Question 1104383
{{{ abs(z1) = abs(a+bi) = sqrt(a^2+b^2) }}}
{{{ abs(z2) = abs(c+di) = sqrt(c^2+d^2) }}}

notice that {{{ abs(z1)^2 = (a+bi)(a-bi) }}}

{{{ abs(z1/z2) = abs((a+bi)/(c+di)) = abs(((a+bi)(c-di)) / (c^2+d^2) ) }}}
= {{{ abs( ((ac+bd) + i(bc-ad)) / (c^2+d^2)) }}}
= {{{ sqrt( ((ac+bd) + i(bc-ad)) * (((ac+bd) - i(bc-ad) )) / (c^2+d^2)^2) }}}
:  :  :
After multiplying out and canceling a 2abcd with a -2abcd in the numerator...
:   : :
= {{{ sqrt( ((c^2*(a^2+b^2)) + (d^2*(a^2+b^2))) / (c^2+d^2)^2 ) }}}
= {{{ sqrt( ((a^2+b^2)(c^2+d^2))/(c^2+d^2)^2) }}}
=  {{{ sqrt ((a^2+b^2)/(c^2+d^2)) }}} =  {{{ sqrt(a^2+b^2) / sqrt(c^2+d^2) }}}
=  {{{ abs(z1) }}} / {{{ abs(z2) }}}
Done.