Polar form

This chapter introduces the polar form of complex numbers, a fundamental representation that significantly simplifies operations like exponentiation and root extraction. Mastery of De Moivre's Theorem and its applications to powers and roots of complex numbers, including roots of unity, is crucial for solving advanced problems in complex analysis often encountered in examinations.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Polar representation | | 2 | De Moivre theorem | | 3 | Powers of complex numbers | | 4 | Roots of complex numbers | | 5 | Roots of unity |

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We begin with Polar representation.

Part 1: Polar representation

Polar Representation

Overview

The polar representation of a complex number converts algebraic form $\qquad z=x+iy$ into a geometric form using its modulus and argument. This is one of the most powerful ways to multiply, divide, take powers, and extract roots of complex numbers. In CMI-style questions, the real skill is not just writing down the formula, but using it cleanly in computation and proof. ---

Learning Objectives

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Core Definitions

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Why Polar Form Works

Since $\qquad x=r\cos\theta,\qquad y=r\sin\theta$ we can rewrite $\qquad z=x+iy=r\cos\theta+ir\sin\theta=r(\cos\theta+i\sin\theta)$ So polar form is just the geometric form of the same complex number. ---

Non-Uniqueness of the Argument

So the polar form is not unique unless we restrict the argument. ---

Principal Argument

Examples:

• $\Arg(1+i)=\dfrac{\pi}{4}$

• $\Arg(-1+i)=\dfrac{3\pi}{4}$

• $\Arg(-1-i)=-\dfrac{3\pi}{4}$

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Multiplication and Division

These are the main reasons polar form is useful. ---

De Moivre's Theorem

This turns powers of complex numbers into straightforward angle multiplication. ---

Roots of a Complex Number

These are $n$ distinct roots, equally spaced on a circle. ---

Minimal Worked Examples

Example 1 Write $z=-1+i\sqrt{3}$ in polar form. First, $\qquad |z|=\sqrt{(-1)^2+(\sqrt{3})^2}=2$ The point lies in the second quadrant, and $\qquad \tan\theta=\dfrac{\sqrt{3}}{-1}$ So the principal argument is $\qquad \Arg z=\dfrac{2\pi}{3}$ Hence $\qquad z=2\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)$ --- Example 2 Compute $(1+i)^8$ using polar form. First, $\qquad 1+i=\sqrt{2}\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$ So $\qquad (1+i)^8=(\sqrt{2})^8\left(\cos 2\pi+i\sin 2\pi\right)=16$ Thus $\qquad (1+i)^8=16$ ---

Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The polar form of $-1+i\sqrt{3}$ is" options=["$2\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)$","$2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)$","$\sqrt{3}\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)$","$2\left(\cos\dfrac{5\pi}{3}+i\sin\dfrac{5\pi}{3}\right)$"] answer="A" hint="Find modulus and quadrant first." solution="The modulus is $2$, and the point lies in the second quadrant with principal argument $\dfrac{2\pi}{3}$. Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="Find the modulus of $(1+i)^6$." answer="8" hint="Use $|z^n|=|z|^n$." solution="We have $|1+i|=\sqrt{2}$. Therefore $\qquad |(1+i)^6|=|1+i|^6=(\sqrt{2})^6=8$ Hence the answer is $\boxed{8}$." ::: :::question type="MSQ" question="Which of the following are true for nonzero complex numbers?" options=["Arguments differ by integer multiples of $2\pi$","Polar form is useful for powers and roots","The argument of every complex number is unique without restriction","If $z=r(\cos\theta+i\sin\theta)$, then $|z|=r$"] answer="A,B,D" hint="Think about non-uniqueness of the argument." solution="1. True. Arguments differ by $2k\pi$.

True. That is one of the main uses of polar form.

False. Only the principal argument is unique after restriction.

