Question 1011591
answer is yes, and here is why:


{{{(a-bi)/(bi-a)= -1 }}}...to eliminate {{{i}}} from denominator, multiply both numerator and denominator by {{{(bi+a)}}}


{{{((a-bi)(bi+a))/((bi-a)(bi+a))= -1 }}}


{{{(abi+a^2-bi*bi-abi)/((bi)^2-a^2)= -1 }}}


{{{(cross(abi)+a^2-b^2*i^2-cross(abi))/(b^2*i^2-a^2)= -1 }}}


{{{(a^2-b^2*(-1))/(b^2*(-1)-a^2))= -1 }}}


{{{(a^2+b^2)/(-b^2-a^2)= -1 }}}


{{{(a^2+b^2)/(-(b^2+a^2))= -1 }}}


{{{cross((a^2+b^2))1/(-cross((a^2+b^2))1)= -1 }}}


{{{1/-1= -1 }}}


{{{-1= -1 }}}