Chapter 1 · The complex plane and why it exists

The hook

Numbers as points, and the question polar form answers

Story

Picture the real number line. One dimension, points strung out left and right, useful for everything until the day you try to take the square root of $-1$ . The number line has no room for that answer. There is nowhere on the line for $i$ to live.

So we add a dimension. Draw a vertical axis through zero. Now every point in the plane is a complex number: its horizontal coordinate is the real part, its vertical coordinate is the imaginary part. The number $3 + 4i$ is the point three to the right and four up. The number $-1$ is one to the left. The number $i$ is one straight up.

This is the complex plane, also called the Argand diagram. It is the picture the rest of the unit will be drawn on.

Story

Once complex numbers are points in a plane, adding them is exactly what you would guess: tip to tail, like vectors. $(1 + 2i) + (3 + i) = 4 + 3i$ . Geometric, easy.

But multiplying them is where the plane earns its keep. Multiply $(1 + i)$ by $(1 + i)$ : you get $2i$ . The first factor lives at distance $\sqrt{2}$ from the origin, at angle $\pi/4$ . The product lives at distance $2$ , at angle $\pi/2$ . The distance was squared. The angle was doubled.

That is not an accident. Multiplication in the complex plane rotates and scales. Multiplying two complex numbers multiplies their distances-from-origin and adds their angles. This is the deep fact the rest of the chapter builds on. It is also the reason that, once you start writing complex numbers in polar form — by distance and angle, rather than by horizontal-and-vertical coordinates — multiplication, powers, and roots become almost transparent.

Why this matters

Every later chapter in Unit 1 assumes you can switch fluently between two pictures of the same complex number: the Cartesian picture ( $x + iy$ , good for addition) and the polar picture ( $r\,e^{i\theta}$ , good for multiplication). The conformal-mapping chapter and the Möbius transformations chapter repeatedly use the geometric picture of modulus, argument, circles, and angles. The chapter on Laurent series and residues computes contour integrals by tracking how the argument changes around a closed curve. If polar form is not second nature by the end of this chapter, the rest of Unit 1 gets harder than it needs to.

And in the engineering-math exam: routine conversion questions, De Moivre computations, and $n$ -th-root problems are common, and the marks lost are usually lost to one specific kind of mistake — picking the wrong angle when the point is in the second or third quadrant. We will name that mistake, draw the picture, and drill the fix.

If the Cartesian-and-polar picture is already familiar, move quickly through the definitions in the next section; the real work in this chapter is the principal-value convention for $\operatorname{Arg}(z)$ , the induction proof of De Moivre, and the $2k\pi/n$ term in the $n$ -th-root formula.

Intuition before formula

The argument is not a number

For $z = x + iy$ with $x, y \in \mathbb{R}$ , four quantities are written exam-style like this:

\operatorname{Re}(z) = x, \qquad \operatorname{Im}(z) = y, \qquad \bar{z} = x - iy, \qquad |z| = \sqrt{x^2 + y^2}.

The first three are routine. The fourth — the modulus — is the distance from the origin to $z$ in the plane. Always non-negative, always real. The identity $|z|^2 = z\bar{z}$ is worth memorising; you will use it more than you would guess.

Why "the argument" needs a careful definition

You have used the argument of a complex number informally — the angle from the positive real axis to the line from the origin to $z$ . That informal definition has a problem: there are infinitely many such angles, all differing by integer multiples of $2\pi$ . The angle $\pi/4$ , the angle $9\pi/4$ , and the angle $-7\pi/4$ all point in the same direction.

So $\arg(z)$ is, strictly, not a number. It is a set:

\arg(z) = \{\,\theta_0 + 2k\pi : k \in \mathbb{Z}\,\}

where $\theta_0$ is any one of the angles. For computation we need a single number, and there is a convention for picking one — the principal value:

\operatorname{Arg}(z) \in (-\pi, \pi].

Capital A for the principal value. This is the convention used in Zill, in NCERT Class 11, and in the engineering-math textbooks at most Indian universities. We use it throughout this chapter and the rest of Unit 1.

