Question Number 140784 by ajfour last updated on 12/May/21
$${Let}\:{the}\:{i}-{j}\:{plane}\:{be}\:{the}\:{complex} \\ $$$$\:{plane},\:{with}\:{basic}\:{operations} \\ $$$$\:\:{ij}=−{i} \\ $$$$\:\:{ji}=−{j} \\ $$$$\:\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\:\:{j}^{\mathrm{2}} =−\mathrm{1} \\ $$$${z}={r}+{xi}+{yj}\:\:\:\:{w}={s}+{pi}+{qj} \\ $$$${zw}={rs}+{pri}+{qrj}+{sxi}−{px}−{qxi} \\ $$$$\:\:\:\:\:\:\:\:\:\:+{syj}−{pyj}−{qy} \\ $$$$\:\:\:=\:\left({rs}−{px}−{qy}\right)+\left({pr}+{sx}−{qx}\right){i} \\ $$$$\:\:\:\:\:\:\:\:+\left({qr}+{sy}−{py}\right){j} \\ $$$${wz}={rs}+{sxi}+{syj}+{pri}−{px}−{pyi} \\ $$$$\:\:\:\:\:\:\:\:\:\:+{qrj}−{qxj}−{qy} \\ $$$$\:\:\:=\:\left({rs}−{px}−{qy}\right)+\left({pr}+{sx}−{py}\right){i} \\ $$$$\:\:\:\:\:\:\:\:\:+\left({qr}+{sy}−{qx}\right){j} \\ $$$$\left({little}\:{difference}..\right) \\ $$$$\Rightarrow\:\:{zw}−{wz}=\:\left({py}−{qx}\right)\left({i}−{j}\right) \\ $$$$\:\:\:=\:\mathrm{0}\:\:{if}\:\:\:{py}−{qx}=\:\mathrm{0} \\ $$$${z}^{\mathrm{2}} =\:\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)+\left(\mathrm{2}{rx}−{xy}\right){i} \\ $$$$\:\:\:\:\:\:\:\:\:+\left(\mathrm{2}{ry}−{xy}\right){j} \\ $$$${And}\:{if}\:\:{y}=\mathrm{0}\:\:\Rightarrow\:\:{z}={r}+{xi} \\ $$$${z}^{\mathrm{2}} =\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)+\mathrm{2}{rxi} \\ $$$$\:\:{either}\:{way}! \\ $$$$\left({z}^{\mathrm{2}} \right){z}={r}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)+\mathrm{2}{r}^{\mathrm{2}} {xi} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+{x}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i}−\mathrm{2}{rx}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\:\left({z}^{\mathrm{2}} \right){z}=\:{r}\left({r}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{2}} \right)+{x}\left(\mathrm{3}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i} \\ $$$${z}\left({z}^{\mathrm{2}} \right)=\:{r}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)+{x}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{r}^{\mathrm{2}} {xi}−\mathrm{2}{rx}^{\mathrm{2}} \\ $$$$\:\:\:\:=\:{r}\left({r}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{2}} \right)+{x}\left(\mathrm{3}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i} \\ $$$$\:{so}\:\:\:{z}\left({z}^{\mathrm{2}} \right)=\left({z}^{\mathrm{2}} \right){z}\:=\:{z}^{\mathrm{3}} \:\:\left({so}\:{far}\:{so}\:{good}\right) \\ $$$$….. \\ $$$$ \\ $$