Question

Please answer number 4

Answer 1

`\large\boxed{(2+3i)/(i(4-5i))+(2)/(i)=(22)/(41)-(75)/(41)i}`

Explanation:

`i=√(-1)\to i^2=-1`

`(2+3i)/(i(4-5i))+(2)/(i)\qquad\text{use distributive property}\\\\=(2+3i)/(4i-5i^2)+(2)/(i)=(2+3i)/(4i-5(-1))+(2)/(i)=(2+3i)/(4i+5)+(2)/(i)\\\\\text{use}\ (a+b)(a-b)=a^2-b^2\\\\=(2+3i)/(4i+5)\cdot(4i-5)/(4i-5)+(2)/(i)\cdot(i)/(i)=((2+3i)(4i-5))/((4i)^2-5^2)+(2i)/(i^2)\\\\\text{use FOIL}\ (a+b)(c+d)=ac+ad+bc+bd\\\\=((2)(4i)+(2)(-5)+(3i)(4i)+(3i)(-5))/(16i^2-25)+(2i)/(-1)`

`=(8i-10+12i^2-15i)/(16(-1)-25)-2i=((-10-12)+(8i-15i))/(-16-25)-2i\\\\=(-22-7i)/(-31)-2i=(-22)/(-41)+(-7i)/(-41)-2i=(22)/(31)+(7)/(41)i-(82)/(41)i\\\\=(22)/(41)-(75)/(41)i`