Question

Simplify 10 over quantity of 8 minus 5 i.

Answer 1

`\large\boxed{(10)/(8-5i)=(80)/(89)+(50)/(89)i}`

Explanation:

`\text{Use}\\\\(a-b)(a+b)=a^2-b^2\\\\i=√(-1)\to i^2=-1\\\\\text{distributive property}\ a(b+c)=ab+ac\\-----------------------------\\\\(10)/(8-5i)=(10)/(8-5i)\cdot(8+5i)/(8+5i)=(10(8+5i))/((8-5i)(8+5i))\\\\=((10)(8)+(10)(5i))/(8^2-(5i)^2)=(80+50i)/(64-5^2i^2)=(80+50i)/(64-25(-1))\\\\=(80+50i)/(64+25)=(80+50i)/(89)=(80)/(89)+(50)/(89)i`

Answer 2

`\frac { 80 + 5i } { 89}`

Explanation:

We are given the following expression to simplify:

10 over quantity of 8 minus 5 i -->
`\frac { 10 } { 8 - 5i }`

`\frac { 10 } { 8 - 5i }` ×
`\frac { ( 8 + 5i ) } { ( 8 + 5i ) }`

`10` ×
`\frac { ( 8 + 5i ) } { 64 + 25 }`

`\frac { 80 + 5i } { 89}`

Therefore, the simplified version of the given expression is
`\frac { 80 + 5i } { 89}`.