True. In polar form, the modulus is exactly $r$.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Find all cube roots of $-8i$." answer="The cube roots are $\sqrt{3}-i$, $2i$, and $-\sqrt{3}-i$" hint="Write $-8i$ in polar form first." solution="We write $\qquad -8i=8\left(\cos\left(-\dfrac{\pi}{2}\right)+i\sin\left(-\dfrac{\pi}{2}\right)\right)$ The cube roots are $\qquad 2\left(\cos\dfrac{-\pi/2+2k\pi}{3}+i\sin\dfrac{-\pi/2+2k\pi}{3}\right),\qquad k=0,1,2$ Now compute: For $k=0$: $\qquad 2\left(\cos\left(-\dfrac{\pi}{6}\right)+i\sin\left(-\dfrac{\pi}{6}\right)\right)=\sqrt{3}-i$ For $k=1$: $\qquad 2\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)=2i$ For $k=2$: $\qquad 2\left(\cos\dfrac{7\pi}{6}+i\sin\dfrac{7\pi}{6}\right)=-\sqrt{3}-i$ Hence all cube roots are $\qquad \boxed{\sqrt{3}-i,\ 2i,\ -\sqrt{3}-i}$." ::: ---

Summary

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Part 2: De Moivre theorem

De Moivre Theorem

Overview

De Moivre's theorem is the main tool for raising complex numbers in polar form to powers and for extracting roots. It converts algebraic exponentiation into angle multiplication. In CMI-style questions, it is used for powers, roots of unity, trigonometric identities, and geometric arrangements of points on a circle. ---

Learning Objectives

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Polar Form

Using Euler notation, this becomes $\qquad z = re^{i\theta}$ ::: ---

De Moivre Theorem

Equivalently, $\qquad (re^{i\theta})^n = r^n e^{in\theta}$ ::: ---

Powers of Complex Numbers

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Roots of Complex Numbers

So there are exactly $n$ distinct $n$th roots. ---

Roots of Unity

These are equally spaced points on the unit circle. ---

Minimal Worked Examples

Example 1 Compute $\qquad (1+i)^4$ First, $\qquad 1+i = \sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$ So by De Moivre, $\qquad (1+i)^4 = (\sqrt{2})^4\left(\cos\pi+i\sin\pi\right)$ $\qquad = 4(-1+0i) = -4$ Hence $\qquad \boxed{(1+i)^4=-4}$ --- Example 2 Find all cube roots of $1$. Write $\qquad 1=\cos(2k\pi)+i\sin(2k\pi)$ So the cube roots are $\qquad 1,\quad \cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3},\quad \cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}$ That is, $\qquad 1,\quad -\frac{1}{2}+\frac{\sqrt{3}}{2}i,\quad -\frac{1}{2}-\frac{\sqrt{3}}{2}i$ ---

Geometry of the Roots

This geometric picture is essential for roots of unity and many CMI-style problems. ---

Trigonometric Consequences

For example, $\qquad (\cos\theta+i\sin\theta)^2 = \cos2\theta + i\sin2\theta$ gives the double-angle identities. ---

Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The value of $(1+i)^4$ is" options=["$4$","$-4$","$4i$","$-4i$"] answer="B" hint="Convert $1+i$ to polar form." solution="We have $\qquad 1+i=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$ Hence $\qquad (1+i)^4 = (\sqrt{2})^4\left(\cos\pi+i\sin\pi\right)=4(-1)= -4$ Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="How many distinct fourth roots does a nonzero complex number have?" answer="4" hint="Use the general $n$th-root formula." solution="A nonzero complex number has exactly $n$ distinct $n$th roots. Therefore it has exactly $\boxed{4}$ distinct fourth roots." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["De Moivre's theorem gives $(re^{i\theta})^n = r^n e^{in\theta}$","The $n$th roots of unity lie on the unit circle","A nonzero complex number has exactly $n$ distinct $n$th roots","The roots of unity are equally spaced in argument"] answer="A,B,C,D" hint="Use the theorem and the standard geometry of roots." solution="1. True. This is De Moivre's theorem in exponential form.

True. Roots of unity have modulus $1$.

True. The general root formula gives exactly $n$ distinct roots.

True. Their arguments differ by $\dfrac{2\pi}{n}$.

