Question 742768
{{{(2-2i)/(3-5i)=((2-2i)/(3-5i))((3+5i)/(3+5i))}}}

Next we multiply out both numerator and denominator:

{{{(2(3+5i)-2i(3+5i))/(3(3+5i)-5i(2+5i))=(6+10i-6i-10i^2)/(3^2-(5i)^2)}}}

Note that 

{{{-(5i)^2=-(-25)=25}}}

and 

{{{-10i^2=-10(-1)=10}}}

So our expression becomes

{{{(16+4i)/(9+25)}}}

Which simplifies to:

{{{16/34+4i/34}}}

or 
{{{8/17+(2/17)i}}}