Insight

The principal value is not arbitrary. It is the unique argument that lies in the half-open interval $(-\pi, \pi]$ . Half-open on purpose: $-\pi$ is excluded so that $\operatorname{Arg}(-1) = \pi$ and not $-\pi$ . There is exactly one allowed answer per non-zero $z$ , and the half-open convention is what makes that uniqueness work.

Some advanced texts use $[0, 2\pi)$ instead. Under that convention, points in the upper half-plane and on the negative real axis carry the same labels as ours; points in the lower half-plane carry labels larger by $2\pi$ — for instance $\operatorname{Arg}(-i) = 3\pi/2$ rather than $-\pi/2$ . If your lecturer uses a different convention, the mathematics is identical; only the labels change.

Exam trap

A common mistake on this material in Sem 3: compute $\arctan(y/x)$ on a calculator and call it $\operatorname{Arg}(z)$ . This is wrong in two of the four quadrants.

The reason: $\arctan$ returns values in $(-\pi/2, \pi/2)$ , which only covers the right half-plane. If $z$ is in the second or third quadrant ( $x < 0$ ), the principal value is $\arctan(y/x)$ plus or minus $\pi$ , not $\arctan(y/x)$ itself.

Which sign? Whichever lands the answer inside $(-\pi, \pi]$ : add $\pi$ if the point is in Q2, subtract $\pi$ if it is in Q3.

Quadrant of $z$	Sign of $x$	Sign of $y$	$\operatorname{Arg}(z)$
Q1	+	+	$\arctan(y/x)$
Q2	−	+	$\arctan(y/x) + \pi$
Q3	−	−	$\arctan(y/x) - \pi$
Q4	+	−	$\arctan(y/x)$

Plus the four axis cases: positive real axis $\operatorname{Arg} = 0$ , positive imaginary $\pi/2$ , negative real $\pi$ , negative imaginary $-\pi/2$ .

Sketch the point. Read the angle off the sketch. Then check your formula. This habit costs ten seconds per problem and saves the marks that would otherwise be donated to the examiner.

Formal statement

Polar form, and why De Moivre is just induction

Every non-zero complex number $z$ can be written in polar form:

Polar form and Euler form

z = r\,(\cos\theta + i\sin\theta) = r\,e^{i\theta}, \qquad r = |z|,\ \ \theta \in \arg(z).

The $e^{i\theta}$ notation is the Euler form, justified by Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ . Both notations name the same number; engineers tend to write $e^{i\theta}$ , and some textbooks use the shorthand $\operatorname{cis}\theta$ for $\cos\theta + i\sin\theta$ .

Multiplication in polar form

In polar form, multiplication takes its cleanest shape. Take $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$ , and multiply them out in trig form:

\begin{aligned} z_1 z_2 &= r_1 r_2 (\cos\theta_1 + i\sin\theta_1)(\cos\theta_2 + i\sin\theta_2) \\[2pt] &= r_1 r_2 \bigl[(\cos\theta_1 \cos\theta_2 - \sin\theta_1 \sin\theta_2) + i(\cos\theta_1 \sin\theta_2 + \sin\theta_1 \cos\theta_2)\bigr] \\[2pt] &= r_1 r_2 \bigl[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\bigr]. \end{aligned}

The last step is just the angle-sum identities for $\cos$ and $\sin$ , read right to left. Now both pieces are visible — moduli multiplied ( $r_1 r_2$ ), arguments added ( $\theta_1 + \theta_2$ ):

Multiplication: moduli multiply, arguments add

z_1 z_2 = r_1 r_2\, e^{i(\theta_1 + \theta_2)}.

De Moivre's theorem

If multiplying adds arguments, then raising to the $n$ -th power multiplies the argument by $n$ . That is the geometric content of De Moivre's theorem:

De Moivre's theorem

(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta), \qquad n \in \mathbb{Z}.

The geometric reason makes this feel inevitable. The proof writes it down so it earns marks.

Proof, for positive integers, by induction on $n$ .

Base case $n = 1$ : $(\cos\theta + i\sin\theta)^1 = \cos(1\cdot\theta) + i\sin(1\cdot\theta)$ . ✓

Inductive step. Assume the result for $n = k$ , that is $(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$ . Then

(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k \cdot (\cos\theta + i\sin\theta).

