Question

12+8i
over
2i

Answer 1

i as in imaginary number i not a variable

Answer 2

\[\frac{ 12+8i }{ 2i }\] This is the problem, right?

Answer 3

yes

Answer 4

whenever you deal with i as an imaginary number this is what it represents \[i=\sqrt{-1}\] which doesn't work because -1 doesn't have a square root. But we can put that aside and treat like a real square root, because either way we don't want it on the bottom of the fraction. In order to get rid of it we can multiply the fraction by \[\frac{ \sqrt{-1} }{ \sqrt{-1} }\] because that equals 1 and multiplying anything by one doesn't change it's value, so now we have\[\frac{ [12+8(\sqrt{-1})] }{ 2(\sqrt{-1})^{2} }\]
Are you following me so far? Everything make sense?

Answer 5

yes

Answer 6

Oh I actually forgot something in my fraction. Here it is corrected.
\[\frac{ [12+8(\sqrt{-1})](\sqrt{-1)} }{ 2(\sqrt{-1})^{2} }\]
Do you see anyway to simplify this?

Answer 7

\[i^{2}\] is -1

Answer 8

Yup and we can multiply the i on the top of the fraction too.
\[\frac{ 12i -8 }{ -2 }\]
and then you can just divide and get
\[-6i+4\]
and so that it is in scientific notation we want to have the imaginary number second so the answer in its correct form is
\[4-6i\]

Answer 9

Thank you!

Answer 10

No problem, glad I could help!