
# 3.4: Complex Numbers

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With all the operations defined in §§1-3, $$E^{n}(n>1)$$ is not yet a field because of the lack of a vector multiplication satisfying the field axioms. We shall now define such a multiplication, but only for $$E^{2} .$$ Thus $$E^{2}$$ will become a field, which we shall call the complex field, $$C .$$

We make some changes in notation and terminology here. Points of$$E^{2},$$ when regarded as elements of $$C,$$ will be called complex numbers (each being an ordered pair of real numbers). We denote them by single letters (preferably z) without a bar or an arrow. For example, $$z=(x, y)$$. We preferably write $$(x, y)$$ for $$\left(x_{1}, x_{2}\right) .$$ If $$z=(x, y),$$ then $$x$$ and $$y$$ are called the real and imaginary parts of $$z,$$ respectively, $$^{1}$$ and $$\overline{z}$$ denotes the complex number $$(x,-y),$$ called the conjugate of $$z$$ (see Figure 5$$)$$.

Complex numbers with vanishing imaginary part, $$(x, 0),$$ are called real points of $$C .$$ For brevity, we simply write $$x$$ for $$(x, 0) ;$$ for example, $$2=(2,0)$$. In particular, $$1=(1,0)=\overline{\theta}_{1}$$ is called the real unit in $$C .$$ Points with vanishing real part, $$(0, y),$$ are called (purely) imaginary numbers. In particular, $$\overline{\theta}_{2}=(0,1)$$ is such a number; we shall now denote it by $$i$$ and call it the imaginary unit in $$C .$$ Apart from these peculiarities, all our former definitions of §§1-3 remain valid in $$E^{2}=C .$$ In particular, if $$z=(x, y)$$ and $$z^{\prime}=\left(x^{\prime}, y^{\prime}\right),$$ we have

$z \pm z^{\prime}=(x, y) \pm\left(x^{\prime}, y^{\prime}\right)=\left(x \pm x^{\prime}, y \pm y^{\prime}\right),$

$\rho\left(z, z^{\prime}\right)=\sqrt{\left(x-x^{\prime}\right)^{2}+\left(y-y^{\prime}\right)^{2}}, \text{ and}$

$|z|=\sqrt{x^{2}+y^{2}}.$

All theorems of §§1-3 are valid.

We now define the new multiplication in $$C,$$ which will make it a field.

Definition

$\text{If } z=(x, y) \text{ and } z^{\prime}=\left(x^{\prime}, y^{\prime}\right), \text{ then } z z^{\prime}=\left(x x^{\prime}-y y^{\prime}, x y^{\prime}+y x^{\prime}\right).$

Theorem $$\PageIndex{1}$$

$$E^{2}=C$$ is a field, with zero element $$0=(0,0)$$ and unity $$1=(1, 0)$$, under addition and multiplication as defined above.

Proof

We only must show that multiplication obeys Axioms 1-6 of the field axioms. Note that for addition, all is proved in Theorem 1 of §§1-3.

Axiom 1 (closure) is obvious from our definition, for if $$z$$ and $$z^{\prime}$$ are in $$C,$$ so is $$z z^{\prime}$$.

To prove commutativity, take any complex numbers

$z=(x, y) \text{ and } z^{\prime}=\left(x^{\prime}, y^{\prime}\right)$

and verify that $$z z^{\prime}=z^{\prime} z .$$ Indeed, by definition,

$z z^{\prime}=\left(x x^{\prime}-y y^{\prime}, x y^{\prime}+y x^{\prime}\right) \text{ and } z^{\prime} z=\left(x^{\prime} x-y^{\prime} y, x^{\prime} y+y^{\prime} x\right);$

but the two expressions coincide by the commutative laws for real numbers. Associativity and distributivity are proved in a similar manner.

Next, we show that $$1=(1,0)$$ satisfies Axiom 4(b), i.e., that $$1 z=z$$ for any complex number $$z=(x, y) .$$ In fact, by definition, and by axioms for $$E^{1}$$,

$1 z=(1,0)(x, y)=(1 x-0 y, 1 y+0 x)=(x-0, y+0)=(x, y)=z.$

It remains to verify Axiom 5(b), i.e., to show that each complex number $$z=(x, y) \neq(0,0)$$ has an inverse $$z^{-1}$$ such that $$z z^{-1}=1 .$$ It turns out that the inverse is obtained by setting

$z^{-1}=\left(\frac{x}{|z|^{2}},-\frac{y}{|z|^{2}}\right).$

In fact, we then get

$z z^{-1}=\left(\frac{x^{2}}{|z|^{2}}+\frac{y^{2}}{|z|^{2}},-\frac{x y}{|z|^{2}}+\frac{y x}{|z|^{2}}\right)=\left(\frac{x^{2}+y^{2}}{|z|^{2}}, 0\right)=(1,0)=1$

since $$x^{2}+y^{2}=|z|^{2},$$ by definition. This completes the proof. $$\square$$

corollary $$\PageIndex{1}$$

$$i^{2}=-1 ; i . e .,(0,1)(0,1)=(-1,0)$$.

