Question 638507
The standard form does not allow for complex numbers to be at the denominator.

The standard form is:

{{{a + bi}}}

We'll have to get the complex number out of the denominator. For this reason, 
we'll have to multiply the numerator and denominator by the conjugate of the 
denominator: {{{(i - 2) }}}.


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


={{{(3i-6 - i^2+2i)/(i^2-2i + 2i-4 )}}}


={{{(5i-6 - (-1))/((-1)-cross(2i) + cross(2i)-4 )}}}


={{{ (5i-6 +1)/(-1-4 )}}}


={{{ (5i-5)/(-5 )}}}

The standard form is:

{{{a + bi= 5i/-5 - 5/-5}}}

={{{ -i +1}}}

={{{1-i }}}