# Complex Numbers and the Complex Plane: Why Imaginary Numbers Became Necessary, and Why They Are Rotations

> From the cubic equations that made imaginary numbers unavoidable to the complex field, Euler's formula, De Moivre's theorem, the n-th roots of unity, and the topology of domains.
> https://rikai.mugen-giken.com/en/mathematics/complex-analysis/complex-numbers

## 0. Key points

- Complex numbers are not "a suspicious symbol $\sqrt{-1}$ handed down from above". Once the plane $\mathbb{R}^2$ is equipped with a suitable product, one obtains, in a single natural way, a field containing the real numbers. Here $i$ is a concrete element of that field, namely $(0,1)$.
- The meaning of complex multiplication is "multiply the absolute values, add the arguments" — that is, **a scaling composed with a rotation**. This single fact supports the geometric intuition behind all of complex analysis.
- If the exponential function is defined by a power series, Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ can be proved by computation alone. The polar form $z = re^{i\theta}$ is a consequence of it.
- De Moivre's theorem shows that a nonzero complex number has exactly $n$ $n$-th roots, and that they form the vertices of a regular $n$-gon centred at the origin.
- The field $\mathbb{C}$ admits no order relation compatible with its arithmetic. Complex numbers are never "larger" or "smaller" than one another; they are always measured by their **absolute value**.
- We may identify $\mathbb{C}$ with the metric space $\mathbb{R}^2$, and there open sets, connectedness and domains are defined. This is the stage on which later chapters place the statement "holomorphic functions are considered on a domain".

## 1. Motivation

### 1.1. Imaginary numbers were unavoidable in the search for real roots

Square roots of negative numbers were born from the attempt to solve $x^2 + 1 = 0$ — so one often reads, but history was not that simple. One can perfectly well end the discussion by saying that $x^2+1=0$ has no real solution. What made it impossible to remain inside the world of the real numbers was the attempt to solve **cubic equations that do have real roots**.

In sixteenth-century Italy, Cardano published in *Ars Magna* (1545) a method for solving the cubic $x^3 = px + q$. In modern notation it reads

$$
x = \sqrt[3]{\frac{q}{2} + \sqrt{\frac{q^2}{4} - \frac{p^3}{27}}} + \sqrt[3]{\frac{q}{2} - \sqrt{\frac{q^2}{4} - \frac{p^3}{27}}}.
$$

Let us substitute $p = 15,\ q = 4$. The equation $x^3 = 15x + 4$ has the root $x = 4$ (indeed $4^3 = 64 = 15\cdot 4 + 4$). But the inner square root of the formula is

$$
\frac{q^2}{4} - \frac{p^3}{27} = 4 - 125 = -121,
$$

so a square root of a negative number appears. To extract the real root $x=4$ from the formula, one has no choice but to pass through $\sqrt{-121}$. It is known that this must happen whenever the cubic has three distinct real roots; the phenomenon is called the **casus irreducibilis**, the irreducible case.

In his *Algebra* (1572) Bombelli verified that if one manipulates this "quantity that ought not to exist" formally, according to the rules of computation, the imaginary parts cancel along the way and the correct real root emerges. We carry this computation through to the end in <Ref to="ex-bombelli" />. Imaginary numbers were not "introduced because we wanted to solve something"; they appeared as **a place one is forced to pass through while solving a problem about real numbers**.

### 1.2. A phenomenon the real world cannot explain

Algebra is not the only motivation for introducing complex numbers. Consider the following two functions.

$$
f(x) = \frac{1}{1-x^2}, \qquad g(x) = \frac{1}{1+x^2} \qquad (x \in \mathbb{R})
$$

Both can be expanded around the origin as geometric series.

$$
f(x) = \sum_{n=0}^{\infty} x^{2n}, \qquad g(x) = \sum_{n=0}^{\infty} (-1)^n x^{2n}
$$

Each has radius of convergence $1$ and diverges for $|x| \ge 1$. For $f$ this is understandable: the denominator vanishes at $x = \pm 1$, so the function itself breaks down there. But $g$ is infinitely differentiable on all of $\mathbb{R}$ and breaks down nowhere. Why, then, does its power series diverge at $|x| = 1$? Surveying the real line alone, the number $1$ carries no special significance.

The answer lies outside the real line. The denominator $1+z^2$ of $g$ vanishes at $z = \pm i$, and these two points lie at distance $1$ from the origin. The radius of convergence $1$ is **the distance to the nearest singularity in the complex plane** (this fact is proved in a later chapter as <Ref to="mathematics/complex-analysis/properties-of-holomorphic-functions#cor-radius" text="radius of convergence and distance to the singularities" />). A question about the real numbers that the language of real numbers cannot answer is answered by the complex numbers — and this happens again and again throughout complex analysis. The computation of the definite integral $\int_{-\infty}^{\infty} \frac{dx}{1+x^2} = \pi$ in an instant by residues ([The residue theorem and its applications to definite integrals](/mathematics/complex-analysis/residue-theorem), <Ref to="mathematics/complex-analysis/residue-theorem#prop-rational-integral" text="improper integrals of rational functions" />) rests on the same pattern.

For power series and radii of convergence themselves see [Series and convergence tests](/mathematics/calculus/series-and-convergence), and for the remainder term in Taylor's theorem see [The mean value theorem and Taylor's theorem](/en/mathematics/calculus/mean-value-and-taylor).

<Aside type="note">
The name "imaginary" was used by Descartes half in mockery and then stuck, while "complex number" is Gauss's term. As the names suggest, more than two hundred years passed before $i$ was accepted as a number that genuinely exists. The matter was settled by the point of view described in the next section — defining complex numbers as points of the plane — due to Hamilton (1837).
</Aside>

## 2. Constructing the complex field

### 2.1. Putting a product on the plane

Defining $i$ as "the number whose square is $-1$" is circular: whether such a number exists is precisely the question. So we start instead from an object whose existence is already established — ordered pairs of real numbers — and define operations on it.

<Definition id="def-complex-number" title="The field of complex numbers">
On the set $\mathbb{C} := \mathbb{R}^2 = \{(a,b) \mid a, b \in \mathbb{R}\}$ define two operations by

$$
\begin{aligned}
(a,b) + (c,d) &:= (a+c,\ b+d), \\
(a,b) \cdot (c,d) &:= (ac - bd,\ ad + bc).
\end{aligned}
$$

Equipped with these operations, $\mathbb{C}$ is called the **field of complex numbers**, and its elements **complex numbers**.

We write $1 := (1,0)$ and $i := (0,1)$. For a complex number $z = (a,b)$ we call $a$ the **real part** and $b$ the **imaginary part**, written $\operatorname{Re} z := a$ and $\operatorname{Im} z := b$. A complex number with $\operatorname{Im} z = 0$ is identified with a **real number**, and a complex number with $\operatorname{Re} z = 0$ and $z \ne 0$ is called **purely imaginary**.
</Definition>

The definition of the product looks abrupt, but it is the unique formula forced by the requirement "write $(a,b)$ as $a + bi$, set $i^2 = -1$, and expand by distributivity". Indeed

$$
(a+bi)(c+di) = ac + adi + bci + bd\,i^2 = (ac - bd) + (ad + bc)i.
$$

Thus the product in <Ref to="def-complex-number" /> is not written down by anticipating the answer. If one tries to put on the plane a product that extends the arithmetic of the real numbers, preserves distributivity, and admits an element whose square is $-1$, then that product is necessarily given by this formula. Rather than starting from an $i$ whose existence is in doubt, we **adopt as the definition the formula that is forced upon us**. That is the intent of <Ref to="def-complex-number" />.

<Theorem id="thm-field" title="Basic properties of the complex field">
For the operations of <Ref to="def-complex-number" /> the following hold.

1. $(\mathbb{C}, +, \cdot)$ is a field. That is, addition and multiplication are commutative and associative, multiplication distributes over addition, there is an additive identity $(0,0)$ and a multiplicative identity $1 = (1,0)$, and every element other than $(0,0)$ has a multiplicative inverse.
2. The map $\iota : \mathbb{R} \to \mathbb{C},\ a \mapsto (a,0)$ is injective and preserves sums and products: $\iota(a+b) = \iota(a) + \iota(b)$ and $\iota(ab) = \iota(a)\iota(b)$.
3. $i^2 = -1$; that is, $(0,1)\cdot(0,1) = (-1,0)$.
4. Every $z = (a,b) \in \mathbb{C}$ can be written $z = \iota(a) + \iota(b)\, i$, and the numbers $a, b$ in this representation are unique.
</Theorem>

<Proof of="thm-field">
**(1) Addition.** Since addition is componentwise, the commutativity and associativity of addition in $\mathbb{R}$ carry over directly; the identity is $(0,0)$ and the inverse of $(a,b)$ is $(-a,-b)$.

**(1) Commutativity of multiplication.** We have $(c,d)\cdot(a,b) = (ca - db,\ cb + da)$, and since multiplication in $\mathbb{R}$ is commutative this equals $(ac-bd,\ ad+bc) = (a,b)\cdot(c,d)$.

