Question

Express 4-3i/(1+i)(2-3i) in the a+bi form

Answer 1

`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`

Answer 2

`(4-3i)/((1+i)(2-3i))=(4-3i)/(2-3i+2i-3i^2)=(4-3i)/(2-i+3)=(4-3i)/(5-i)*(5+i)/(5+i)=(20+4i-15i-3i^2)/(5^2-i^2)\\\\=(20-11i+3)/(25+1)=(23-11i)/(26)=(23)/(26)-(11)/(26)i`