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

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