**(1) Associativity of multiplication.** Computing from the left,

$$
\begin{aligned}
\bigl((a,b)(c,d)\bigr)(e,f) &= (ac-bd,\ ad+bc)(e,f) \\
&= \bigl((ac-bd)e - (ad+bc)f,\ (ac-bd)f + (ad+bc)e\bigr) \\
&= (ace - bde - adf - bcf,\ acf - bdf + ade + bce).
\end{aligned}
$$

Computing from the right,

$$
\begin{aligned}
(a,b)\bigl((c,d)(e,f)\bigr) &= (a,b)(ce-df,\ cf+de) \\
&= \bigl(a(ce-df) - b(cf+de),\ a(cf+de) + b(ce-df)\bigr) \\
&= (ace - adf - bcf - bde,\ acf + ade + bce - bdf).
\end{aligned}
$$

The first components agree and the second components agree, each being a rearrangement of the same terms.

**(1) Distributivity.**

$$
\begin{aligned}
(a,b)\bigl((c,d)+(e,f)\bigr) &= (a,b)(c+e,\ d+f) \\
&= \bigl(a(c+e) - b(d+f),\ a(d+f) + b(c+e)\bigr) \\
&= (ac - bd + ae - bf,\ ad + bc + af + be) \\
&= (ac-bd,\ ad+bc) + (ae-bf,\ af+be) \\
&= (a,b)(c,d) + (a,b)(e,f).
\end{aligned}
$$

**(1) Identity and inverses.** Since $(a,b)(1,0) = (a\cdot 1 - b\cdot 0,\ a\cdot 0 + b\cdot 1) = (a,b)$, the element $1=(1,0)$ is a multiplicative identity. If $(a,b) \ne (0,0)$ then $a^2+b^2 > 0$, and

$$
(a,b) \cdot \left( \frac{a}{a^2+b^2},\ \frac{-b}{a^2+b^2} \right)
= \left( \frac{a^2 + b^2}{a^2+b^2},\ \frac{-ab + ba}{a^2+b^2} \right) = (1,0),
$$

so an inverse exists. All the field axioms have now been verified.

**(2)** The equality $\iota(a) = \iota(b)$ means $(a,0)=(b,0)$, that is $a=b$, so $\iota$ is injective. For sums, $(a,0)+(b,0) = (a+b,0)$; for products, $(a,0)(b,0) = (ab - 0,\ 0 + 0) = (ab, 0)$. Both are preserved.

**(3)** Substituting into the definition, $(0,1)(0,1) = (0\cdot 0 - 1 \cdot 1,\ 0\cdot 1 + 1 \cdot 0) = (-1, 0)$. Since $(-1,0) = \iota(-1)$ is $-1$ under the identification, $i^2 = -1$ holds.

**(4)** We have $\iota(b)\,i = (b,0)(0,1) = (b\cdot 0 - 0 \cdot 1,\ b \cdot 1 + 0 \cdot 0) = (0,b)$, so $\iota(a) + \iota(b)i = (a,0)+(0,b) = (a,b)$. Uniqueness follows because $(a,b)=(a',b')$ means $a=a'$ and $b=b'$.
</Proof>

By (2) we may regard $\mathbb{R}$ as a subfield of $\mathbb{C}$, and by (4) we shall from now on write $(a,b)$ as $a+bi$. As a vector space over $\mathbb{R}$, the field $\mathbb{C}$ is two-dimensional with basis $\{1, i\}$.

<Aside type="caution">
The notation $i = \sqrt{-1}$ is convenient, but it breaks down if the symbol $\sqrt{\ \cdot\ }$ is extended to negative numbers without restriction. Indeed, granting $\sqrt{a}\sqrt{b} = \sqrt{ab}$ one can derive the falsehood

$$
-1 = i^2 = \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1.
$$

The identity $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is a rule valid only for $a, b \ge 0$. In this article $i$ is defined as $(0,1)$ following <Ref to="def-complex-number" />, and $\sqrt{\ \cdot\ }$ is used only for nonnegative real numbers.
</Aside>

### 2.2. Arithmetic, and an answer to the motivating question

Division is multiplication by the inverse. Using the inverse computed in the proof of <Ref to="thm-field" />, for $c+di \ne 0$ we get

$$
\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{(ac+bd) + (bc-ad)i}{c^2+d^2}.
$$

The operation of "multiplying by the conjugate" to make the denominator real is an application of the identity $z\bar z = |z|^2$, which we shall see in $\S 3$.

<Example id="ex-bombelli" title="Bombelli's computation: through the imaginary and back to a real root">
Let us apply Cardano's formula to the cubic $x^3 = 15x + 4$ of $\S 1$. With $p=15,\ q=4$,

$$
x = \sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}} = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}.
$$

Now let us guess a cube root of $2+11i$. Since $(2+i)^2 = 4 + 4i + i^2 = 3 + 4i$,

$$
(2+i)^3 = (2+i)(3+4i) = 6 + 8i + 3i + 4i^2 = 6 + 11i - 4 = 2 + 11i,
$$

so $2+i$ is one cube root of $2+11i$. Likewise $(2-i)^2 = 3-4i$ gives $(2-i)^3 = 2-11i$. Therefore

$$
x = (2+i) + (2-i) = 4.
$$

The imaginary parts $+i$ and $-i$ cancel exactly, and the real root $x=4$ appears. The $i$ that showed up along the way leaves no trace in the final answer, yet **one cannot arrive there without passing through it**. This was Bombelli's discovery.
</Example>

## 3. The complex plane, conjugation and absolute value

By <Ref to="def-complex-number" /> the set $\mathbb{C}$ is the plane itself, so a complex number $z = a+bi$ may be drawn as the point with coordinates $(a,b)$. This plane is called the **complex plane** or the **Gaussian plane**, its horizontal axis the **real axis** and its vertical axis the **imaginary axis**.

<Definition id="def-conjugate" title="Complex conjugate and absolute value">
For $z = a+bi$ with $a,b\in\mathbb{R}$ set

$$
\bar{z} := a - bi, \qquad |z| := \sqrt{a^2+b^2}.
$$

We call $\bar z$ the **complex conjugate** of $z$, and $|z| \ (\ge 0)$ the **absolute value** (or modulus) of $z$.
</Definition>

Geometrically $\bar z$ is the reflection of $z$ in the real axis, and $|z|$ is the Euclidean distance from the origin to $z$. When $z$ is real, $|z|$ agrees with the absolute value of a real number.

<Figure caption="The point z = a + bi in the complex plane. Here r is the distance from the origin and θ is the angle measured from the positive real axis. The lower point is the conjugate a − bi, the mirror image of z in the real axis.">
<svg viewBox="0 0 420 300" width="100%" role="img" aria-label="The complex plane with a point z, its polar representation and its conjugate">
  <line x1="20" y1="150" x2="400" y2="150" stroke="currentColor" stroke-width="1" opacity="0.55" />
  <line x1="120" y1="15" x2="120" y2="290" stroke="currentColor" stroke-width="1" opacity="0.55" />
  <line x1="120" y1="150" x2="255" y2="60" stroke="var(--sl-color-accent)" stroke-width="2" />
  <line x1="255" y1="60" x2="255" y2="150" stroke="currentColor" stroke-width="1" stroke-dasharray="4 3" opacity="0.7" />
  <line x1="120" y1="60" x2="255" y2="60" stroke="currentColor" stroke-width="1" stroke-dasharray="4 3" opacity="0.7" />
  <line x1="120" y1="150" x2="255" y2="240" stroke="currentColor" stroke-width="1.5" stroke-dasharray="5 4" opacity="0.8" />
  <path d="M 160 150 A 40 40 0 0 0 153.3 127.8" fill="none" stroke="var(--sl-color-accent)" stroke-width="1.5" />
  <circle cx="255" cy="60" r="4.5" fill="var(--sl-color-accent)" />
  <circle cx="255" cy="240" r="4.5" fill="currentColor" opacity="0.8" />
  <circle cx="120" cy="150" r="3" fill="currentColor" />
  <text x="266" y="55" font-size="15" fill="currentColor">z = a + bi</text>
  <text x="266" y="248" font-size="15" fill="currentColor">a − bi (conjugate)</text>
  <text x="106" y="167" font-size="14" fill="currentColor">O</text>
  <text x="185" y="98" font-size="15" fill="var(--sl-color-accent)">r</text>
  <text x="167" y="143" font-size="14" fill="var(--sl-color-accent)">θ</text>
  <text x="184" y="168" font-size="14" fill="currentColor">a</text>
  <text x="262" y="112" font-size="14" fill="currentColor">b</text>
  <text x="338" y="140" font-size="14" fill="currentColor">real axis</text>
  <text x="128" y="26" font-size="14" fill="currentColor">imaginary axis</text>
</svg>
</Figure>

<Proposition id="prop-conjugate" title="Basic laws of conjugation and absolute value">
Let $z, w \in \mathbb{C}$.