The first factor has modulus $1$ and argument $k\theta$ ; the second has modulus $1$ and argument $\theta$ . By the multiplication formula just proved, the product has modulus $1$ and argument $k\theta + \theta$ :

(\cos\theta + i\sin\theta)^{k+1} = \cos\!\bigl((k+1)\theta\bigr) + i\sin\!\bigl((k+1)\theta\bigr).

By induction, the result holds for every positive integer $n$ . ∎

The case $n = 0$ . Both sides equal $1$ directly: the left side is $(\cos\theta + i\sin\theta)^0 = 1$ , the right side is $\cos 0 + i\sin 0 = 1$ .

The case of negative $n$ . Write $n = -m$ with $m$ a positive integer. The reciprocal $(\cos\theta + i\sin\theta)^{-1}$ equals $\cos(-\theta) + i\sin(-\theta)$ — verify by direct multiplication, $(\cos\theta + i\sin\theta)(\cos\theta - i\sin\theta) = \cos^2\theta + \sin^2\theta = 1$ . Apply the positive case to $m$ , and the result holds for $-m$ .

Insight

De Moivre is the theorem that lets you compute $(1+i)^{10}$ without expanding it. Convert: $1+i = \sqrt{2}\,e^{i\pi/4}$ . Raise: $(\sqrt{2})^{10}\, e^{i\,10\pi/4} = 32\,e^{i\,5\pi/2} = 32\,e^{i\pi/2} = 32i$ . Done. The brute-force binomial expansion is eleven terms and an almost guaranteed sign error. The polar route is three lines of arithmetic.

That is why high-power computation questions — anything of the form "compute $z^n$ " with $n \geq 5$ — are usually best attacked through polar form.

Interactive playground

Try it: the Argand diagram and the roots of unity

To solve $z^n = w$ for a non-zero complex $w = R\,e^{i\varphi}$ :

The

n

-th roots of a complex number

z_k = R^{1/n}\,\exp\!\left(i\,\frac{\varphi + 2k\pi}{n}\right), \qquad k = 0, 1, 2, \ldots, n-1.

That is $n$ distinct roots, all with the same modulus $R^{1/n}$ , sitting at equal angular spacing $2\pi/n$ around a circle of that radius. The $2k\pi/n$ term is the one students forget under exam pressure — without it you get one root and miss the other $n-1$ .

When $w = 1$ (so $R = 1$ , $\varphi = 0$ ), these are the $n$ -th roots of unity:

The

n

-th roots of unity

\omega_k = \exp\!\left(\frac{2k\pi i}{n}\right), \qquad k = 0, 1, \ldots, n-1.

Equally spaced on the unit circle, starting at $1$ . For $n = 3$ : an equilateral triangle. $n = 4$ : a square. $n = 6$ : a regular hexagon.

We write $\omega_k = e^{2k\pi i/n}$ for the $k$ -th root, with $\omega_0 = 1$ . The first non-trivial one, $\omega_1 = e^{2\pi i/n}$ , is a primitive $n$ -th root of unity, because every other root is a power of it: $\omega_k = (\omega_1)^k$ . Some books write this primitive root simply as $\omega$ ; here we keep the subscripted form $\omega_k$ to hold the whole family of roots in view.

Two reflexes to build below. First, switching between Cartesian and polar at speed — click a point, read off its modulus and principal argument, and check the picture against the numbers. Second, the shape of the $n$ -th roots of unity. Use the toggle to explore both modes.

The Argand diagram

Free point

n

-th roots of unity · n = 5

Insight

A subtle thing the playground reveals. Click a point in the second or third quadrant — say $z = -1 + i$ . The principal argument the playground reads off is $3\pi/4 \approx 2.356$ , not the $-\pi/4 \approx -0.785$ you would get from a careless $\arctan(y/x)$ . The picture and the formula agree. They have to. The formula is wrong in two quadrants on its own.

That visible agreement — picture matches formula — is the thing the click-to-place mode is for. Click around. Confirm. Build the reflex.

This reflex matters most in the chapter on Laurent series and residues, where contour integrals are evaluated by tracking how the argument changes as a curve is traversed. Wrong principal value, wrong integral.

Retrieval gauntlet

Test yourself before moving on

Do not peek. Commit to an answer before you reveal anything. Committing — even to a wrong answer — is what builds the memory. Recognising a correct answer once you see it only feels like learning.