Hence the correct answer is $\boxed{A,B,C,D}$." ::: :::question type="SUB" question="Find all solutions of $z^3=1$." answer="$1,\ -\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i,\ -\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i$" hint="Use the cube roots of unity." solution="We write $\qquad 1=\cos(2k\pi)+i\sin(2k\pi)$ So the cube roots are $\qquad z_k = \cos\frac{2k\pi}{3} + i\sin\frac{2k\pi}{3}$ for $\qquad k=0,1,2$ Thus the three roots are $\qquad z_0 = 1$ $\qquad z_1 = \cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3} = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$ $\qquad z_2 = \cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3} = -\frac{1}{2}-\frac{\sqrt{3}}{2}i$ Hence all solutions are $\qquad \boxed{1,\ -\frac{1}{2}+\frac{\sqrt{3}}{2}i,\ -\frac{1}{2}-\frac{\sqrt{3}}{2}i}$" ::: ---

Summary

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Part 3: Powers of complex numbers

Powers of Complex Numbers

Overview

Powers of complex numbers become simple and structured once the number is written in polar form. In CMI-style questions, this topic is often not about long algebraic expansion. Instead, it is about understanding how the modulus and argument behave under repeated powers, and then deciding when the set $\qquad \{z^k \mid k \text{ is a positive integer}\}$ is finite or infinite. ---

Learning Objectives

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Polar Form

The same number also has the exponential form $\qquad z = re^{i\theta}$ but for calculations in this topic, the trigonometric form is often the safest. ---

De Moivre's Theorem

This is the main tool of the topic. ---

What Powers Do to Modulus and Argument

So repeated powers do two things:

the modulus is repeatedly multiplied by $r$,

the argument is repeatedly added by $\theta$.

This completely explains when the powers repeat and when they do not. ---

The Set of Powers

For a complex number $z$, define $\qquad P(z) = \left|\{z^k \mid k \text{ is a positive integer}\}\right|$ that is, the number of distinct positive integer powers of $z$. This number may be finite or infinite. ---

First Classification

This is the basic classification structure. ---

Root of Unity Criterion

So the finite case occurs exactly when the argument eventually comes back to a multiple of $2\pi$. ---

Why Rational Multiple of $2\pi$ Matters

Suppose $\qquad z = \cos \theta + i\sin \theta$ with $|z|=1$. If $\qquad \theta = \dfrac{2\pi p}{q}$ where $p,q$ are integers and $\gcd(p,q)=1$, then $\qquad z^q = \cos(2\pi p) + i\sin(2\pi p) = 1$ So the powers repeat with period $q$, and hence $P(z)$ is finite. If instead $\qquad \dfrac{\theta}{2\pi}\notin \mathbb{Q}$ then no positive multiple of $\theta$ can equal an integer multiple of $2\pi$, so the powers never repeat, and $P(z)$ is infinite. ---

Exact Value of $P(z)$ in the Root-of-Unity Case

This is because $q$ is the smallest positive integer such that $\qquad z^q = 1$ and the powers $\qquad z, z^2, \dots, z^q$ then cycle periodically. ---

Standard Examples

The last example is important because $\qquad e^i = \cos 1 + i\sin 1$ and the angle $1$ is not a rational multiple of $2\pi$, since that would imply $\pi$ is rational. ---

Primitive Roots of Unity

For such a number, $\qquad P(z)=n$ ::: Examples:

• $i$ is a primitive $4$th root of unity

• $\cos \dfrac{2\pi}{3} + i\sin \dfrac{2\pi}{3}$ is a primitive $3$rd root of unity

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Minimal Worked Examples

Example 1 Find $P(i)$. Since $\qquad i = \cos \dfrac{\pi}{2} + i\sin \dfrac{\pi}{2}$ we get $\qquad i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1$ and then the powers repeat. So $\qquad P(i)=4$ --- Example 2 Find $P\!\left(\dfrac{1+i\sqrt{3}}{2}\right)$. We have $\qquad \dfrac{1+i\sqrt{3}}{2} = \cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3}$ So its argument is $\qquad \dfrac{\pi}{3} = \dfrac{2\pi}{6}$ and the corresponding order is $6$. Hence $\qquad P\!\left(\dfrac{1+i\sqrt{3}}{2}\right)=6$ --- Example 3 Show that $P(e^i)$ is infinite. If $P(e^i)$ were finite, then $e^i$ would be a root of unity, so there would exist a positive integer $n$ such that $\qquad (e^i)^n = 1$ That would mean $\qquad e^{in} = 1$ so $n$ would be an integer multiple of $2\pi$, which would imply $\pi$ is rational. This is impossible. Therefore $\qquad P(e^i)\text{ is infinite}$ ---

