Question 186575
Perform the complex number operations and write the answer in the a + bi form.

2i + (3 - i) / (2 + 5i) 
;
assume it's
2i + {{{(3-i)/(2+5i)}}}
over a single denominator
{{{(2i(2+5i) + (3-i))/(2+5i)}}} = {{{((4i + 10i^2) + (3-i))/(2+5i)}}} = {{{((4i + 10(-1)) + (3-i))/(2+5i)}}} = {{{(4i - 10 + 3 - i)/(2+5i)}}} = {{{(3i - 7)/( 2+5i)}}}
Multiply by the conjugate of the denominator over itself:
{{{(3i - 7)/(2+5i)}}} *{{{(2-5i)/(2-5i)}}} = {{{(6i - 15i^2-14 + 35i)/(4 - 25i^2)}}} = {{{(6i - 15(-1)-14 + 35i)/(4 - 25(-1))}}} = {{{(6i + 15 -14 + 35i)/(4 + 25)}}} = {{{(1 + 41i)/29}}}