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already done :D

2*2sqrt(2)(cos(-pi/6+5pi/4)+isin(-pi/6+5pi/4))

Then I just calculate that?

Mmm ok ya, looks good!

so for theta I would put 13pi/12?

and r is 4sqrt(2)

Yup, but my teacher wants full form unfortunately :(

Yes that looks correct! :)

awesome! Thanks for the help! Can I have help with one more question really quickly?

sure

z^2 so (sqrt(3)-i)^2. I think it is 3+i, but I have a feeling I did it wrong.

Oh I guess NOT converting to polar is probably better :) lol
is that what you did?

yup

oh crap, sorry, I forgot to say I needed to write in a+bi form

my bad

Is that all I need to do? I thought there was some formula or something. I did (a)^+(b+^2

(b)^2*

\[\Large\rm (a+bi)(a-bi)\ne(a+bi)^2\]

Oh, so I would leave my results as i^2-sqrt(3)i-sqrt(3)i+3?

No no :) We can simplify further.
I just wanted to show the FOIL process

i^2? or if you mean root, i

-2sqrt(3)i

woops!

-1?

\[\Large\rm \sqrt{-1}\cdot\sqrt{-1}=-1\]Yah that looks better :)
I made a silly mistake lol

\[\Large\rm z^2=3-2\sqrt3 i-1\]

Simplify furtherrrrr :U

-2sqrt(3)i+2

maybe we write the a first,
a+(b)i = 2 + (-2sqrt3)i
ok looks good! \c:/

Awesome, I presume it's the exact same if I were to do w^4? Thanks for the help, I learned a lot :D

64(cis(5pi))?

ah sorry i ran away for a bit hehe

That is fine :)

|dw:1434425783548:dw|

So if THAT direction is -3,
which direction is positive 3?
maybe you can label it on the graph.

|dw:1434425967838:dw|

|dw:1434426009450:dw|Think about your regular x-y plane.
if i go 3 in this direction,

|dw:1434426045344:dw|Then the opposite (negative) of that is to go 3 in the other direction, ya?

yea

|dw:1434426087433:dw|So we want to go 3 in the opposite direction

Oh, gotcha

4pi/3?

|dw:1434426162247:dw|

Awesome, once again thanks!

np