Algebraic Shortcuts

Sometimes a complex number is not already in visible polar form. Then use algebra first. Example Find the least positive $n$ such that $\qquad \left(\dfrac{1+i}{1-i}\right)^n = 1$ Simplify first: $\qquad \dfrac{1+i}{1-i} = \dfrac{(1+i)^2}{1+1} = \dfrac{1+2i+i^2}{2} = i$ So the problem becomes: $\qquad i^n=1$ The least positive such $n$ is $\qquad 4$ ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The value of $P(i)$ is" options=["$1$","$2$","$4$","Infinite"] answer="C" hint="Compute the first few powers of $i$." solution="We have $\qquad i,\ -1,\ -i,\ 1,\ i,\dots$ So the powers repeat after $4$ distinct values. Hence $\qquad P(i)=4$ Therefore the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Find $P\!\left(\cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3}\right)$." answer="6" hint="Write $\dfrac{\pi}{3}$ as a reduced multiple of $2\pi$." solution="We have $\qquad \dfrac{\pi}{3} = \dfrac{2\pi}{6}$ So $\qquad z = \cos \dfrac{2\pi}{6} + i\sin \dfrac{2\pi}{6}$ The denominator in reduced form is $6$, so the order is $6$. Therefore $\qquad P(z)=6$ Hence the answer is $\boxed{6}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$P(0)=1$","If $z\ne 0$ and $|z|\ne 1$, then $P(z)$ is infinite","If $z$ is a root of unity, then $P(z)$ is finite","If $|z|=1$, then $P(z)$ is always finite"] answer="A,B,C" hint="Separate the zero case, the non-unit-modulus case, and the root-of-unity case." solution="1. True.

True, because the moduli $|z|^k$ are all distinct when $|z|\ne 1$ and $z\ne 0$.

True, because powers repeat once some power becomes $1$.

False, because points on the unit circle with irrational argument multiple of $2\pi$ have infinitely many distinct powers.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Explain why, for a nonzero complex number $z$, the set of powers $\{z^k \mid k \text{ is a positive integer}\}$ is finite if and only if $z$ is a root of unity." answer="Because finiteness forces two powers to be equal, giving a positive power equal to $1$, and conversely a root of unity makes the powers periodic." hint="Use repetition of powers in one direction and periodicity in the other." solution="Suppose first that the set of powers of $z$ is finite. Then among the infinitely many numbers $\qquad z,\ z^2,\ z^3,\dots$ two of them must be equal, say $\qquad z^m = z^n$ with $m>n\ge 1$. Since $z\ne 0$, we can divide by $z^n$ and get $\qquad z^{m-n}=1$ So $z$ is a root of unity. Conversely, if $z$ is a root of unity, then for some positive integer $t$, $\qquad z^t=1$ Hence $\qquad z^{k+t}=z^k$ for every positive integer $k$, so the powers repeat periodically. Therefore only finitely many distinct powers occur. Thus, for $z\ne 0$, the set of positive integer powers is finite if and only if $z$ is a root of unity." ::: ---

Summary

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Part 4: Roots of complex numbers

Roots of Complex Numbers

Overview

The roots of a complex number are most naturally understood in polar form. Once a complex number is written as $\qquad z = r(\cos \theta + i\sin \theta)=re^{i\theta}$ its $n$th roots can be described exactly, geometrically and algebraically. In CMI-style problems, this topic is not only about applying De Moivre’s theorem; it is also about understanding:

• how many roots a complex number has

• where those roots lie geometrically

• how conjugation interacts with roots

• how to solve equations where the right-hand side is real, such as

$\qquad z^n = z+\overline z$ Your PYQs show both kinds of testing:

• a conceptual MSQ on roots and modulus

• a heavy SUB problem built around

$\qquad z^n=z+\overline z$ So these notes are built to cover both the basic root formula and the more advanced argument-based analysis needed for such equations. ---

