Question

Please solve this question.​

Answer 1

`\left((1)/(1+2i)+(3)/(1-i)\right)\left((3-2i)/(1+3i)\right)=\\\\\left((1-2i)/((1+2i)(1-2i))+(3(1+i))/((1-i)(1+i))\right)\left(((3-2i)(1-3i))/((1+3i)(1-3i))\right)=\\\\\left((1-2i)/(1+4)+(3+3i)/(1+1)\right)\left((3-9i-2i-6)/(1+9)\right)=\\\\\left((1-2i)/(5)+(3+3i)/(2)\right)\left((-3-11i)/(10)\right)=\\\\\left((2(1-2i))/(10)+(5(3+3i))/(10)\right)\left((-3-11i)/(10)\right)=`

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