Question

Find indicated power using DeMovire's theorem
Write in a+bi form.

Answer 1

\[(3\sqrt{3}+3i)^{-5}\]

Answer 2

I already solved it but the numbers are pretty outrageous so I just want to check that I didn't do it wrong o.0

Answer 3

@rvc

Answer 4

@rvc it doesn't let me msg you since you've not fanned me. Thank you anyway:) do you know someone who does know?

Answer 5

0_o

Answer 6

Okie I check :O

Answer 7

Yay! thank you :)

Answer 8

So before I turn it into a+bi,
I'm getting something like this,\[\Large\rm 6^{-5}\left[\cos\left(-\frac{5\pi}{6}\right)+\mathcal i \sin\cos\left(-\frac{5\pi}{6}\right)\right]\]

Answer 9

That what you're getting? :O
These numbers are not outrageous, so I'm thinking you got something else? 0_o

Answer 10

oh gosh...so when I was finding r and got to where i'm mean to square 36 i wrote it like
(36)^(1/2) and put the answer as 18 because I divided it instead of sq rooting it *facepalm. epic fail.

Answer 11

XD

Answer 12

How did you find your r?
Using the weird square root and all that jazz?

Answer 13

I have different method :O
Maybe too crazy for you though.

Answer 14

Um I found it by doing
\[r= ((3\sqrt{3})^{2}+(3)^{2})^{1/2}\]

Answer 15

haha hrm, well you can try teaching to me, but it might just confuse me more xD

Answer 16

I mean, this method is pretty straight forward I think

Answer 17

Hmm yah maybe I don't wanna confuse you :d

Answer 18

\[z ^{-5}=-3888(-\sqrt{3})+(-3888i)\] is the final answer then?

Answer 19

lol ,thanks. I appreciate that ;P

Answer 20

Hmm I don't understand your answer....

Answer 21

It was a -5 exponent, so our radial length is really really really short.
6^(-5)

Answer 22

So each component should be multiplied by, \(\Large\rm \frac{1}{6^5}=\frac{1}{7776}\)

Answer 23

right, so then I got
-7776(-sq3/2) + i(-1/2)
so when i factor it in it gets divided by two. yeah?

Answer 24

\[\Large\rm =6^{-5}\left[\cos\left(-\frac{5\pi}{6}\right)+\mathcal i \sin\cos\left(-\frac{5\pi}{6}\right)\right]\]\[\Large\rm =6^{-5}\left[-\frac{\sqrt3}{2}-\frac{1}{2}\mathcal i\right]\]Again, I'm confused about your radius.
You have -7776. Your radius should be a positive length.

Answer 25

You should have 1/7776 for your radius.

Answer 26

Looks good besides that though.

Answer 27

Isn't when something raised to a negative power it's still negative? o.0

Answer 28

ooh or when it's multiplied in it will become positive?

Answer 29

Dat Miriam :O She so silly.
Go plant another light bulb!

Did you get this for your radius \(\Large\rm 6^{-5}\), yes no?

Answer 30

lol xD
Yes I did.

Answer 31

So then,\[\Large\rm 6^{-5}=\frac{1}{6^5}=\frac{1}{7776}\]Just forgetting your exponent rules maybe?

Answer 32

but.. I don't want it under a 1.

Answer 33

No? Why not?