Learning Objectives

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Polar Form

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De Moivre's Theorem

This is the main tool for finding roots. ---

$n$th Roots of a Nonzero Complex Number

So a nonzero complex number has exactly $n$ distinct $n$th roots. ---

Why There Are Exactly $n$

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Geometric Picture

This is one of the most important geometric interpretations in the topic. ---

Roots of Unity

These lie on the unit circle and form a regular $n$-gon. ---

Important Facts About Roots of Unity

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Sum and Product of the Roots

Why? If one root is $w_0$, then all roots are $\qquad w_0,\ w_0\omega,\ w_0\omega^2,\dots,w_0\omega^{n-1}$ where $\omega$ is a primitive $n$th root of unity. Their sum is $\qquad w_0(1+\omega+\omega^2+\cdots+\omega^{n-1})=0 $ ::: This follows from the polynomial $\qquad w^n-z=0$ and Vieta’s formula. ::: ---

Conjugation and Roots

So roots of conjugate numbers are conjugates of roots. ---

Minimal Worked Examples

Example 1 Find all cube roots of $-8$. Write $\qquad -8 = 8e^{i(\pi+2m\pi)}$ So the cube roots are $\qquad 2e^{i(\pi+2k\pi)/3}, \qquad k=0,1,2 $ Hence the roots are $\qquad 2e^{i\pi/3},\quad 2e^{i\pi},\quad 2e^{i5\pi/3} $ or in trigonometric form, $\qquad 1+\sqrt3\,i,\quad -2,\quad 1-\sqrt3\,i $ --- Example 2 Find all fourth roots of $-16$. Write $\qquad -16 = 16e^{i(\pi+2m\pi)}$ Hence the fourth roots are $\qquad 2e^{i(\pi+2k\pi)/4}, \qquad k=0,1,2,3 $ that is, $\qquad 2e^{i\pi/4},\quad 2e^{i3\pi/4},\quad 2e^{i5\pi/4},\quad 2e^{i7\pi/4} $ ---

Solving Equations of the Form $z^n = z+\overline z$

Let $\qquad z = re^{i\theta}$ with $r\ge0$. Then $\qquad z+\overline z = 2r\cos\theta$ and $\qquad z^n = r^n e^{in\theta}$ So the equation becomes $\qquad r^n e^{in\theta} = 2r\cos\theta$ ::: ---

Zero Solution

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Nonzero Solutions

For $z\ne0$, divide by $r$: $\qquad r^{n-1} e^{in\theta} = 2\cos\theta$ Since the right side is real, we must have $\qquad \sin(n\theta)=0$ So $\qquad n\theta = k\pi \quad\text{for some integer }k $ Hence $\qquad \theta = \dfrac{k\pi}{n} $ Substituting back, $\qquad r^{n-1}(-1)^k = 2\cos\left(\dfrac{k\pi}{n}\right)$ Therefore nonzero solutions occur exactly when $\qquad 2(-1)^k\cos\left(\dfrac{k\pi}{n}\right) > 0 $ and then $\qquad r = \left( 2(-1)^k\cos\left(\dfrac{k\pi}{n}\right) \right)^{1/(n-1)} $ ::: So each valid integer $k$ gives one nonzero solution $\qquad z = \left( 2(-1)^k\cos\left(\dfrac{k\pi}{n}\right) \right)^{1/(n-1)} e^{ik\pi/n} $ ::: ---

Advanced Count for $z^n=z+\overline z$

This count includes the solution $\qquad z=0$. :::

