Complex numbers/Field/Fact/Proof

Proof

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

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

so ${}1$ 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

{}{\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

{}{\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

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

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