Question

Please solve this question.​

Answer 1

$\left(\dfrac{1}{1+2i}+\dfrac{3}{1-i}\right)\left(\dfrac{3-2i}{1+3i}\right)=\\\\\left(\dfrac{1-2i}{(1+2i)(1-2i)}+\dfrac{3(1+i)}{(1-i)(1+i)}\right)\left(\dfrac{(3-2i)(1-3i)}{(1+3i)(1-3i)}\right)=\\\\\left(\dfrac{1-2i}{1+4}+\dfrac{3+3i}{1+1}\right)\left(\dfrac{3-9i-2i-6}{1+9}\right)=\\\\\left(\dfrac{1-2i}{5}+\dfrac{3+3i}{2}\right)\left(\dfrac{-3-11i}{10}\right)=\\\\\left(\dfrac{2(1-2i)}{10}+\dfrac{5(3+3i)}{10}\right)\left(\dfrac{-3-11i}{10}\right)=$

$\dfrac{2-4i+15+15i}{10}\cdot\dfrac{-3-11i}{10}=\\\\\dfrac{17+11i}{10}\cdot\dfrac{-3-11i}{10}=\\\\\dfrac{-51-187i-33i+121}{100}=\\\\\dfrac{70-220i}{100}=\\\\\dfrac{70}{100}-\dfrac{220i}{100}=\\\\\boxed{\dfrac{7}{10}-\dfrac{11}{5}i}$