Quick Check

• for $\qquad n=3$, total solutions $=3$

• for $\qquad n=4$, total solutions $=4$

• for $\qquad n=5$, total solutions $=7$

This matches direct computation. ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="A nonzero complex number has exactly how many distinct $n$th roots?" options=["$1$","$n$","$n^2$","$2n$"] answer="B" hint="Use the general root formula." solution="If $z\ne0$, then its $n$th roots are $\qquad r^{1/n}e^{i(\theta+2k\pi)/n}$ for $k=0,1,\dots,n-1$. These are exactly $n$ distinct roots. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="How many $12$th roots of unity have positive real part?" answer="5" hint="Place the $12$ roots equally on the unit circle." solution="The $12$th roots of unity have arguments $\qquad \dfrac{2k\pi}{12},\ k=0,1,\dots,11$. Positive real part means the cosine is positive. This happens for angles in the right half-plane, excluding the imaginary axis. The valid indices are $\qquad k=0,1,2,10,11$. So the number of such roots is $\boxed{5}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["Every nonzero complex number has exactly $n$ distinct $n$th roots","All $n$th roots of a fixed nonzero complex number have the same modulus","Every $n$th root of a positive real number must itself be real","If $w^n=z$, then $\overline{w}^{\,n}=\overline z$"] answer="A,B,D" hint="Check root formula and conjugation." solution="1. True.

True, all roots have modulus $|z|^{1/n}$.

False. For example, a positive real number has many non-real roots when $n>2$.

True, because conjugation respects products and powers.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Find all cube roots of $-8$." answer="$1+\sqrt3\,i,\ -2,\ 1-\sqrt3\,i$" hint="Write $-8$ in polar form and use the root formula." solution="Write $\qquad -8 = 8e^{i(\pi+2m\pi)}$. The cube roots are $\qquad 2e^{i(\pi+2k\pi)/3}$ for $k=0,1,2$. Thus the roots are $\qquad 2e^{i\pi/3}=1+\sqrt3\,i$ $\qquad 2e^{i\pi}=-2$ $\qquad 2e^{i5\pi/3}=1-\sqrt3\,i$ Hence all cube roots are $\qquad \boxed{1+\sqrt3\,i,\ -2,\ 1-\sqrt3\,i}$." ::: ---

Summary

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Part 5: Roots of unity

Roots of Unity

Overview

Roots of unity are the complex solutions of the equation $z^n=1$. They combine algebra, geometry, trigonometry, and complex numbers in one framework. In CMI-style problems, they appear in geometric series, regular polygons on the complex plane, trigonometric sums, and symmetric expressions such as sums and products of vertices. ---

Learning Objectives

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Definition

If we write $\qquad \omega = e^{2\pi i/n}$, then all $n$th roots of unity are $\qquad 1,\omega,\omega^2,\dots,\omega^{n-1}$ with $\qquad \omega^n=1$. ::: ---

Geometric Interpretation

The arguments are $\qquad 0,\ \dfrac{2\pi}{n},\ \dfrac{4\pi}{n},\ \dots,\ \dfrac{2\pi(n-1)}{n}$ ---

Fundamental Algebraic Facts

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Sum of All Roots

Reason 1: Geometric series $\qquad 1+\omega+\omega^2+\cdots+\omega^{n-1} =\dfrac{\omega^n-1}{\omega-1} =\dfrac{1-1}{\omega-1}=0$ provided $\omega\ne1$. Reason 2: Geometry These vertices are symmetrically placed around the origin, so their vector sum is zero. ---

Sum of Powers

This is one of the highest-yield facts in the topic. ---

Product of All $n$th Roots of Unity

Equivalently, $\qquad \prod_{z^n=1} z = (-1)^{n+1}$ which is the same parity value. ::: ---

Primitive Roots of Unity

If $\omega=e^{2\pi i/n}$, then $\omega^k$ is primitive if and only if $\qquad \gcd(k,n)=1$. ::: ---

Factorization

Also, $\qquad 1+x+x^2+\cdots+x^{n-1} = \dfrac{x^n-1}{x-1}$ so the non-real or nontrivial roots often come from this factor. ::: ---

Trigonometric Sums from Roots of Unity

For example, if $\qquad \sum_{k=0}^{n-1} e^{ik\theta}=0$, then taking real parts gives $\qquad \sum_{k=0}^{n-1}\cos(k\theta)=0$ whenever the complex sum is zero. ---

Regular Polygon in the Complex Plane

This model is extremely useful.