1. $\overline{z+w} = \bar z + \bar w$, $\overline{zw} = \bar z\,\bar w$, $\overline{\bar z} = z$.
2. $\operatorname{Re} z = \dfrac{z + \bar z}{2}$, $\operatorname{Im} z = \dfrac{z - \bar z}{2i}$. In particular $z \in \mathbb{R} \iff \bar z = z$.
3. $z\bar z = |z|^2$. In particular $z^{-1} = \dfrac{\bar z}{|z|^2}$ when $z \ne 0$.
4. $|zw| = |z|\,|w|$, and $\left| \dfrac{z}{w} \right| = \dfrac{|z|}{|w|}$ when $w \ne 0$.
5. $|z| = 0 \iff z = 0$, and $|\bar z| = |z|$.
</Proposition>

<Proof of="prop-conjugate">
Write $z = a+bi$ and $w = c+di$ with $a,b,c,d\in\mathbb{R}$.

**(1)** For sums, $\overline{(a+c)+(b+d)i} = (a+c) - (b+d)i = (a-bi)+(c-di)$. For products, $zw = (ac-bd)+(ad+bc)i$ gives $\overline{zw} = (ac-bd) - (ad+bc)i$, while on the other hand

$$
\bar z\,\bar w = (a-bi)(c-di) = (ac - bd) + \bigl(a(-d) + (-b)c\bigr)i = (ac-bd) - (ad+bc)i,
$$

and the two agree. Finally $\overline{\bar z} = \overline{a-bi} = a+bi = z$.

**(2)** From $z+\bar z = (a+bi)+(a-bi) = 2a$ we get $\operatorname{Re} z = (z+\bar z)/2$, and from $z - \bar z = 2bi$ we get $\operatorname{Im} z = b = (z-\bar z)/(2i)$. Hence $\bar z = z$ is equivalent to $2bi = 0$, that is $b=0$, which is equivalent to $z \in \mathbb{R}$.

**(3)** $z\bar z = (a+bi)(a-bi) = a^2 - (bi)^2 = a^2 + b^2 = |z|^2$, where we used $i^2=-1$ (<Ref to="thm-field" /> (3)). If $z \ne 0$ then $|z|^2 \ne 0$, so dividing both sides by $|z|^2$ gives $z \cdot \bar z /|z|^2 = 1$, that is $z^{-1} = \bar z/|z|^2$.

**(4)** Using (3) and (1),

$$
|zw|^2 = (zw)\overline{(zw)} = z w \bar z \bar w = (z\bar z)(w\bar w) = |z|^2|w|^2
$$

(along the way we used commutativity of multiplication, <Ref to="thm-field" /> (1)). Both sides are nonnegative reals, so taking square roots gives $|zw| = |z||w|$. For quotients, apply the first part to $z = (z/w)\cdot w$ to get $|z| = |z/w||w|$ and divide by $|w| \ne 0$.

**(5)** $|z| = 0$ is equivalent to $a^2+b^2 = 0$, and since squares of real numbers are nonnegative this is equivalent to $a=b=0$. Also $|\bar z| = \sqrt{a^2 + (-b)^2} = \sqrt{a^2+b^2} = |z|$.
</Proof>

<Example id="ex-two-squares" title="The two-squares identity given by multiplicativity of |zw| = |z||w|">
Squaring both sides of <Ref to="prop-conjugate" /> (4) and writing things out in components, we obtain for $z=a+bi$ and $w=c+di$

$$
(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2.
$$

This is the **Brahmagupta–Fibonacci identity**, which asserts that a product of two sums of two squares is again a sum of two squares.

Let us check a concrete case. Take $z = 2+3i$ and $w = 1+4i$, so that $|z|^2 = 13$ and $|w|^2 = 17$, and $zw = (2\cdot 1 - 3\cdot 4) + (2\cdot 4 + 3 \cdot 1)i = -10 + 11i$. Then

$$
13 \cdot 17 = 221 = (-10)^2 + 11^2 = 100 + 121 = 221,
$$

as it should be. If instead of $w$ we multiply by $\bar w = 1-4i$, we get $z\bar w = (2+12) + (-8+3)i = 14 - 5i$, giving the alternative representation $221 = 14^2+5^2 = 196+25$. In number theory this multiplicativity of complex numbers is what one uses to show that the set of numbers representable as a sum of two squares is closed under multiplication.
</Example>

<Theorem id="thm-triangle" title="The triangle inequality">
Let $z, w \in \mathbb{C}$.

1. $|\operatorname{Re} z| \le |z|$ and $|\operatorname{Im} z| \le |z|$.
2. $|z + w| \le |z| + |w|$. Equality holds if and only if $z = 0$, or there exists a real number $t \ge 0$ with $w = tz$.
3. $\bigl|\,|z| - |w|\,\bigr| \le |z - w|$ (the reverse triangle inequality).
</Theorem>

<Proof of="thm-triangle">
**(1)** Writing $z=a+bi$, we have $|a| = \sqrt{a^2} \le \sqrt{a^2+b^2} = |z|$ (using $b^2 \ge 0$ and monotonicity of $\sqrt{\ \cdot\ }$). The imaginary part is handled in the same way.

**(2)** Using <Ref to="prop-conjugate" /> (3), (1) and (2) in turn,

$$
\begin{aligned}
|z+w|^2 &= (z+w)\overline{(z+w)} = (z+w)(\bar z + \bar w) \\
&= z\bar z + z\bar w + w \bar z + w\bar w \\
&= |z|^2 + \bigl(z\bar w + \overline{z \bar w}\bigr) + |w|^2 \\
&= |z|^2 + 2\operatorname{Re}(z\bar w) + |w|^2
\end{aligned}
$$

(here we used $w\bar z = \overline{\bar w z} = \overline{z\bar w}$). Now by (1) together with <Ref to="prop-conjugate" /> (4) and (5),

$$
\operatorname{Re}(z\bar w) \le |z\bar w| = |z||\bar w| = |z||w|,
$$

whence

$$
|z+w|^2 \le |z|^2 + 2|z||w| + |w|^2 = (|z|+|w|)^2,
$$

and taking nonnegative square roots gives $|z+w| \le |z|+|w|$.

Let us examine equality. Equality in the estimate above holds precisely when $\operatorname{Re}(z\bar w) = |z\bar w|$, that is, when $\zeta := z\bar w$ satisfies $\operatorname{Re}\zeta = |\zeta|$. Writing $\zeta = \alpha + \beta i$ this reads $\alpha = \sqrt{\alpha^2+\beta^2}$; squaring both sides shows that $\beta = 0$ together with $\alpha \ge 0$ is necessary and sufficient. In other words, the condition is that $z\bar w$ be a nonnegative real number.

If $z = 0$ this holds automatically. If $z \ne 0$ and $z\bar w = s$ with $s \ge 0$ real, then $\bar w = s/z$, and taking conjugates, by <Ref to="prop-conjugate" /> (1) and (3),

$$
w = \overline{\left(\frac{s}{z}\right)} = \frac{s}{\bar z} = \frac{s z}{\bar z z} = \frac{s}{|z|^2}\, z,
$$

so setting $t := s/|z|^2 \ge 0$ we get $w = tz$. Conversely, if $w = tz$ with $t\ge 0$ then $z\bar w = t z \bar z = t|z|^2 \ge 0$, so equality holds.

**(3)** Applying (2) to $|z| = |(z-w) + w| \le |z-w| + |w|$ gives $|z| - |w| \le |z-w|$. Exchanging $z$ and $w$ gives $|w| - |z| \le |w-z| = |z-w|$ (that $|-\zeta| = |\zeta|$ follows from <Ref to="prop-conjugate" /> (4) with $|-1|=1$). Combining the two yields the claim.
</Proof>

The triangle inequality guarantees that $d(z,w) := |z-w|$ is a metric on $\mathbb{C}$. Indeed $d(z,w) = 0 \iff z=w$ (<Ref to="prop-conjugate" /> (5)), $d(z,w)=d(w,z)$, and $d(z,w) \le d(z,\zeta)+d(\zeta,w)$. This metric is exactly the metric of the Euclidean plane $\mathbb{R}^2$, so arguments about limits and continuity in $\mathbb{C}$ coincide completely with those in $\mathbb{R}^2$. The framework of $\varepsilon$-$\delta$ arguments can be taken over unchanged from [Limits and continuity](/en/mathematics/calculus/limits-and-continuity).

