[tex]4- \frac{\big{3i}}{\big{(1+i)(2-3i)}} =4- \frac{\big{3i}}{\big{2-3i+2i-3\cdot(-1)}} =\\\\=4- \frac{\big{3i(5+i)}}{\big{(5-i)\cdot(5+i)}} =4- \frac{\big{15i+3\cdot(-1)}}{\big{25-(-1)}} =\\\\=4- \frac{\big{15i-3}}{\big{26}} =4- \frac{\big{15i}}{\big{26}}+\frac{\big{3}}{\big{26}}=4\frac{\big{3}}{\big{26}}- \frac{\big{15}}{\big{26}}i[/tex]
