Question

Question #8380d

Answer 1

Did you mean to put the negative sign after the divide sign?
If so, `(4-3i+2+6i)/-((3-i)(3-i))=-3/10-3/5i`.

Explanation:

First, simplify the numerator by combining the integer terms and combining the `i` terms: `4-3i+2+6i=3i+6`.

Secondly, the denominator: `-((3-i)(3-i))=-(9-3i-3i+i^2)`. Since `i=sqrt(-1)`, `i^2=(sqrt(-1))^2=-1`.
Plugging that into the denominator and combining like terms, we get `-(9-6i+(-1))=-(8-6i)=-8+6i=6i-8`

Our fraction is now `(3i+6)/(6i-8)`

This may be okay for your teacher, but the standard form for complex numbers is `a+-bi`, so let's get our answer in that format by multiplying the numerator and denominator by the conjugate of the denominator: `((3i+6)(6i+8))/((6i-8)(6i+8))`
We get `(18i^2+24i+36i+48)/(-36-64)=(30+60i)/-100` and put it in the form of `a+-bi` then reduce it:
`30/-100-(60i)/-100=-3/10-3/5i`

Hope that helps! Have fun; imaginary numbers are cool!