Question

Simplify (-4+2i)/5i and write in standard form.

Answer 1

I ended up with the answer: \[\frac{ -2+4i}{ 5 }\] It was incorrect, though. I'm not really sure where I went wrong here.

Answer 2

you just mixed up a sign, I think...

how did you start?

Answer 3

I think I started off by separating the whole fraction into two fractions: \[\frac{ -4 }{ 5i } + \frac{ 2i }{ 5i }\]

Answer 4

It's been a bit since I tried to the problem, so I'm trying to remember what to do. I know that the i in the denominator can't be there.

Answer 5

ah..

What I would have done is to multiply by what my teacher used to call a "well chosen "1""

Multiply by (i / i) which is just one, right?

Answer 6

Hm. I don't really get that. It's been a while since I've taken a traditional math class. But, I think I found what I did wrong from separating the fractions.

Answer 7

\[\frac{ 2i }{ 5i } \] The i cancels out there and it's just 2/5 ; So wouldn't the answer be: \[\frac{ 2 }{ 5 } - \frac{ 4 }{ 5i }\] ?

Answer 8

\[\frac{ -4 + 2i }{ 5i }\frac{ i }{ i } = \frac{ -4i + 2(i)(i) }{ -5 } = \frac{ -4i-2}{ -5 }\]

Answer 9

Wow, I hate imaginary numbers. :/

Answer 10

cheer up... they aren't really there ;)

Answer 11

The trick is often to multiply the top and bottom by something to eliminate the i on the bottom.

For a problem with something like (5 + 2i) on the bottom, multiply the top and bottom by (5 - 2i)... then the bottom becomes 25 +10i - 10i -4i^2 = 25 +4 = 29

So you wipe out the i term on the bottom by multiplying by what's called the complex conjugate... just reverse the sign on the i term, in essence.

Answer 12

btw, the final step to my equations above needs to be to factor out and cancel the negatives on top and bottom...

(-4i-2) / (-5) = (-1)(4i+2) / (-1)(5) = (4i + 2) / 5

Answer 13

Thank you. I'll try to remember this, hopefully I'll be able to do similar problems in time for the midterm next week. ;O

Answer 14

study & practice!!!

Good luck!!