Let \widetilde{\C} represent the
extended complex plane excluding zero, \widetilde{\C} := \C \cup \{\infty\} \smallsetminus \{0\}, and \varphi the
bijective function from \C to \widetilde{\C} such that \varphi(z)=1/z. One has identities :\varphi(zt)=\varphi(z)\varphi(t), and :\varphi(z+t)=\varphi(z)\parallel \varphi(t) This implies immediately that \widetilde{\C} is a
field where the parallel operator takes the place of the addition, and that this field is
isomorphic to \C. The following properties may be obtained by translating through \varphi the corresponding properties of the complex numbers.
Field properties As for any field, (\widetilde{\C}, \,\parallel\,, \,\cdot\,) satisfies a variety of basic identities. It is
commutative under parallel and multiplication: :\begin{align} a \parallel b &= b \parallel a \\[3mu] ab &= ba \end{align} It is
associative under parallel and multiplication: :\begin{align} &(a \parallel b) \parallel c = a \parallel (b \parallel c) = a \parallel b \parallel c = \frac{1}{\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} = \frac{a b c}{a b + a c + b c}, \\ &(ab) c = a (b c) = a b c. \end{align} Both operations have an
identity element; for parallel the identity is \infty while for multiplication the identity is : :\begin{align} &a \parallel \infty = \infty \parallel a = \frac1{\dfrac1a + 0} = a, \\ &1 \cdot a = a \cdot 1 = a. \end{align} Every element a of \widetilde{\C} has an
inverse under parallel, equal to -a, the additive inverse under addition. (But has no inverse under parallel.) :a \parallel (-a) = \frac1{\dfrac1a - \dfrac1a} = \infty. The identity element \infty is its own inverse, \infty \parallel \infty = \infty. Every element a \neq \infty of \widetilde{\C} has a
multiplicative inverse {{nobr|a^{-1} = 1/a:}} :a\cdot\frac1a = 1. Multiplication is
distributive over parallel: : k (a \parallel b) = \frac{k}{\dfrac1a + \dfrac1b} = \frac{1}{\dfrac1{ka} + \dfrac1{kb}} = ka \parallel kb.
Repeated parallel Repeated parallel is equivalent to division, : \underbrace{a \parallel a \parallel \cdots \parallel a}_{n\text{ times}} = \frac1{\underbrace{\dfrac1a + \dfrac1a + \cdots + \dfrac1a}_{n\text{ times}}} = \frac an. Or, multiplying both sides by , : n (\underbrace{a \parallel a \parallel \cdots \parallel a}_{n\text{ times}}) = a. Unlike for
repeated addition, this does not commute: :\frac ab \neq \frac ba \quad \text{implies}\quad \underbrace{a \parallel a \parallel \cdots \parallel a}_{b\text{ times}} \,\neq\, \underbrace{b \parallel b \parallel \cdots \parallel b}_{a\text{ times}}~\!.
Binomial expansion Using the distributive property twice, the product of two parallel binomials can be expanded as :\begin{align} (a \parallel b) (c \parallel d) &= a(c \parallel d) \parallel b(c \parallel d) \\[3mu] &= ac \parallel ad \parallel bc \parallel bd. \end{align} The square of a binomial is :\begin{align} (a \parallel b)^2 &= a^2 \parallel ab \parallel ba \parallel b^2 \\[3mu] &= a^2 \parallel \tfrac12ab \parallel b^2. \end{align} The cube of a binomial is :(a \parallel b)^3 = a^3 \parallel \tfrac13a^2b \parallel \tfrac13ab^2 \parallel b^3. In general, the th power of a binomial can be expanded using
binomial coefficients which are the reciprocal of those under addition, resulting in an analog of the
binomial formula: :(a \parallel b)^n = \frac{a^n}{\binom n0} \parallel \frac{a^{n-1}b}{\binom n1} \parallel \cdots \parallel \frac{a^{n-k}b^k}{\binom nk} \parallel \cdots \parallel \frac{b^n}{\binom nn}.
Logarithm and exponential The following identities hold: : \frac{1}{\log(ab)} = \frac{1}{\log(a)}\parallel\frac{1}{\log(b)}, : \exp\left(\frac{1}{a\parallel b}\right) = \exp\left(\frac{1}{a}\right)\exp\left(\frac{1}{b}\right)
Factoring parallel polynomials As with a
polynomial under addition, a parallel polynomial with coefficients a_k in \widetilde\C (with can be
factored into a product of monomials: :\begin{align} &a_0x^n \parallel a_1x^{n-1} \parallel \cdots \parallel a_n =a_0(x \parallel -r_1)(x \parallel -r_2)\cdots(x \parallel -r_n) \end{align} for some roots r_k (possibly repeated) in \widetilde\C. Analogous to polynomials under addition, the polynomial equation :(x \parallel -r_1)(x \parallel -r_2)\cdots(x \parallel -r_n) = \infty implies that x = r_k for some .
Quadratic formula A
linear equation can be easily solved via the parallel inverse: :\begin{align} ax\parallel b &= \infty \\[3mu] \implies x &= -\frac ba. \end{align} To solve a parallel
quadratic equation,
complete the square to obtain an analog of the
quadratic formula : \begin{align} ax^2\parallel bx \parallel c &= \infty \\[5mu] x^2\parallel \frac{b}{a}x &= - \frac{c}{a} \\[5mu] x^2\parallel \frac{b}{a}x\parallel \frac{4b^2}{a^2} &= \left(-\frac{c}{a}\right) \parallel \frac{4b^2}{a^2} \\[5mu] \left(x\parallel \frac{2b}{a}\right)^2 &= \frac{b^2 \parallel -\tfrac14ac}{\tfrac14a^2} \\[5mu] \implies x &= \frac{(-b) \parallel \pm\sqrt{b^2 \parallel -\tfrac14ac} }{\tfrac12a}. \end{align}
Including zero The
extended complex numbers including zero, \overline{\mathbb{C}} := \C \cup \infty, is no longer a field under parallel and multiplication, because has no inverse under parallel. (This is analogous to the way \bigl(\overline{\mathbb{C}}, {+}, {\cdot} \bigr) is not a field because \infty has no additive inverse.) For every non-zero , :a \parallel 0 = \frac1{\dfrac1a + \dfrac10} = 0 The quantity 0 \parallel (-0) = 0 \parallel 0 can either be left undefined (see
indeterminate form) or defined to equal .
Precedence In the absence of parentheses, the parallel operator is defined as
taking precedence over addition or subtraction, similar to multiplication. ==Applications==