Answer :

[tex]\frac{4-3i}{(1+i)(2-3i)}=\frac{4-3i}{2-3i+2i-3i^2}=\frac{4-3i}{2-i+3}=\frac{4-3i}{5-i}\times\frac{5+i}{5+i}=\frac{20+4i-15i-3i^2}{5^2-i^2}\\\\=\frac{20-11i+3}{25+1}=\frac{23-11i}{26}=\frac{23}{26}-\frac{11}{26}i[/tex]
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kate200468
[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]

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