Quiz 1 · principal value, second quadrant

Compute $\operatorname{Arg}(-\sqrt{3} + i)$ , the principal value of the argument.

Quiz 2 · De Moivre as a tool

What is $(1 + i)^8$ ?

Quiz 3 · counting roots of unity

How many distinct sixth roots of unity satisfy $\operatorname{Re}(z) > 0$ ?

Try first. Write

z = -1 - i\sqrt{3}

in polar form and in Euler form, using the principal value of the argument. Compute on paper before you reveal.

Modulus: $|z| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$ .

For the argument, the point is in the third quadrant ( $x < 0$ , $y < 0$ ). The reference angle is $\arctan(\sqrt{3}/1) = \pi/3$ . In Q3 the principal value is $-(\pi - \pi/3) = -2\pi/3$ . Check: $-2\pi/3 \in (-\pi, \pi]$ , and $\cos(-2\pi/3) = -1/2$ , $\sin(-2\pi/3) = -\sqrt{3}/2$ , so $2\bigl(\cos(-2\pi/3) + i\sin(-2\pi/3)\bigr) = -1 - i\sqrt{3}$ . ✓

Polar form: $z = 2\bigl(\cos(-2\pi/3) + i\sin(-2\pi/3)\bigr)$ .
Euler form: $z = 2\,e^{-i\,2\pi/3}$ .

Prove: let $\omega_1 = e^{2\pi i/n}$ and $\omega_k = (\omega_1)^k = e^{2k\pi i/n}$ be the $k$ -th $n$ -th root of unity. Show that

\omega_0 + \omega_1 + \omega_2 + \cdots + \omega_{n-1} = 0 \quad \text{for every integer } n \geq 2.

Work the three sub-steps below. Attempt each on paper before revealing it.

Step 1 — recognise the structure. The expression

\omega_0 + \omega_1 + \cdots + \omega_{n-1}

is a sum you have seen in a different context — not from complex analysis. What is it?

It is a geometric series. Each term is $(\omega_1)^k$ , so the sum is $\sum_{k=0}^{n-1} (\omega_1)^k$ with first term $1$ , common ratio $\omega_1$ , and $n$ terms. The standard formula gives

S = \frac{1 - (\omega_1)^n}{1 - \omega_1} \qquad \text{provided } \omega_1 \neq 1.

Step 2 — use the defining property of $\omega_1$ . What is

(\omega_1)^n

? Use the definition

\omega_1 = e^{2\pi i/n}

directly.

By the definition,

(\omega_1)^n = \bigl(e^{2\pi i/n}\bigr)^n = e^{2\pi i} = \cos 2\pi + i\sin 2\pi = 1.

So $(\omega_1)^n = 1$ — which is the whole point: $\omega_1$ is, by construction, an $n$ -th root of $1$ .

Step 3 — close the argument. Substitute Step 2 into Step 1. What do you get? And check that the assumption

\omega_1 \neq 1

is legitimate.

Substituting $(\omega_1)^n = 1$ into the formula:

S = \frac{1 - 1}{1 - \omega_1} = \frac{0}{1 - \omega_1} = 0.

The division is legitimate because $\omega_1 = e^{2\pi i/n}$ with $n \geq 2$ , so $2\pi/n$ is not a multiple of $2\pi$ , so $\omega_1 \neq 1$ . ∎

Why this matters: this identity unlocks a wide family of exam problems. "Show that $\cos(2\pi/5) + \cos(4\pi/5) + \cos(6\pi/5) + \cos(8\pi/5) = -1$ " is just the real-part identity that falls out of $\omega_0 + \omega_1 + \omega_2 + \omega_3 + \omega_4 = 0$ for $n = 5$ . The complex version is the master key; the trigonometric identities are corollaries. This proof structure recurs in the series and residues work later in the unit.

Exam translator

How this concept appears on exams

Exam trap

In standard Indian engineering-math exams, and likely in MCC201A unless your class notes differ, the material in this chapter shows up in three forms.

CO-1 form — state, define, identify. "Define the modulus and the principal value of the argument of a complex number. Find $\operatorname{Arg}(z)$ for the following points." Roughly 4–6 marks; budget about 5 minutes. Mistakes that cost marks: wrong quadrant adjustment on $\operatorname{Arg}$ , treating $\arctan(y/x)$ as if it were $\operatorname{Arg}$ , and slipping out of the $(-\pi, \pi]$ convention mid-paper.