## 4. The polar form and Euler's formula

### 4.1. Defining the exponential by a power series

The key to understanding complex multiplication geometrically is the exponential function. We define it by the same series as in the real case:

$$
\exp(z) := \sum_{n=0}^{\infty} \frac{z^n}{n!} \qquad (z \in \mathbb{C}).
$$

This series converges absolutely for every $z$. Indeed $\sum_{n\ge 0} |z^n/n!| = \sum_{n \ge 0} |z|^n/n!$ is a series of nonnegative terms, and it converges by the ratio test (<Ref to="mathematics/calculus/series-and-convergence#thm-ratio" text="d'Alembert's ratio test" />), since $|z|^{n+1}/(n+1)! \div |z|^n/n! = |z|/(n+1) \to 0$. An absolutely convergent series has a sum unchanged by rearrangement, and two absolutely convergent series may be multiplied by the Cauchy product ([Series and convergence tests](/mathematics/calculus/series-and-convergence)). We use both facts below.

<Lemma id="lem-exp-addition" title="The exponential law">
For all $z, w \in \mathbb{C}$ we have $\exp(z+w) = \exp(z)\exp(w)$. In particular $\exp(z) \ne 0$ and $\exp(z)^{-1} = \exp(-z)$.
</Lemma>

<Proof of="lem-exp-addition">
Since $\sum z^n/n!$ and $\sum w^n/n!$ both converge absolutely, their Cauchy product converges to the product $\exp(z)\exp(w)$. The $n$-th term of the Cauchy product is

$$
\sum_{k=0}^{n} \frac{z^k}{k!}\cdot\frac{w^{n-k}}{(n-k)!}
= \frac{1}{n!}\sum_{k=0}^{n} \frac{n!}{k!\,(n-k)!} z^k w^{n-k}
= \frac{1}{n!}\sum_{k=0}^{n} \binom{n}{k} z^k w^{n-k}
= \frac{(z+w)^n}{n!},
$$

the last equality being the binomial theorem, which holds verbatim because $\mathbb{C}$ is a commutative ring (<Ref to="thm-field" />). Therefore

$$
\exp(z)\exp(w) = \sum_{n=0}^{\infty} \frac{(z+w)^n}{n!} = \exp(z+w).
$$

Setting $w = -z$ gives $\exp(z)\exp(-z) = \exp(0) = 1$, so $\exp(z)$ is nonzero with inverse $\exp(-z)$.
</Proof>

<Theorem id="thm-euler" title="Euler's formula">
For every real number $\theta$,

$$
\exp(i\theta) = \cos\theta + i \sin\theta.
$$

In particular $|\exp(i\theta)| = 1$, and taking $\theta = \pi$ gives $\exp(i\pi) + 1 = 0$.
</Theorem>

<Proof of="thm-euler">
The powers of $i$ cycle with period $4$: $i^0=1,\ i^1=i,\ i^2=-1,\ i^3=-i$, and in general $i^{2m} = (i^2)^m = (-1)^m$ and $i^{2m+1} = (-1)^m i$.

Since the series for $\exp(i\theta)$ converges absolutely, we are permitted to regroup it into its even-indexed and odd-indexed terms.

$$
\begin{aligned}
\exp(i\theta) &= \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!}
= \sum_{m=0}^{\infty} \frac{(i\theta)^{2m}}{(2m)!} + \sum_{m=0}^{\infty} \frac{(i\theta)^{2m+1}}{(2m+1)!} \\
&= \sum_{m=0}^{\infty} \frac{(-1)^m \theta^{2m}}{(2m)!} + i\sum_{m=0}^{\infty} \frac{(-1)^m \theta^{2m+1}}{(2m+1)!}.
\end{aligned}
$$

The two series on the right are precisely the Maclaurin expansions of $\cos\theta$ and $\sin\theta$ for real $\theta$ (this follows from the estimate $|\theta|^{N+1}/(N+1)! \to 0$ for the remainder term in Taylor's theorem; see [The mean value theorem and Taylor's theorem](/en/mathematics/calculus/mean-value-and-taylor)). Hence $\exp(i\theta) = \cos\theta + i\sin\theta$.

As for the absolute value, by <Ref to="def-conjugate" /> we have $|\exp(i\theta)|^2 = \cos^2\theta + \sin^2\theta = 1$, so $|\exp(i\theta)| = 1$. For $\theta=\pi$ we have $\cos\pi = -1$ and $\sin\pi = 0$, so $\exp(i\pi) = -1$.
</Proof>

<Aside type="tip">
If one adopts the convention of **defining** $\cos$ and $\sin$ by the two series above (rather than geometrically, by angles), then Euler's formula becomes an identity obtained purely by rearranging series. In that case one must in turn redefine what $\pi$ is, starting from the series: one sets $\pi$ to be twice the smallest positive zero of $\cos$. Either route leaves everything below unchanged.
</Aside>

From now on we write $\exp(z)$ as $e^z$. From <Ref to="lem-exp-addition" /> and <Ref to="thm-euler" />, for real $x$ and $y$,

$$
e^{x+iy} = e^x(\cos y + i \sin y), \qquad |e^{x+iy}| = e^x.
$$

### 4.2. The polar form

<Definition id="def-polar" title="Argument and polar form">
Let $z \ne 0$ and put $|z| = r$. If a real number $\theta$ satisfies $z = r e^{i\theta}$, then $\theta$ is called an **argument** of $z$, written $\theta \in \arg z$. The representation $z = re^{i\theta}$ is called the **polar form** of $z$. The element of $\arg z$ satisfying $-\pi < \theta \le \pi$ is called the **principal value** of the argument and is written $\operatorname{Arg} z$.
</Definition>

Let us verify that an argument exists at all, and how many there are.

<Remark id="rem-polar-exists">
For $z \ne 0$ the polar form exists, and the argument is unique up to integer multiples of $2\pi$.

**Existence.** Put $r=|z| > 0$ and $u := z/r$, so that $|u| = 1$ (<Ref to="prop-conjugate" /> (4)) and hence $u = a+bi$ satisfies $a^2+b^2=1$; in particular $a \in [-1,1]$. Since $\cos$ is continuous and strictly decreasing on $[0,\pi]$ with $\cos 0 = 1$ and $\cos\pi=-1$, the intermediate value theorem gives a unique $\theta_0 \in [0,\pi]$ with $\cos\theta_0 = a$ ([Limits and continuity](/en/mathematics/calculus/limits-and-continuity)). Then $\sin\theta_0 = \sqrt{1-a^2} = |b|$ (because $\sin \ge 0$ on $[0,\pi]$). Taking $\theta := \theta_0$ if $b \ge 0$ and $\theta := -\theta_0$ if $b < 0$, we get $\cos\theta = a$ and $\sin\theta = b$, so $u = e^{i\theta}$ and therefore $z = re^{i\theta}$.

**Uniqueness.** If $e^{i\alpha} = e^{i\beta}$ then <Ref to="lem-exp-addition" /> gives $e^{i(\alpha-\beta)} = 1$, that is $\cos(\alpha-\beta) = 1$ and $\sin(\alpha-\beta)=0$. Since $\cos$ has period $2\pi$ and $\cos t < 1$ on $(0,2\pi)$ (it decreases strictly from $1$ to $-1$ on $(0,\pi)$ and increases strictly from $-1$ to $1$ on $(\pi,2\pi)$), the equation $\cos t = 1$ forces $t \in 2\pi\mathbb{Z}$. Hence $\alpha - \beta \in 2\pi\mathbb{Z}$.
</Remark>

## 5. De Moivre's theorem and the n-th roots of unity

<Theorem id="thm-demoivre" title="De Moivre's theorem">
Let $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$ with $r_1, r_2 > 0$ and $\theta_1,\theta_2 \in \mathbb{R}$.

1. $z_1 z_2 = r_1 r_2\, e^{i(\theta_1+\theta_2)}$ and $\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2} e^{i(\theta_1-\theta_2)}$. That is, in a product of complex numbers **the absolute values are multiplied and the arguments are added**.
2. For $r > 0$ and $\theta \in \mathbb{R}$, and for every integer $n$, we have $\bigl(re^{i\theta}\bigr)^n = r^n e^{in\theta}$.

