2-5i/3i

- anonymous

2-5i/3i

- schrodinger

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- anonymous

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

- anonymous

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

- anonymous

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## More answers

- zepdrix

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

- zepdrix

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

- anonymous

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

- zepdrix

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

- anonymous

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

- anonymous

I see thank you so much

- anonymous

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

- anonymous

not after you cancel the common factor of 3

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes please

- anonymous

no actually that can't be it

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

now i am confused @satellite73 @zepdrix

- anonymous

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

- anonymous

yes

- anonymous

break it in to two pieces is all

- anonymous

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

- anonymous

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

- anonymous

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

- beginnersmind

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.

- anonymous

thank you so much

- beginnersmind

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

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