# Complex numbers/Field/Fact/Proof

Proof

The field properties for the addition are clear, since the corresponding properties hold for ${\displaystyle {}\mathbb {R} }$. We have

${\displaystyle {}1\cdot (a+b{\mathrm {i} })=a+b{\mathrm {i} }\,,}$

so ${\displaystyle {}1}$ is the neutral element for the multiplication. The commutativity of the multiplication follows directly from its formula. To show associativity of the multiplication we compute

{\displaystyle {}{\begin{aligned}{\left((a+b{\mathrm {i} })(c+d{\mathrm {i} })\right)}(e+f{\mathrm {i} })&={\left(ac-bd+(bc+ad){\mathrm {i} }\right)}(e+f{\mathrm {i} })\\&=(ac-bd)e-(bc+ad)f+{\left((ac-bd)f+(bc+ad)e\right)}{\mathrm {i} }\\&=ace-bde-bcf-adf+{\left(acf-bdf+bce+ade\right)}{\mathrm {i} }.\end{aligned}}}

We also get

{\displaystyle {}{\begin{aligned}(a+b{\mathrm {i} }){\left((c+d{\mathrm {i} })(e+f{\mathrm {i} })\right)}&=(a+b{\mathrm {i} }){\left(ce-df+(cf+de){\mathrm {i} }\right)}\\&=a(ce-df)-b(cf+de)+{\left(b(ce-df)+a(cf+de)\right)}{\mathrm {i} }\\&=ace-adf-bcf+-bde+{\left(bce-bdf+acf+ade\right)}{\mathrm {i} }.\end{aligned}}}

Suppose now that

${\displaystyle {}a+b{\mathrm {i} }\neq 0\,.}$

Then at least one of the numbers ${\displaystyle {}a}$ or ${\displaystyle {}b}$ is different from ${\displaystyle {}0}$ and therefore ${\displaystyle {}a^{2}+b^{2}>0}$. Hence ${\displaystyle {}{\frac {a}{a^{2}+b^{2}}}-{\frac {b}{a^{2}+b^{2}}}{\mathrm {i} }}$ is a complex number and

${\displaystyle {}{\left(a+b{\mathrm {i} }\right)}{\left({\frac {a}{a^{2}+b^{2}}}-{\frac {b}{a^{2}+b^{2}}}{\mathrm {i} }\right)}={\frac {1}{a^{2}+b^{2}}}{\left(a+b{\mathrm {i} }\right)}{\left(a-b{\mathrm {i} }\right)}={\frac {1}{a^{2}+b^{2}}}{\left(a^{2}+b^{2}\right)}=1\,,}$

so every element ${\displaystyle {}\neq 0}$ has an inverse with respect to the multiplication. The distributivity law follows from

{\displaystyle {}{\begin{aligned}(a+b{\mathrm {i} }){\left(c+d{\mathrm {i} }+e+f{\mathrm {i} }\right)}&=(a+b{\mathrm {i} }){\left((c+e)+(d+f){\mathrm {i} }\right)}\\&=a(c+e)-b(d+f)+(a(d+f)+b(c+e)){\mathrm {i} }\\&=ac+ae-bd-bf+(ad+af+bc+be){\mathrm {i} }\\&=ac-bd+(ad+bc){\mathrm {i} }+ae-bf+(af+be){\mathrm {i} }\\&=(a+b{\mathrm {i} }){\left(c+d{\mathrm {i} }\right)}+(a+b{\mathrm {i} }){\left(e+f{\mathrm {i} }\right)}.\end{aligned}}}