Immediate consequences

Since $\qquad \sum_{k=0}^{n-1}\omega^k=0$, we get $\qquad \sum_{k=0}^{n-1} z_k = nc$. So the sum of the vertices equals $n$ times the center. ---

Sum of Squares of Vertices

Therefore:

• if $n\nmid 1$ and $n\nmid 2$, then both extra sums vanish, so

$\qquad \sum z_k^2 = nc^2$. For a regular pentagon, this gives $\qquad \sum_{k=1}^{5} z_k^2 = 5c^2$. This directly matches the PYQ style. ---

Product of Vertices of a Regular $n$-gon

This follows from the factorization $\qquad \prod_{k=0}^{n-1}(x-a\omega^k)=x^n-a^n$ by setting $\qquad x=c,\ a=-r\alpha$. ::: This is exactly the structure behind product questions for regular polygons. ---

Minimal Worked Examples

Example 1 Let $\omega=e^{2\pi i/5}$. Find $\qquad 1+\omega+\omega^2+\omega^3+\omega^4$. Using the standard identity, $\qquad 1+\omega+\omega^2+\omega^3+\omega^4=0$. --- Example 2 Find the product of all fourth roots of unity. The roots are $\qquad 1,\ i,\ -1,\ -i$ Their product is $\qquad 1\cdot i\cdot (-1)\cdot (-i)=-1$ which agrees with $\qquad (-1)^{4-1}=-1$. ---

Common Patterns in Questions

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $\omega$ is a primitive $7$th root of unity, then $1+\omega+\omega^2+\cdots+\omega^6$ equals" options=["$1$","$-1$","$0$","$7$"] answer="C" hint="Use the geometric-series identity for roots of unity." solution="Since $\omega\ne1$ and $\omega^7=1$, $\qquad 1+\omega+\omega^2+\cdots+\omega^6=\dfrac{\omega^7-1}{\omega-1}=0$. Therefore the correct option is $\boxed{C}$." ::: :::question type="NAT" question="If $\omega=e^{2\pi i/5}$, find $\omega^5+\omega^{10}+\omega^{15}$." answer="3" hint="Reduce each exponent modulo $5$." solution="Since $\omega^5=1$, we get $\qquad \omega^{10}=(\omega^5)^2=1,\qquad \omega^{15}=(\omega^5)^3=1$. Hence $\qquad \omega^5+\omega^{10}+\omega^{15}=1+1+1=3$. So the answer is $\boxed{3}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["All $n$th roots of unity lie on the unit circle","If $\omega$ is a primitive $n$th root of unity, then $\omega^k$ is primitive exactly when $\gcd(k,n)=1$","For every positive integer $m$, $\sum_{k=0}^{n-1}\omega^{mk}=0$","The sum of all $n$th roots of unity is $0$ for $n>1$"] answer="A,B,D" hint="Be careful when $n$ divides $m$ in the power-sum identity." solution="1. True. Each $n$th root has modulus $1$.

True. This is the standard primitive-root criterion.

False. If $n\mid m$, then every term is $1$, so the sum is $n$.

True for $n>1$, by the geometric-series identity.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Prove that if $\omega$ is a primitive $n$th root of unity, then $1+\omega+\omega^2+\cdots+\omega^{n-1}=0$." answer="Use the geometric-series formula." hint="Multiply the sum by $\omega-1$." solution="Let $\qquad S=1+\omega+\omega^2+\cdots+\omega^{n-1}$. Then $\qquad (\omega-1)S = (\omega+\omega^2+\cdots+\omega^n)-(1+\omega+\cdots+\omega^{n-1}) = \omega^n-1$. Since $\omega$ is an $n$th root of unity, $\qquad \omega^n=1$. So $\qquad (\omega-1)S=0$. Because $\omega$ is primitive, in particular $\omega\ne1$, so $\omega-1\ne0$. Hence $\qquad S=0$. Therefore $\qquad 1+\omega+\omega^2+\cdots+\omega^{n-1}=0$." ::: ---

Summary

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="What is the value of $(1+i\sqrt{3})^6$?" options=["$-64$","$64$","$64i$","$-64i$"] answer="$64$" hint="Convert the complex number to polar form before applying De Moivre's Theorem." solution="First, convert $1+i\sqrt{3}$ to polar form. The modulus is $r = |1+i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2$. The argument is $\theta = \operatorname{atan2}(\sqrt{3}, 1) = \pi/3$. So, $1+i\sqrt{3} = 2(\cos(\pi/3) + i\sin(\pi/3))$.
Applying De Moivre's Theorem:

