Question 1026604
{{{abs(1/z) = abs(1/(a-ib) - 1/(a-ic)) = abs((i(b-c))/((a-ib)(a-ic))) = abs(b-c)/(abs(a-ib)abs(a-ic))) }}}

==> {{{abs(z) = (abs(a-ib)abs(a-ic))/abs(b-c) = sqrt(a^2+b^2)sqrt(a^2+c^2)/abs(b-c)}}}