Question Number 16917 by tawa tawa last updated on 28/Jun/17
$$\mathrm{If}\:\:\:\mathrm{x}\:+\:\mathrm{iy}\:=\:\frac{\mathrm{1}}{\mathrm{a}\:+\:\mathrm{ib}} \\ $$$$\mathrm{prove}\:\mathrm{that}\::\:\:\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \right)\:=\:\mathrm{1} \\ $$
Answered by Tinkutara last updated on 28/Jun/17
$${x}\:+\:{iy}\:=\:\frac{\mathrm{1}}{{a}\:+\:{ib}} \\ $$$${x}\:−\:{iy}\:=\:\frac{\mathrm{1}}{{a}\:−\:{ib}} \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} } \\ $$
Commented by tawa tawa last updated on 28/Jun/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{but}\:\mathrm{sir}\:\mathrm{why}\:\:\:\left(\mathrm{x}\:−\:\mathrm{iy}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{a}\:−\:\mathrm{ib}\right)} \\ $$
Commented by Tinkutara last updated on 28/Jun/17
$$\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rule}\:\mathrm{for}\:\mathrm{complex}\:\mathrm{numbers}:\:\mathrm{If} \\ $$$${z}\:=\:\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }\:\mathrm{then}\:\bar {{z}}\:=\:\frac{\bar {{z}}_{\mathrm{1}} }{\bar {{z}}_{\mathrm{2}} },\:\mathrm{where}\:\bar {{z}}\:\mathrm{denotes} \\ $$$$\mathrm{the}\:\mathrm{conjugate}\:\mathrm{of}\:{z}. \\ $$
Commented by tawa tawa last updated on 28/Jun/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$