrac{\pi}{6} - (-\tfrac{5\pi}{6}) = \tfrac{\pi}{6} + \tfrac{5\pi}{6} = \pi$. Answer: $2\,\text{cis}\,\pi = -2$.

5. By the product rule, $|z_1 z_2| = |z_1||z_2|$ and $\arg(z_1 z_2) = \arg z_1 + \arg z_2$. Then by the quotient rule applied to $(z_1 z_2)/z_3$: modulus $= \tfrac{|z_1 z_2|}{|z_3|} = \tfrac{|z_1||z_2|}{|z_3|}$; argument $= \arg(z_1 z_2) - \arg z_3 = \arg z_1 + \arg z_2 - \arg z_3$ (mod $2\pi$). $\blacksquare$

Q1 (2 marks): Moduli divide: $\tfrac{8}{2} = 4$ [1]. Arguments subtract: $\tfrac{5\pi}{12} - \tfrac{\pi}{4} = \tfrac{5\pi}{12} - \tfrac{3\pi}{12} = \tfrac{\pi}{6}$. Answer: $4\,\text{cis}\tfrac{\pi}{6}$ [1].

Q2 (3 marks): $|z| = \sqrt{(-\sqrt 3)^2 + 1^2} = 2$; $z$ is in the second quadrant ($a < 0, b > 0$). Reference angle $\tan^{-1}\tfrac{1}{\sqrt 3} = \tfrac{\pi}{6}$, so $\arg z = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$. Hence $z = 2\,\text{cis}\tfrac{5\pi}{6}$ [1]. Reciprocal: $\tfrac{1}{z} = \tfrac{1}{2}\,\text{cis}(-\tfrac{5\pi}{6})$, in principal range [1]. Cartesian: $\tfrac{1}{2}\!\left(\cos(-\tfrac{5\pi}{6}) + i\sin(-\tfrac{5\pi}{6})\right) = \tfrac{1}{2}\!\left(-\tfrac{\sqrt 3}{2} - \tfrac{i}{2}\right) = -\tfrac{\sqrt 3}{4} - \tfrac{i}{4}$ [1].

Q3 (3 marks): Quotient: modulus $\tfrac{2}{4} = \tfrac{1}{2}$; argument $\tfrac{\pi}{3} - (-\tfrac{2\pi}{3}) = \pi$. Answer: $\tfrac{1}{2}\,\text{cis}\,\pi = -\tfrac{1}{2}$ [1]. Geometric effect of $z \mapsto \tfrac{z}{z_2}$: shrink $z$ by the factor $|z_2| = 4$ (giving $\tfrac{|z|}{4}$) and rotate by $-\arg z_2 = \tfrac{2\pi}{3}$, i.e., anticlockwise by $\tfrac{2\pi}{3}$ [1]. Justification: $\tfrac{z}{z_2} = z \cdot \tfrac{1}{z_2}$ and $\tfrac{1}{z_2} = \tfrac{1}{4}\,\text{cis}\tfrac{2\pi}{3}$, so multiplication by $\tfrac{1}{z_2}$ scales by $\tfrac{1}{4}$ and rotates by $\tfrac{2\pi}{3}$ [1].

Boss battle · The Polar Divider

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Five timed questions on polar division, the reciprocal rule, and the shrink-counter-rotate interpretation. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering quick polar-division questions. Lighter alternative to the boss.

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Module overview · Extension 2 · Checkpoint 1