CO-3 form — convert and compute. "Express $(1+i\sqrt{3})^7$ in $a + ib$ form" or "find all complex $z$ such that $z^4 = -16$ ." Roughly 5–8 marks; budget about 7 minutes. Mistakes that cost marks: attempting it by binomial expansion instead of polar form (and slipping on term 4 of 8); and producing only one of the $n$ roots instead of writing them all out as a $k$ -indexed family.

CO-3 form, identity flavour. "Prove that the $n$ -th roots of unity sum to zero" or "evaluate $\sum_{k=0}^{n-1} \cos(2k\pi/n)$ ." Roughly 5–7 marks; budget about 6 minutes. This is the territory of the scaffolded proof you just worked through, and that proof is the model answer.

Exam trap

Three patterns that commonly cost marks on this material.

Quadrant sign error on $\operatorname{Arg}$ . Trusting $\arctan(y/x)$ outside the right half-plane. The fix: sketch the point, read the rough angle, then plug into the formula. Ten seconds, two marks.
$n$ -th roots given as a single answer. "Find all complex $z$ with $z^4 = -16$ " wants four answers, written as $z_k$ for $k = 0, 1, 2, 3$ . Writing only one of them — usually the principal root — costs the other three marks. The phrase "all complex $z$ " is the tell.
Forgetting the $2k\pi/n$ term. When applying the $n$ -th-root formula, the $\varphi/n$ part is the "obvious" piece — the extra $2k\pi/n$ shift is what produces the other $n-1$ roots. Dropping it is one of the most common slips on $n$ -th-root questions.

Six things to carry out of this chapter

The complex plane lets you treat complex numbers as points. $z = x + iy$ is the point $(x, y)$ ; its modulus is its distance from the origin; its argument is the angle from the positive real axis.
The argument is multi-valued; the principal value is the unique representative in $(-\pi, \pi]$ . $\arctan(y/x)$ is not $\operatorname{Arg}(z)$ outside the right half-plane — apply the four-quadrant correction.
Polar form makes multiplication transparent. $z_1 z_2 = r_1 r_2\, e^{i(\theta_1 + \theta_2)}$ : moduli multiply, arguments add. This is the reason De Moivre's theorem is true.
De Moivre's theorem. $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ for every integer $n$ . Proved by induction on positive $n$ ; the cases $n = 0$ and $n < 0$ follow directly.
The $n$ -th roots of $w = R\,e^{i\varphi}$ are $z_k = R^{1/n}\,e^{i(\varphi + 2k\pi)/n}$ for $k = 0, 1, \ldots, n-1$ . Equally spaced on a circle of radius $R^{1/n}$ . The $2k\pi$ term is the one to never forget.
The $n$ -th roots of unity sum to zero. $\omega_0 + \omega_1 + \cdots + \omega_{n-1} = 0$ for $\omega_k = e^{2k\pi i/n}$ and $n \geq 2$ . Geometric series; uses $(\omega_1)^n = 1$ . The master identity behind many exam problems.

Why this matters

What comes next. Chapter 2, Complex functions, limits, and what differentiability means in 2D, builds on the picture of complex numbers as points in a plane and asks what it means for a function to take a plane to a plane. That sets up the direction-of-approach idea that drives the Cauchy-Riemann chapter.

Before you open Chapter 2: attempt 3–4 fresh problems from Zill, Complex Analysis, §§1.1–1.4 that mix the four ideas — modulus, principal argument, polar form, $n$ -th roots. The "Powers and Roots" exercises in §1.4 are the highest-value drill. Re-reading this chapter is a trap; retrieval beats recognition every time.

Same material, another voice

If a different explanation would help, these two are worth your time — both free, both from MIT OpenCourseWare:

Read: Orloff, MIT 18.04, Topic 1 — Complex algebra and the complex plane.
Watch: Herbert Gross, Calculus Revisited, Lecture 1 — The Complex Numbers (a friendly chalkboard introduction).

✓ Chapter complete. Your progress, and every quiz answer, is saved on this computer — revisit any time.