Question 827926
the quotient of {{{(2-3i)}}} and {{{ (1+i) }}}: {{{(2-3i)/(1+i)}}}


{{{ (2-3i)/(1+i) }}} ...........both numerator and denominator multiply by {{{ (1-i) }}}


{{{((2-3i)(1-i))/((1+i)(1-i)) }}}


{{{((2-3i)(1-i))/(1^2-i^2) }}} ...since {{{i=sqrt(-1)}}}=> {{{i^2=(sqrt(-1))^2=-1}}}, so we have


{{{((2-3i)(1-i))/(1-(-1)) }}}


{{{((2-3i)(1-i))/(1+1) }}}


{{{((2-3i)(1-i))/2 }}} ..................now do numerator


{{{((2*1-3i*1+2(-i)-3i(-I)))/2 }}}


{{{(2-3i-2i+3i^2)/2 }}}


{{{(2-5i+3(-1))/2 }}}


{{{(2-5i-3)/2 }}}


{{{(-1-5i)/2 }}}


{{{-1/2-5i/2 }}}