In particular, taking $r = 1$ in 2 and rewriting in terms of trigonometric functions, $(\cos\theta + i \sin\theta)^n = \cos n\theta + i \sin n\theta$ for every real $\theta$ and every integer $n$.
</Theorem>

<Proof of="thm-demoivre">
**(1)** By <Ref to="lem-exp-addition" /> we have $e^{i\theta_1}e^{i\theta_2} = e^{i(\theta_1+\theta_2)}$, and since the real factors $r_1,r_2$ commute and can be pulled out, $z_1z_2 = r_1r_2 e^{i(\theta_1+\theta_2)}$. For the quotient, <Ref to="lem-exp-addition" /> gives $(e^{i\theta_2})^{-1} = e^{-i\theta_2}$, so

$$
\frac{z_1}{z_2} = r_1 e^{i\theta_1}\cdot \frac{1}{r_2}e^{-i\theta_2} = \frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}.
$$

**(2)** First we treat $n \ge 0$ by induction. For $n=0$ both sides equal $1$. Assuming the claim for $n$, apply (1) with $z_1 = r^n e^{in\theta}$ and $z_2 = re^{i\theta}$:

$$
(re^{i\theta})^{n+1} = (re^{i\theta})^n \cdot re^{i\theta} = r^n e^{in\theta}\cdot re^{i\theta} = r^{n+1}e^{i(n+1)\theta}.
$$

For $n < 0$ put $m := -n > 0$; then by the case already proved and <Ref to="lem-exp-addition" />,

$$
(re^{i\theta})^{n} = \bigl((re^{i\theta})^{m}\bigr)^{-1} = \bigl(r^m e^{im\theta}\bigr)^{-1} = r^{-m}e^{-im\theta} = r^n e^{in\theta}.
$$

Finally, setting $r=1$ and applying <Ref to="thm-euler" /> to both sides gives the trigonometric form.
</Proof>

What (1) says is that the map $z \mapsto z_2 z$, "multiplication by $z_2$", is a scaling by the factor $r_2$ about the origin composed with a rotation by $\theta_2$. Whereas multiplication of real numbers was a stretching of the number line, multiplication of complex numbers is a **rotation-and-scaling** of the plane. This fact reappears in a later chapter in the form of conformality: the derivative of a holomorphic function behaves locally like a rotation-and-scaling ([Holomorphic functions and the Cauchy–Riemann equations](/en/mathematics/complex-analysis/holomorphic-functions), [Conformal mappings and the Riemann mapping theorem](/mathematics/complex-analysis/conformal-mapping), <Ref to="mathematics/complex-analysis/conformal-mapping#thm-conformality" text="local behaviour of angles under holomorphic maps" />).

<Example id="ex-polar-power" title="Computing powers via the polar form">
Let us compute $z^{10}$ for $z = -1 + \sqrt{3}\,i$. First, $|z| = \sqrt{(-1)^2 + (\sqrt3)^2} = \sqrt{4} = 2$. Next, $z/2 = -\frac12 + \frac{\sqrt3}{2}i$, and $\cos\theta = -\frac12,\ \sin\theta = \frac{\sqrt3}{2}$ is satisfied by $\theta = \frac{2\pi}{3}$, so $z = 2e^{2\pi i/3}$. By <Ref to="thm-demoivre" /> (2),

$$
z^{10} = 2^{10} e^{20\pi i/3} = 1024\, e^{i(20\pi/3 - 6\pi)} = 1024\, e^{2\pi i/3}
$$

(we used $e^{-6\pi i} = 1$). Therefore

$$
z^{10} = 1024\left(-\frac12 + \frac{\sqrt3}{2}i\right) = -512 + 512\sqrt{3}\, i.
$$

Expanding $(-1+\sqrt3 i)^{10}$ by the binomial theorem would mean computing $11$ terms; with the polar form two lines suffice.
</Example>

<Example id="ex-cos5theta" title="Generating multiple-angle formulas wholesale">
Expanding the left-hand side of <Ref to="thm-demoivre" /> (2) by the binomial theorem and comparing real parts expresses $\cos n\theta$ as a polynomial in $\cos\theta$ and $\sin\theta$. Let us do this for $n=5$. Abbreviating $c=\cos\theta$ and $s=\sin\theta$,

$$
(c+is)^5 = c^5 + 5c^4(is) + 10c^3(is)^2 + 10c^2(is)^3 + 5c(is)^4 + (is)^5,
$$

and, noting that $i^2=-1,\ i^3=-i,\ i^4=1,\ i^5=i$, extracting the real part alone gives

$$
\cos 5\theta = c^5 - 10c^3s^2 + 5cs^4.
$$

Substituting $s^2 = 1-c^2$,

$$
\begin{aligned}
\cos 5\theta &= c^5 - 10c^3(1-c^2) + 5c(1-c^2)^2 \\
&= c^5 - 10c^3 + 10c^5 + 5c(1 - 2c^2 + c^4) \\
&= c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5 \\
&= 16c^5 - 20c^3 + 5c.
\end{aligned}
$$

Let us check this. For $\theta = 0$ the right-hand side is $16-20+5 = 1 = \cos 0$, and for $\theta=\pi$ it is $-16+20-5 = -1 = \cos\pi$; both agree. This polynomial is nothing other than the fifth Chebyshev polynomial $T_5$.
</Example>

<Theorem id="thm-nth-roots" title="The n-th roots of a complex number">
Let $n$ be a positive integer and let $w = \rho e^{i\varphi}$, with $\rho > 0$ and $\varphi\in\mathbb{R}$, be a nonzero complex number. Then the equation $z^n = w$ has exactly $n$ solutions, given by

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

where $\rho^{1/n}$ denotes the positive $n$-th root of the positive real number $\rho$. These points are equally spaced on the circle of radius $\rho^{1/n}$ centred at the origin, and for $n \ge 3$ they form the vertices of a regular $n$-gon.

In particular, for $w=1$ the set of all solutions $\mu_n = \{1, \zeta, \zeta^2, \ldots, \zeta^{n-1}\}$, where $\zeta := e^{2\pi i/n}$, is a cyclic group of order $n$ under multiplication, and for $n \ge 2$

$$
1 + \zeta + \zeta^2 + \cdots + \zeta^{n-1} = 0.
$$
</Theorem>

<Proof of="thm-nth-roots">
**Each $z_k$ is a solution.** By <Ref to="thm-demoivre" /> (2),

$$
z_k^n = \rho\,\exp\!\left(i\, n\cdot \frac{\varphi+2\pi k}{n}\right) = \rho\, e^{i(\varphi + 2\pi k)} = \rho\, e^{i\varphi}e^{2\pi i k} = w,
$$

where at the last step we used $e^{2\pi i k} = (\cos 2\pi + i \sin 2\pi)^k = 1$ (<Ref to="thm-euler" />).

**They are distinct.** Suppose $z_k = z_l$ for $0 \le k < l \le n-1$. By the uniqueness part of <Ref to="rem-polar-exists" />,

$$
\frac{\varphi+2\pi l}{n} - \frac{\varphi+2\pi k}{n} = \frac{2\pi(l-k)}{n} \in 2\pi\mathbb{Z},
$$

that is $(l-k)/n \in \mathbb{Z}$, which is impossible since $0 < l-k \le n-1$. Hence $z_0,\ldots,z_{n-1}$ are pairwise distinct.

**There are no other solutions.** If $z^n = w \ne 0$ then $z \ne 0$, so by <Ref to="rem-polar-exists" /> we may write $z = re^{i\theta}$ with $r>0$. By <Ref to="thm-demoivre" /> (2) we get $r^n e^{in\theta} = \rho e^{i\varphi}$, and taking absolute values of both sides (using $|e^{it}|=1$ from <Ref to="thm-euler" />) gives $r^n = \rho$. Since $t \mapsto t^n$ is strictly increasing on the positive reals, $r = \rho^{1/n}$ is determined. From the remaining equation $e^{in\theta} = e^{i\varphi}$, the uniqueness part of <Ref to="rem-polar-exists" /> yields an integer $k$ with $n\theta - \varphi = 2\pi k$, so $\theta = (\varphi+2\pi k)/n$. Replacing $k$ by its remainder modulo $n$ changes $\theta$ only by an integer multiple of $2\pi$ and leaves $z$ unchanged, so $z \in \{z_0,\ldots,z_{n-1}\}$.

**Geometric arrangement.** The modulus $|z_k| = \rho^{1/n}$ does not depend on $k$, and the difference of the arguments of consecutive points $z_k, z_{k+1}$ is constantly $2\pi/n$. Hence the $n$ points lie equally spaced on one circle and, for $n\ge3$, form the vertices of a regular $n$-gon.

**It is a cyclic group.** For $w=1$ we may take $\varphi=0$ and $\rho=1$, giving $z_k = e^{2\pi i k/n} = \zeta^k$ (<Ref to="thm-demoivre" /> (2)). Since $\zeta^n = 1$, the set $\mu_n$ is closed under multiplication, and $(\zeta^k)^{-1} = \zeta^{n-k} \in \mu_n$, so it is a group; it is cyclic of order $n$ with generator $\zeta$.