"
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:::question type="NAT" question="If $z^3 = -8i$, and $z_0$ is the root with the smallest positive argument, what is the real part of $z_0$?" answer="0" hint="First, express $-8i$ in polar form. Then, use the formula for $n$-th roots of complex numbers to find $z_0$." solution="First, express $-8i$ in polar form. The modulus is $r = |-8i| = 8$. The argument is $\theta = -\pi/2$ or $3\pi/2$. For finding roots, it's often convenient to use $\theta = 3\pi/2$ to ensure positive arguments for $k=0, 1, \dots, n-1$ initially, or to adjust later.
So, $-8i = 8(\cos(3\pi/2) + i\sin(3\pi/2))$.
The $n$-th roots are given by $z_k = r^{1/n}\left(\cos\left(\frac{\theta+2k\pi}{n}\right) + i\sin\left(\frac{\theta+2k\pi}{n}\right)\right)$.
Here, $n=3$, $r=8$, $\theta=3\pi/2$.
$z_k = 8^{1/3}\left(\cos\left(\frac{3\pi/2+2k\pi}{3}\right) + i\sin\left(\frac{3\pi/2+2k\pi}{3}\right)\right)$
$z_k = 2\left(\cos\left(\frac{3\pi/2+2k\pi}{3}\right) + i\sin\left(\frac{3\pi/2+2k\pi}{3}\right)\right)$
For $k=0$: $z_0 = 2\left(\cos\left(\frac{3\pi/2}{3}\right) + i\sin\left(\frac{3\pi/2}{3}\right)\right) = 2(\cos(\pi/2) + i\sin(\pi/2)) = 2(0+i) = 2i$. The argument is $\pi/2$. This is the smallest positive argument.
For $k=1$: $z_1 = 2\left(\cos\left(\frac{3\pi/2+2\pi}{3}\right) + i\sin\left(\frac{3\pi/2+2\pi}{3}\right)\right) = 2\left(\cos\left(\frac{7\pi/2}{3}\right) + i\sin\left(\frac{7\pi/2}{3}\right)\right) = 2(\cos(7\pi/6) + i\sin(7\pi/6))$.
For $k=2$: $z_2 = 2\left(\cos\left(\frac{3\pi/2+4\pi}{3}\right) + i\sin\left(\frac{3\pi/2+4\pi}{3}\right)\right) = 2\left(\cos\left(\frac{11\pi/2}{3}\right) + i\sin\left(\frac{11\pi/2}{3}\right)\right) = 2(\cos(11\pi/6) + i\sin(11\pi/6))$.
The root with the smallest positive argument is $z_0 = 2i$. The real part of $z_0$ is $0$."
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:::question type="MCQ" question="Let $\omega$ be a primitive $5$-th root of unity. Which of the following is equal to $1 + \omega + \omega^2 + \omega^3 + \omega^4$?" options=["$5$","$1$","$0$","$-1$"] answer="$0$" hint="Recall the property of the sum of $n$-th roots of unity." solution="The sum of the $n$-th roots of unity is $0$ for $n>1$. In this case, $n=5$, which is greater than $1$. Therefore, $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$."
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:::question type="NAT" question="What is the principal argument (in radians, to two decimal places) of the complex number $z = -1-i$?" answer="-2.36" hint="Identify the quadrant of the complex number in the Argand plane and calculate the argument relative to the negative real axis." solution="The complex number $z = -1-i$ is in the third quadrant of the Argand plane.
The modulus is $r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.
The reference angle $\alpha = \arctan\left(\left|\frac{-1}{-1}\right|\right) = \arctan(1) = \pi/4$.
Since $z$ is in the third quadrant, the principal argument $\theta$ (which must satisfy $-\pi < \theta \le \pi$) is given by $\theta = -\pi + \alpha$ or $\theta = \pi - (\pi-\alpha)$ (if using positive reference).
Using $\theta = -\pi + \pi/4 = -3\pi/4$.
In decimal form, $-3\pi/4 \approx -3 \times 3.14159265 / 4 \approx -2.35619$.
Rounding to two decimal places, the principal argument is $-2.36$ radians."
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