Question

Multiplying complex numbers (HELP!!)

Answer 1

\[\left( \sqrt{3 }i \right) \left( \sqrt{-3}i \right) - \sqrt{3}i (2-\sqrt{-3})\]

Answer 2

Please show work :(

Answer 3

\[\Large \left( \sqrt{3 }\;i \right) \color{royalblue}{\left( \sqrt{-3}\;i \right)} - \sqrt{3}\;i (2-\sqrt{-3})\]

\[\Large \left( \sqrt{3 }\;i \right) \color{royalblue}{\left( \sqrt{3}\;i^2 \right)} - \sqrt{3}\;i (2-\sqrt{-3})\]

Answer 4

Do you understand what I did in that step there?

Answer 5

Oh because you took out the negative in the radical and added that i to the other one

Answer 6

Multiplied the i's together :) Yes.

Answer 7

If we multiply the first 2 brackets together we get,
\[\Large 3\;i^3 - \sqrt{3}\;i (2-\sqrt{-3})\](multiply the 3's together and then the i's with one another)

Answer 8

Ok, i follow. and the last term (before any multiplication) in the equation becomes minus radical 3i?

Answer 9

ya that's a good idea probably.\[\Large 3\;i^3 - \sqrt{3}\;i (2-\sqrt{3}\;i)\]

Answer 10

Distributing the sqrt3i to each term in the brackets is going to be a little messy.
Do you think you can do it? :D

Answer 11

Ill work it right now. Thanks so much btw!! @.@

Answer 12

...and i already have a question... when distributing, does the two multiply the three (becoming radical 6i) or does it become 2 radical 3i

Answer 13

\[\Large\bf\sf \sqrt2\cdot \sqrt3\quad=\quad \sqrt{2\cdot3}\quad=\quad \sqrt6\]

\[\Large\bf\sf 2\cdot \sqrt3\quad\ne\quad \sqrt6\]\[\Large\bf\sf 2\cdot \sqrt3\quad=\quad 2\sqrt3\]

Answer 14

Can more be done?

Answer 15

Did you distribute the thing to the second term as well?
Yes, lots more can be done ^^

Answer 16

\[3i ^{3}-2\sqrt{3}i + \sqrt{6}i ^{2}\]

Answer 17

9 my bad

Answer 18

Ah good fix.

Answer 19

then 3

Answer 20

-3 times -2 so +6?

Answer 21

\[\Large 3i ^{3}-2\sqrt{3}i + 3i ^{2}\]

Answer 22

just -3....final answer

Answer 23

whut? XD

Answer 24

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

Answer 25

Oh for the last term? Ah good.

Answer 26

so 3 times -1 equals -3?

Answer 27

Ya we can simplify our last term using that,
\[\Large 3i ^{3}-2\sqrt{3}i -3\]

Answer 28

Do you know how to simplify i^3?

Answer 29

sorta... i^2 times i equals -1 times i so it simplifies to negative i?

Answer 30

yesss good.\[\Large -3i-2\sqrt{3}i -3\]

Answer 31

And from here, I'm not really sure how we want to leave our answer.
Does your teacher want it in the form `a+bi` ?

Answer 32

I would assume so yes...is this the answer? No more could be done?

Answer 33

Well that is a fine answer, yes.
But if your teacher wants it in the form a+bi,
then we should rewrite the real part first, and also factor an i out of each imaginary part.\[\Large -3+(-3-2\sqrt3)i\]See how it's in the form,\[\Large a+(b)i\]now?

Answer 34

I do! Oh my god you dont know how much this helped. You are a god among men! A thousand thank yous! <3

Answer 35

lol np \c:/