Question 33906
I dont think the second term is 1/(1/2i)
But anyway i am solving it 
{{{(3+2i)/(1+2i)}}} - {{{1/(1/2i)}}}
{{{(3+2i)(1-2i)/((1)^2 - (2i)^2)}}} - {{{(2i)}}}
{{{(3 - 6i +2i - 4i^2)/(1+4)}}}-{{{(2i)}}}
{{{(3-4i+4)/5}}}- {{{(2i)}}}
{{{((7-4i))/5}}} -{{{(2i)}}}
{{{((7 - 4i)- (10i))/5}}}
{{{((7-14i))/5}}}
{{{(7/5)}}}+{{{(-14/5)i}}}