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multiply by the conjugate

multiply by 1 using the conjugate*

* top n bottom by 3-i

So 9i?

6(3-i) / (3+i)(3-i)

Is that all that I have to do? I'm not really sure how to do these sorts of problems

You need to multiply out the brackets and simplify if possible

Just like normal except you have i's to deal with...

6(3-i)? Wouldn't the bottom cancel itself out?

It is a kind of trick to get rid of the i on the bottom.....

Similar to how you get rid of a rdical, same sort of idea...

So confused

Multiply it out and you will see...

9-i/9

which you could divide 9 by 9 to get one, so -i?

6(3-i) is 18-6i
Where do you get 9-i from?

Oh wow opps. haha I added.

Ok, now what do you get for the bottom?

9?

You have (3+i)(3-i)
Tell me how it comes to 9

You multiply and the +i and -i cancel out? i'm really sorry, this is new to me haha

(3+i)(3-i) is 3*3 + 3*i -3*i - i^2 (using FOIL?)

So the 3i's cancel out and leave 9-i^2
So what is i^2 ?

-1

Right, so what is the bottom ( hint-it is not 9)

8?

9-(-1)

10

Ok so finally you have 18-6i/10 which you can divide by 2 to get
9-3i/5 as a final answer.

So, procedure for these is *by conjugate top and bottom then simplify

thank you so much!!!

ur welcome