Proof

By definition, $$(0,1)(0,1)=(0 \cdot 0-1 \cdot 1,0 \cdot 1+1 \cdot 0)=(-1,0)$$.

Thus $$C$$ has an element $$i$$ whose square is $$-1,$$ while $$E^{1}$$ has no such element, by Corollary 2 in Chapter $$2,8\{1-4 .$$ This is no contradiction since that corollary holds in ordered fields only. It only shows that $$C$$ cannot be made an ordered field.

However, the "real points" in $$C$$ form a subfield that can be ordered by setting

$(x, 0)<\left(x^{\prime}, 0\right) \text{ iff } x<x^{\prime} \text{ in } E^{1} .$

Then this subfield behaves exactly like $$E^{1} .$$ Therefore, it is customary not to distinguish between "real points in $$C^{\prime \prime}$$ and "real numbers," identifying $$(x, 0)$$ with $$x .$$ With this convention, $$E^{1}$$ simply is a subset $$($$and a subfield $$)$$ of $$C .$$ Henceforth, we shall simply say that "x is real" or "x $$\in E^{1 \prime}$$ instead of "$$x =(x, 0)$$ is a real point." We then obtain the following result.

Theorem $$\PageIndex{2}$$

Every $$z \in C$$ has a unique representation as

$z=x+y i,$

where $$x$$ and $$y$$ are real and $$i=(0,1) .$$ Specifically,

$z=x+y i \text{ iff } z=(x, y).$

Proof

By our conventions, $$x=(x, 0)$$ and $$y=(y, 0),$$ so

$x+y i=(x, 0)+(y, 0)(0,1).$

Computing the right-hand expression from definitions, we have for any $$x, y \in E^{1}$$ that

$x+y i=(x, 0)+(y \cdot 0-0 \cdot 1, y \cdot 1+0 \cdot 1)=(x, 0)+(0, y)=(x, y).$

Thus $$(x, y)=x+y i$$ for any $$x, y \in E^{1} .$$ In particular, if $$(x, y)$$ is the given number $$z \in C$$ of the theorem, we obtain $$z=(x, y)=x+y i,$$ as required.

To prove uniqueness, suppose that we also have

$z=x^{\prime}+y^{\prime} i \text{ with } x^{\prime}=\left(x^{\prime}, 0\right) \text{ and } y^{\prime}=\left(y^{\prime}, 0\right).$

Then, as shown above, $$z=\left(x^{\prime}, y^{\prime}\right) .$$ since also $$z=(x, y),$$ we have $$(x, y)= \left(x^{\prime}, y^{\prime}\right),$$ i.e., the two ordered pairs coincide, and so $$x=x^{\prime}$$ and $$y=y^{\prime}$$ after all. $$\square$$

Geometrically, instead of Cartesian coordinates $$(x, y),$$ we may also use polar coordinates $$r, \theta,$$ where

$r=\sqrt{x^{2}+y^{2}}=|z|$

and $$\theta$$ is the (counterclockwise) rotation angle from the $$x$$-axis to the directed line $$\overrightarrow{0 z} ;$$ see Figure $$6 .$$ Clearly, $$z$$ is uniquely determined by $$r$$ and $$\theta$$ but $$\theta$$ is not uniquely determined by $$z ;$$ indeed, the same point of $$E^{2}$$ results if $$\theta$$ is replaced by $$\theta+2 n \pi(n=1,2, \ldots)$$. (If $$z=0,$$ then $$\theta$$ is not defined at all.) The values $$r$$ and $$\theta$$ are called, respectively, the modulus and argument of $$z=(x, y) .$$ By elementary trigonometry, $$x=r \cos \theta$$ and $$y=r \sin \theta .$$ Substituting in $$z=x+y i,$$ we obtain the following corollary.

corollary $$\PageIndex{2}$$

$z=r(\cos \theta+i \sin \theta)(\text{trigonometric or polar form of }z).$