**The sum is $0$.** For $n \ge 2$ we have $\zeta \ne 1$ (since $2\pi/n \notin 2\pi\mathbb{Z}$, by <Ref to="rem-polar-exists" />). The formula for the sum of a geometric progression in a field (that $\mathbb{C}$ is a field is <Ref to="thm-field" />) gives

$$
\sum_{k=0}^{n-1}\zeta^k = \frac{\zeta^n - 1}{\zeta - 1} = \frac{1-1}{\zeta-1} = 0.
$$
</Proof>

<Figure caption="The fifth roots of unity. The powers of ζ = exp(2πi/5) are equally spaced on the unit circle and form the vertices of a regular pentagon. The five points sum to 0.">
<svg viewBox="0 0 340 340" width="100%" role="img" aria-label="The fifth roots of unity on the unit circle, forming a regular pentagon">
  <line x1="20" y1="170" x2="320" y2="170" stroke="currentColor" stroke-width="1" opacity="0.4" />
  <line x1="170" y1="20" x2="170" y2="320" stroke="currentColor" stroke-width="1" opacity="0.4" />
  <circle cx="170" cy="170" r="110" fill="none" stroke="currentColor" stroke-width="1.5" opacity="0.65" />
  <polygon points="280,170 204,65.4 81,105.3 81,234.7 204,274.6" fill="none" stroke="var(--sl-color-accent)" stroke-width="2" />
  <circle cx="280" cy="170" r="5" fill="var(--sl-color-accent)" />
  <circle cx="204" cy="65.4" r="5" fill="var(--sl-color-accent)" />
  <circle cx="81" cy="105.3" r="5" fill="var(--sl-color-accent)" />
  <circle cx="81" cy="234.7" r="5" fill="var(--sl-color-accent)" />
  <circle cx="204" cy="274.6" r="5" fill="var(--sl-color-accent)" />
  <circle cx="170" cy="170" r="3" fill="currentColor" />
  <text x="288" y="166" font-size="15" fill="currentColor">1</text>
  <text x="212" y="56" font-size="15" fill="currentColor">ζ</text>
  <text x="52" y="100" font-size="15" fill="currentColor">ζ²</text>
  <text x="52" y="244" font-size="15" fill="currentColor">ζ³</text>
  <text x="212" y="288" font-size="15" fill="currentColor">ζ⁴</text>
  <text x="178" y="186" font-size="14" fill="currentColor">O</text>
  <text x="196" y="150" font-size="13" fill="var(--sl-color-accent)">2π/5</text>
</svg>
</Figure>

<Example id="ex-fifth-roots" title="Computing cos(2π/5) from the fifth roots of unity">
Let $\zeta = e^{2\pi i/5}$. By <Ref to="thm-nth-roots" /> we have $1+\zeta+\zeta^2+\zeta^3+\zeta^4 = 0$. Using <Ref to="thm-euler" /> together with $\overline{\zeta^k} = \zeta^{-k} = \zeta^{5-k}$,

$$
\zeta + \zeta^4 = 2\cos\frac{2\pi}{5}, \qquad \zeta^2 + \zeta^3 = 2\cos\frac{4\pi}{5}.
$$

Put $u := \zeta+\zeta^4$ and $v := \zeta^2+\zeta^3$. Their sum is

$$
u + v = (\zeta+\zeta^2+\zeta^3+\zeta^4) = -1.
$$

Their product, using $\zeta^5=1$, is

$$
uv = (\zeta+\zeta^4)(\zeta^2+\zeta^3) = \zeta^3 + \zeta^4 + \zeta^6 + \zeta^7 = \zeta^3+\zeta^4+\zeta+\zeta^2 = -1.
$$

Hence $u$ and $v$ are the two roots of $t^2 + t - 1 = 0$, namely $t = \dfrac{-1\pm\sqrt5}{2}$. Since $2\pi/5 = 72^\circ$ is an angle in the first quadrant, $u = 2\cos(2\pi/5) > 0$, and therefore

$$
u = \frac{-1+\sqrt5}{2}, \qquad \cos\frac{2\pi}{5} = \frac{\sqrt5 - 1}{4}.
$$

Numerically, $(\sqrt5-1)/4 = (2.2360679\ldots-1)/4 = 0.3090169\ldots$, which agrees with $\cos 72^\circ = 0.309016\ldots$. That the value can be written using only the square root $\sqrt5$ is the algebraic reason why the regular pentagon is constructible with ruler and compass.
</Example>

<Aside type="tip">
Since $z^n = w$ has $n$ solutions, the symbol $\sqrt[n]{w}$ does not determine a single value in the complex world: "$n$-th root" is essentially multivalued. If one tries to specify the choice continuously, going once around the origin moves the value onto a different branch — a phenomenon that leads to the logarithm and to branch points, treated in later chapters.
</Aside>

## 6. Topology of the complex plane: open sets and domains

As we saw in $\S 3$, the metric on $\mathbb{C}$ coincides with the Euclidean metric on $\mathbb{R}^2$. In order to say precisely what $\Omega$ means when later chapters speak of "a function holomorphic on $\Omega$", we fix here the necessary topological vocabulary.

<Definition id="def-open-set" title="Discs, open sets, closed sets">
For $a \in \mathbb{C}$ and $r > 0$ the set

$$
D(a,r) := \{ z \in \mathbb{C} : |z - a| < r \}
$$

is called the **open disc** of radius $r$ centred at $a$. For a set $S \subseteq \mathbb{C}$:

- A point $a \in S$ is an **interior point** of $S$ if there exists $r>0$ with $D(a,r) \subseteq S$.
- $S$ is **open** if every point of $S$ is an interior point of $S$.
- $S$ is **closed** if the complement $\mathbb{C}\setminus S$ is open.
- $S$ is **bounded** if there exists $M>0$ such that $|z| \le M$ for all $z \in S$.
- A point $a \in \mathbb{C}$ is a **boundary point** of $S$ if for every $r>0$ the disc $D(a,r)$ contains both a point of $S$ and a point of $\mathbb{C}\setminus S$. The set of all boundary points is written $\partial S$.
</Definition>

<Definition id="def-domain" title="Connectedness and domains">
An open set $\Omega \subseteq \mathbb{C}$ is **connected** if, whenever $\Omega = U \cup V$ with $U \cap V = \varnothing$ and $U, V$ both open, necessarily $U = \varnothing$ or $V = \varnothing$.

A nonempty connected open set is called a **domain**.
</Definition>

The definition of connectedness is a negative statement — "cannot be split in two" — and as such is awkward to use directly. For open sets it can be restated concretely as follows.

<Proposition id="prop-polygonal" title="Connectedness and polygonal connectedness of open sets">
Let $\Omega \subseteq \mathbb{C}$ be a nonempty open set. The following two conditions are equivalent.

1. $\Omega$ is connected in the sense of <Ref to="def-domain" />.
2. For all $p, q \in \Omega$ there exists a polygonal path starting at $p$, ending at $q$, all of whose segments are contained in $\Omega$.
</Proposition>

<Proof of="prop-polygonal">
**(2) $\Rightarrow$ (1).** We prove the contrapositive. Suppose $\Omega = U \cup V$ with $U\cap V = \varnothing$ and $U,V$ open and both nonempty. Choose $p \in U$ and $q \in V$, and let $p = p_0, p_1, \ldots, p_m = q$ be the polygonal path given by (2). Since $p_0 \in U$ and $p_m \in V$, there is an index $j$ with $p_j \in U$ and $p_{j+1}\in V$.

Write the segment as $\gamma(t) := p_j + t(p_{j+1}-p_j)$ for $t\in[0,1]$, and set $S := \{t \in [0,1] : \gamma(t) \in U\}$ and $t^* := \sup S$. Since $0 \in S$ we have $S \ne \varnothing$, and $S \subseteq [0,1]$ is bounded above, so the supremum exists. For $c := \gamma(t^*) \in \Omega = U \cup V$ we rule out two cases.

Case $c \in U$. Since $U$ is open there is $\varepsilon>0$ with $D(c,\varepsilon)\subseteq U$. Because $|\gamma(t)-\gamma(t^*)| = |t-t^*|\,|p_{j+1}-p_j|$, we have $\gamma(t) \in U$ whenever $|t - t^*|$ is small enough. If $t^* = 1$ then $c = p_{j+1} \in U \cap V = \varnothing$, a contradiction. If $t^* < 1$ then some $t \in S$ exceeds $t^*$, contradicting that $t^*$ is the supremum.

Case $c \in V$. Similarly take $\varepsilon>0$ with $D(c,\varepsilon)\subseteq V$; then there is $\delta>0$ such that $\gamma(t)\in V$ for all $t\in[0,1]$ with $|t-t^*| < \delta$, hence $\gamma(t)\notin U$ (as $U\cap V=\varnothing$). But then $S \cap (t^*-\delta, t^*] = \varnothing$, so $\sup S \le t^* - \delta < t^*$, contradicting the definition of $t^*$. (Note that $t^*>0$ follows from $0 \in S$ and the openness of $U$: since $\gamma(0)=p_j\in U$, nearby values of $t$ also lie in $S$.)

Both cases are contradictory, so no such splitting $U, V$ exists and $\Omega$ is connected.

**(1) $\Rightarrow$ (2).** Fix $a \in \Omega$ and set

$$
U := \{ z\in\Omega : a \text{ and } z \text{ can be joined by a polygonal path inside } \Omega\}, \qquad V := \Omega\setminus U.
$$

Since $a \in U$ (via the path of length $0$), we have $U \ne \varnothing$.

$U$ is open. Given $z \in U$, openness of $\Omega$ provides $r>0$ with $D(z,r)\subseteq\Omega$. A disc is convex, so for every $w \in D(z,r)$ the segment $[z,w]$ lies in $D(z,r)\subseteq\Omega$. Appending this segment to a polygonal path from $a$ to $z$ produces a polygonal path from $a$ to $w$, so $w \in U$. Hence $D(z,r)\subseteq U$.

