Question 697390
I bet you meant (2-3i)/(1+i) = {{{(2-3i)/(1+i)}}}
{{{1-i}}} is the conjugate of {{{1+i}}}
When you multiply conjugate complex numbers, you get a real number.
{{{(1+i)(1-i)=1^2-i^2=1-(-1)-2}}}
That is very useful when trying to get rid of complex denominators.
You just multiply numerator and denominator times the conjugate of the denominator:
{{{(2-3i)/(1+i) = (2-3i)(1-i)/(1+i)/(1-i)=(2-2i-3i+3i^2)/(1^2-i^2)=(2-5i+3(-1))/(1-(-1))=(2-3-5i)/(1+1)=(-1-5i)/2=-1/2-(5/2)i}}}
 
NOTE:
2-3i/1+i = {{{2-3i/1+i=2-3i+i=2-2i}}} is a different (easier) problem.
If want to write {{{(2-3i)/(1+i)}}} ,
and you are not able to draw a long horizontal line with {{{2-3i}}} above it and {{{1+i}}} below it,
you must write out the brackets around {{{2-3i}}} and around {{{1+i}}} that the long horizontal fraction bar implies.