Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

3.4: Complex Numbers

  • Page ID
    19038
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    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\()\).

    Screen Shot 2019-05-29 at 11.58.53 PM.png

    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.

    Screen Shot 2019-05-30 at 2.22.11 PM.png

    corollary \(\PageIndex{2}\)

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