Question

2-5i/3i

Answer 1

\[\frac{ 2-5i }{ 3i }\]

Answer 2

I understand you are supposed to multiply by the conjugate, but what is the conjugate of 3i

Answer 3

@Hero @nincompoop

Answer 4

The conjugate of \(\large\rm 0+3i\) is \(\large\rm 0-3i\)

Answer 5

So therefore the conjugate of \(\large\rm 3i\) is simply \(\large\rm -3i\)
k? :)

Answer 6

don't be a slave to method
you can multiply by \(i\) top and bottom if you like
or \(-i\)

Answer 7

Ya that'll save you a couple steps :D
Seems like a good idea

Answer 8

either way the denominator will become a real number, either 3 or -3 depending on which you pick

Answer 9

I see thank you so much

Answer 10

Wait, I get a different answer from when i multiply -3i to when i multiply from just i

Answer 11

not after you cancel the common factor of 3

Answer 12

which is why multiplying by \(-3i\) is silly in this case

Answer 13

ohhh so should the final answer be -2i -5

Answer 14

idk i didn't do it
want me to check?

Answer 15

yes please

Answer 16

no actually that can't be it

Answer 17

\[\frac{ 2-5i }{ 3i }\times \frac{i}{i}\]
\[=\frac{2i+5}{-3}\]

Answer 18

i got -6i -15 but then factored out 3

Answer 19

yes but my answer has to be in complex number form i'm sorry i forgot to mention that

Answer 20

now i am confused @satellite73 @zepdrix

Answer 21

you mean in the form \(a+bi\)?

Answer 22

yes

Answer 23

break it in to two pieces is all

Answer 24

for example
\[\frac{7+8i}{5}=\frac{7}{5}+\frac{8}{5}i\]

Answer 25

oh i understand now can i get back to you in a minute with my answer?

Answer 26

is this correct?
\[\frac{ 2i }{ -3 }+\frac{ 5 }{ -3 }\]

Answer 27

Yes, although you might want to write it as

\[-\frac{ 5 }{ 3 } -\frac{ 2 }{ 3 }i\]
or even

\[\frac{ -5-2i }{ 3 }\]

but it's just cosmetical at this point.

Answer 28

thank you so much

Answer 29

No problem, satellite did all the work though. ;)