Question

Write your answer as a complex number in standard form, I need help please

Answer 1

10. -0.12+0.16i


11. 1.5+1.5i

Answer 2

10.
`(-3+4i)/(25)`

11.
`(3+3i)/(2)`

Explanation:

A complex number is a number made of two parts: a real number and an imaginary number. We write them as a+bi.

• a is a real number
• bi is imaginary

We never use i as a variable in math because i is a symbol we use for the square root of -1. We need to remember how powers of i work because we will be multiplying them and exponents may be possible.

`i^1 = i\\i^2 = -1\\i^3 = -i\\i^4 = 1`

Notice that
`i^2` and
`i^4` both give real number values of -1, and 1. Since we cannot combine real numbers and imaginary numbers with operations like
`4-3i` ≠
`i`, we will use exponents of imaginary numbers to convert and properties of algebra to make it possible.

For
`(i)/(4-3i)` we will multiply the entire equation by the identity or 1. But we will choose what 1 we use. We choose
`(4+3i)/(4+3i) = 1` and multiply.

`(i)/(4-3i) * (4+3i)/(4+3i) = (i(4+3i))/((4-3i)(4+3i) =(4i+3i^2)/(16-9i^2)`

Notice
`i^2=-1`. We replace it and simplify the integers. Then write it in a+bi order.

`(4i+3i^2)/(16-9i^2)=(4i+3(-1))/(16-9(-1))= (4i-3)/(16+9)=(-3+4i)/(25)`.

We repeat the steps for the second problem.

For
`(-3+3i)/(2i)` we will multiply the entire equation by the identity or 1. But we will choose what 1 we use. We choose
`(2i)/(2i) = 1` and multiply.

`(-3+3i)/(2i)* (2i)/(2i) = ((-3+3i)(2i))/((2i)(2i)) =(-6i+6i^2)/(4i^2)`

Notice
`i^2=-1`. We replace it and simplify the integers. Then write it in a+bi order.

`(-6i+6i^2)/(4i^2)=(-6i+6(-1))/(4(-1))=(-6i-6)/(-4)=(-6-6i)/(-4)=  (-2(3+3i))/(-2(2))=(3+3i)/(2)`.