$V$ is open. Given $z\in V$, take $D(z,r)\subseteq\Omega$ in the same way. If some $w\in D(z,r)$ belonged to $U$, then appending the segment $[w,z]$ to a polygonal path from $a$ to $w$ would give $z \in U$, a contradiction. Hence $D(z,r)\subseteq V$.

Thus $U, V$ split $\Omega$ into open sets with $U\cap V=\varnothing$ and $U\ne\varnothing$, so (1) forces $V=\varnothing$; that is, every point of $\Omega$ can be joined to $a$ by a polygonal path. For two points $p,q\in\Omega$, concatenate a polygonal path from $p$ to $a$ (traverse the path from $a$ to $p$ backwards) with one from $a$ to $q$.
</Proof>

<Example id="ex-topology" title="Deciding openness, closedness and being a domain">
**(a) The annulus $A = \{ z : 1 < |z| < 2 \}$ is a domain.** Let us show it is open. For $a \in A$ put $r := \min\{\,|a|-1,\ 2-|a|\,\} > 0$. If $|z-a| < r$ then by <Ref to="thm-triangle" /> (3) we have $\bigl||z|-|a|\bigr| \le |z-a| < r$, hence

$$
|z| > |a| - r \ge |a| - (|a|-1) = 1, \qquad |z| < |a| + r \le |a| + (2-|a|) = 2,
$$

so $z \in A$, that is $D(a,r)\subseteq A$. Connectedness is checked with <Ref to="prop-polygonal" />: any two points of $A$ can be joined by first moving radially along a segment to a point on the circle $|z| = 3/2$, and then following a polygonal path inscribed in that circle (each segment can be taken inside $A$).

**(b) The closed disc $\overline{D} = \{ z : |z| \le 1\}$ is closed but not open.** That the complement $\{|z| > 1\}$ is open follows from the same estimate as in (a). It is not open because $1 \in \overline{D}$ is not an interior point: for any $r > 0$, taking $z := 1 + \min\{r,1\}/2$ gives $|z - 1| < r$ and $|z| > 1$, so $D(1,r) \not\subseteq \overline D$.

**(c) The set $H = \{ z : \operatorname{Re} z \ne 0 \}$ is open but not a domain.** Let us show it is open. For $a \in H$ put $r := |\operatorname{Re} a| > 0$. If $|z-a| < r$ then by <Ref to="thm-triangle" /> (1)

$$
|\operatorname{Re} z - \operatorname{Re} a| = |\operatorname{Re}(z-a)| \le |z-a| < r = |\operatorname{Re} a|.
$$

If $\operatorname{Re} z = 0$ the left-hand side would equal $|\operatorname{Re} a|$, contradicting this inequality. Hence $\operatorname{Re} z \ne 0$, that is $D(a,r)\subseteq H$. But $H = \{\operatorname{Re} z > 0\} \cup \{\operatorname{Re} z < 0\}$ is a disjoint union of two nonempty open sets, so $H$ is not connected in the sense of <Ref to="def-domain" />. Indeed there is no polygonal path inside $H$ joining $-1$ to $1$, since such a path would have to cross the imaginary axis.

**(d) The punctured disc $D(0,1)\setminus\{0\}$ is a domain.** Let us show it is open. For $0 < |a| < 1$ put $r := \min\{\,|a|,\ 1-|a|\,\} > 0$. If $|z-a| < r$ then, as in (a), <Ref to="thm-triangle" /> (3) gives $|z| > |a| - r \ge 0$ and $|z| < |a| + r \le 1$; in particular $|z| > 0$, that is $z \ne 0$. As for connectedness: unless two points lie on the same ray from the origin they can be joined directly by a polygonal path avoiding the origin, and if they do lie on the same ray it suffices to step slightly to the side once. This set is the standard stage on which later chapters consider <Ref to="mathematics/complex-analysis/properties-of-holomorphic-functions#thm-laurent" text="the Laurent expansion" /> and residues.
</Example>

<Aside type="note">
Why insist on "domains" (connected open sets) rather than merely open sets? Because on a disconnected open set a function may be prescribed independently on each connected component. For instance the assertion "if the derivative vanishes identically then the function is constant" fails on a disconnected open set: one may take a different constant on each component. Theorems of the type "information on a part determines the whole", such as <Ref to="mathematics/complex-analysis/properties-of-holomorphic-functions#thm-identity" text="the identity theorem" /> and analytic continuation, all use connectedness in an essential way ([Powerful properties of holomorphic functions](/mathematics/complex-analysis/properties-of-holomorphic-functions)).
</Aside>

<Remark id="rem-completeness">
Since the metric on $\mathbb{C}$ coincides with the Euclidean metric on $\mathbb{R}^2$ ($\S 3$), the topological properties of $\mathbb{R}^2$ become properties of $\mathbb{C}$ verbatim. In particular the following hold.

- **Completeness**: every Cauchy sequence $(z_n)$ of complex numbers converges. This follows because the real and imaginary parts are each Cauchy sequences in $\mathbb{R}$ (by <Ref to="thm-triangle" /> (1)), together with the completeness of $\mathbb{R}$.
- **The Bolzano–Weierstrass theorem**: every bounded sequence of complex numbers has a convergent subsequence.
- **The Heine–Borel theorem**: a set $K\subseteq\mathbb{C}$ is compact (every open cover has a finite subcover) if and only if $K$ is closed and bounded.

In each case the proof is the standard one for $\mathbb{R}^2$ (split into real and imaginary parts and apply the result for $\mathbb{R}$), so we leave it to the textbooks listed in the references.
</Remark>

## 7. Exercises

<Exercise id="exr-no-order" difficulty="Standard">
Show that there is no total order $\le$ on $\mathbb{C}$ compatible with the field operations — that is, no total order such that $x \le y$ implies $x+z \le y+z$ for every $z$, and such that $0 \le x$ and $0\le y$ imply $0 \le xy$. Consequently an inequality "$z < w$" has no meaning unless $z$ and $w$ are real.

<Solution>
Suppose such an order $\le$ existed.

**Lemma: if $x \ne 0$ then $0 < x^2$.** Since the order is total, either $0 \le x$ or $x \le 0$. In the case $0 \le x$, apply the compatibility condition ($0\le x,\ 0\le y \Rightarrow 0 \le xy$) with $y=x$ to get $0 \le x^2$. In the case $x \le 0$, apply the translation condition with $z = -x$ to get $x + (-x) \le 0 + (-x)$, that is $0 \le -x$, and then the same compatibility condition gives $0 \le (-x)(-x) = x^2$. In either case $0 \le x^2$, and if $x \ne 0$ then $x^2 \ne 0$ (a field has no zero divisors, <Ref to="thm-field" />), so $0 < x^2$.

Applying the lemma with $x=1$ gives $0 < 1^2 = 1$, and applying it with $x=i$ gives $0 < i^2 = -1$ (<Ref to="thm-field" /> (3)). Adding $1$ to both sides of the latter gives $1 < 0$, contradicting $0 < 1$. Hence no such order exists.

Because of this result, the "size" of a complex number is always measured by the absolute value $|z|$, which is real. Looking back at the places where inequalities occur from $\S 3$ onwards, one finds that both sides of every inequality sign are real numbers.
</Solution>
</Exercise>

<Exercise id="exr-parallelogram" difficulty="Easy">
Show that for all $z, w \in \mathbb{C}$

$$
|z+w|^2 + |z-w|^2 = 2\bigl(|z|^2 + |w|^2\bigr),
$$

and state the geometric meaning of this identity.

<Solution>
We expand both terms using <Ref to="prop-conjugate" /> (3) and (1). As shown in the proof of <Ref to="thm-triangle" />,

$$
|z+w|^2 = |z|^2 + 2\operatorname{Re}(z\bar w) + |w|^2,
$$

and replacing $w$ by $-w$, using $\operatorname{Re}(z\overline{(-w)}) = -\operatorname{Re}(z\bar w)$ and $|-w| = |w|$,

$$
|z-w|^2 = |z|^2 - 2\operatorname{Re}(z\bar w) + |w|^2.
$$

Adding the two identities, the terms in $\operatorname{Re}(z\bar w)$ cancel and we obtain $|z+w|^2+|z-w|^2 = 2|z|^2+2|w|^2$.

Geometrically, in the parallelogram with vertices $0,\ z,\ z+w,\ w$, the quantities $|z+w|$ and $|z-w|$ are the lengths of the two diagonals, while $|z|$ and $|w|$ are the lengths of the two pairs of sides. The identity therefore states the **parallelogram law**: the sum of the squares of the two diagonals of a parallelogram equals the sum of the squares of its four sides. In the theory of normed spaces it is known that this identity holds precisely when the norm is induced by an inner product.
</Solution>
</Exercise>

<Exercise id="exr-quartic" difficulty="Standard">
Find all solutions of the equation $z^4 = -4$, and use the result to factor the real polynomial $x^4+4$ into a product of two quadratics with real coefficients.

