Question

Simplify (Are all division)

2+4i
3i

3+2i
4+i

2i^11

Answer 1

`We\ know:\ i=√(-1)\to i^2=-1`

`(2+4i)/(3i)=(2+4i)/(3i)\cdot(3i)/(3i)=(6i+12i^2)/(9i^2)=(6i+12(-1))/(9(-1))\\\\=(6i-12)/(-9)=(2i-4)/(-3)=(4)/(3)-(2)/(3)i`

`(3+2i)/(4+i)=(3+2i)/(4+i)\cdot(4-i)/(4-i)=((3+2i)(4-i))/(4^2-i^2)\\\\=((3)(4)+(3)(-i)+(2i)(4)+(2i)(-i))/(16-(-1))=(12-3i+8i-2i^2)/(16+1)\\\\=(12+5i-2(-1))/(17)=(12+5i+2)/(17)=(14+5i)/(17)=(14)/(17)+(5)/(17)i`

`2i^(11)=2i^(10+1)=2i^(10)i^1=2i^(2\cdot5)i=2i(i^2)^5=2i(-1)^5=2i(-1)=-2i`

`Used:\ (a+b)(a-b)=a^2-b^2`