<Solution>
Since $-4 = 4e^{i\pi}$, applying <Ref to="thm-nth-roots" /> with $n=4,\ \rho=4,\ \varphi=\pi$ gives solutions with modulus $4^{1/4} = \sqrt2$:

$$
z_k = \sqrt{2}\,\exp\!\left(i\,\frac{\pi + 2\pi k}{4}\right)\qquad (k=0,1,2,3).
$$

Writing these out with <Ref to="thm-euler" />, and using $\cos(\pi/4)=\sin(\pi/4)=\frac{\sqrt2}{2}$,

$$
z_0 = \sqrt2\left(\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}i\right) = 1+i,\quad z_1 = -1+i,\quad z_2 = -1-i,\quad z_3 = 1-i
$$

(each of $z_1,z_2,z_3$ is $z_0$ rotated successively by $\pi/2$, <Ref to="thm-demoivre" /> (1)). As a check, $(1+i)^2 = 2i$, so $(1+i)^4 = (2i)^2 = -4$; it is indeed a solution.

Now the factorization. The complex roots of $x^4+4$ are the four numbers above, and pairing conjugate roots produces quadratics with real coefficients.

$$
\bigl(x-(1+i)\bigr)\bigl(x-(1-i)\bigr) = (x-1)^2 - i^2 = x^2-2x+2
$$

$$
\bigl(x-(-1+i)\bigr)\bigl(x-(-1-i)\bigr) = (x+1)^2 - i^2 = x^2+2x+2
$$

Hence $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. As a check,

$$
(x^2+2-2x)(x^2+2+2x) = (x^2+2)^2 - (2x)^2 = x^4+4x^2+4-4x^2 = x^4+4,
$$

which agrees. (This is the case $b=1$ of Sophie Germain's identity $a^4+4b^4 = (a^2-2ab+2b^2)(a^2+2ab+2b^2)$.)
</Solution>
</Exercise>

<Exercise id="exr-dirichlet" difficulty="Hard">
Let $\theta$ be a real number that is not an integer multiple of $2\pi$, and let $n$ be a positive integer. Express $\sum_{k=0}^{n-1} e^{ik\theta}$ in closed form using the sum of a geometric progression, and deduce from it

$$
\sum_{k=0}^{n-1}\cos k\theta, \qquad \sum_{k=0}^{n-1}\sin k\theta.
$$

<Solution>
By hypothesis $e^{i\theta}\ne 1$ (uniqueness in <Ref to="rem-polar-exists" />). Since $\mathbb{C}$ is a field (<Ref to="thm-field" />) we may use the formula for the sum of a geometric progression, and $(e^{i\theta})^k = e^{ik\theta}$ by <Ref to="thm-demoivre" /> (2), so

$$
S := \sum_{k=0}^{n-1} e^{ik\theta} = \frac{e^{in\theta}-1}{e^{i\theta}-1}.
$$

The standard move is to factor out half-angles from numerator and denominator. By <Ref to="lem-exp-addition" /> and <Ref to="thm-euler" />,

$$
e^{i\alpha}-1 = e^{i\alpha/2}\bigl(e^{i\alpha/2}-e^{-i\alpha/2}\bigr) = e^{i\alpha/2}\cdot 2i\sin\frac{\alpha}{2}
$$

(we used $e^{i\beta}-e^{-i\beta} = (\cos\beta+i\sin\beta)-(\cos\beta-i\sin\beta) = 2i\sin\beta$). Applying this with $\alpha = n\theta$ in the numerator and $\alpha=\theta$ in the denominator, and noting $\sin(\theta/2)\ne0$ (since $\theta\notin2\pi\mathbb{Z}$),

$$
S = \frac{e^{in\theta/2}\cdot 2i\sin(n\theta/2)}{e^{i\theta/2}\cdot 2i \sin(\theta/2)}
= e^{i(n-1)\theta/2}\,\frac{\sin(n\theta/2)}{\sin(\theta/2)}.
$$

Since $\sin(n\theta/2)/\sin(\theta/2)$ is real, reading off real and imaginary parts with <Ref to="thm-euler" /> gives

$$
\sum_{k=0}^{n-1}\cos k\theta = \cos\frac{(n-1)\theta}{2}\cdot\frac{\sin(n\theta/2)}{\sin(\theta/2)}, \qquad
\sum_{k=0}^{n-1}\sin k\theta = \sin\frac{(n-1)\theta}{2}\cdot\frac{\sin(n\theta/2)}{\sin(\theta/2)}.
$$

Let us check this. For $n=2$ the first right-hand side is

$$
\cos\frac{\theta}{2}\cdot\frac{\sin\theta}{\sin(\theta/2)} = \cos\frac\theta2\cdot\frac{2\sin(\theta/2)\cos(\theta/2)}{\sin(\theta/2)} = 2\cos^2\frac\theta2 = 1+\cos\theta,
$$

which agrees with the left-hand side $\cos 0 + \cos\theta = 1 + \cos\theta$. Handling sums of trigonometric functions directly is laborious, but collecting them into complex exponentials reduces everything to a single geometric sum. The kernel $\sin(n\theta/2)/\sin(\theta/2)$ obtained here is the main part of the Dirichlet kernel of Fourier series.
</Solution>
</Exercise>

## References

- L. V. Ahlfors, *Complex Analysis*, 3rd ed., McGraw-Hill, 1979 — Chapter 1 (the algebra and geometry of complex numbers, the topology of the complex plane). The standard English textbook.
- E. M. Stein and R. Shakarchi, *Complex Analysis* (Princeton Lectures in Analysis II), Princeton University Press, 2003 — Chapter 1. A concise account of the topology of the complex plane and an introduction to holomorphic functions.
- R. Remmert, *Theory of Complex Functions*, Graduate Texts in Mathematics 122, Springer, 1991 — a reference with a detailed account of the historical background of complex numbers (Cardano, Bombelli, Gauss, Hamilton).
- Takagi Teiji, *Kaiseki Gairon* (A Course of Analysis), revised 3rd ed., Iwanami Shoten, 1983 (in Japanese) — complex numbers and elementary functions treated by power series.
- Jimbo Michio, *Fukuso Kansū Nyūmon* (Introduction to Complex Functions), Iwanami Shoten, 2003 (in Japanese) — an introductory text in Japanese, giving a unified account from the complex plane and the polar form through to residue calculus.

## Appendix: Is there anything beyond the complex numbers?

**Why the road stops at $\mathbb{C}$.** Having extended the reals to the complex numbers, it is natural to ask whether one can extend further. Hamilton in fact worked on that problem for more than ten years and in 1843 discovered the four-dimensional **quaternions** $\mathbb{H}$ — at the cost of abandoning commutativity of multiplication ($ij = -ji$). For the eight-dimensional octonions, associativity is lost as well.

This is no accident. By Frobenius's theorem (1878), the only finite-dimensional associative division algebras over $\mathbb{R}$ are $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$, and among these only $\mathbb{R}$ and $\mathbb{C}$ are commutative. In other words, as long as one demands a commutative field containing $\mathbb{R}$ and finite-dimensional over $\mathbb{R}$, nothing strictly larger than $\mathbb{C}$ exists. The complex numbers are special precisely because they are a dead end.

**The fundamental theorem of algebra.** The motive for introducing $\mathbb{C}$ was "to solve $x^2+1=0$", but what we obtained was far more. The **fundamental theorem of algebra** (Gauss, in his 1799 dissertation and elsewhere) asserts that every polynomial of degree at least $1$ with complex coefficients has a root in $\mathbb{C}$. Adjoining the single element $i$ makes every algebraic equation solvable. One of the shortest proofs of this theorem uses <Ref to="mathematics/complex-analysis/properties-of-holomorphic-functions#thm-liouville" text="Liouville's theorem" /> (a bounded entire function is constant), and is available once the machinery of complex analysis is in place ([Cauchy's integral theorem and integral formula](/mathematics/complex-analysis/cauchy-integral-theorem), [Powerful properties of holomorphic functions](/mathematics/complex-analysis/properties-of-holomorphic-functions)).

**A bridge to the next chapter.** The stage is set. In the next chapter we consider, for a function $f : \Omega \to \mathbb{C}$ defined on a domain $\Omega$ of the complex plane, the derivative

$$
f'(z_0) = \lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h} \qquad (h \in \mathbb{C}).
$$

The form is the same as for real functions, but the decisive difference is that $h$ may approach $0$ **from every direction in the complex plane**. The consequence is <Ref to="mathematics/complex-analysis/holomorphic-functions#thm-cr-necessary" text="the Cauchy–Riemann equations" />, and from there begins a phenomenon impossible in real analysis: once differentiable, differentiable infinitely often ([Holomorphic functions and the Cauchy–Riemann equations](/en/mathematics/complex-analysis/holomorphic-functions)).