Question

Complex numbers!

Answer 1

took me forever and I still have no clue.

Answer 2

or should i just guess

Answer 3

lol

Answer 4

\[\Large \left(-1+i \sqrt3\right)^{1/3}\]Factoring out a 2 from each term,\[\Large 2^{1/3}\left(-\frac{1}{2}+i\frac{\sqrt3}{2}\right)^{1/3}\]These values come frommmmm an angle of 2pi/3 or 120 degrees,\[\Large 2^{1/3}\left(\cos120+i \sin120\right)^{1/3}\]Using Euler's Formula we can write it like this,\[\Large 2^{1/3}\left(e^{120i}\right)^{1/3} \qquad=\qquad 2^{1/3}e^{40i}\]
Going back to the form we had before,\[\Large 2^{1/3}e^{40i}=2^{1/3}\left(\cos40^o+i \sin40^o\right)\]

Hmm I wouldn't really know how to explain this to you :(
I don't know much about complex numbers.

There's probably a much easier way than the approach I took.

Answer 5

Did you take your Trig Exam today? :)

Answer 6

it was horrible omg

Answer 7

i had to take both a midterm and final

Answer 8

oh boy :(

Answer 9

at the same time

Answer 10

i answered almost everything except for vectors

Answer 11

oh good :O

Answer 12

not really :/

Answer 13

i tried filling up everything and not leaving unanswered stuff

Answer 14

OH btw is the answer D x3

Answer 15

D? Yah that's what I came up with D:

Answer 16

But I dunno if I can explain it in a clear way.
There is some formula Louv.... something?
That allows you to do these a bit easier... I forget what it's called...

Answer 17

Oh oh `DeMoivre's Theorem`, does that sound familiar?

Answer 18

yeah

Answer 19

idk that theorem

Answer 20

i have to use it to change this complex number : (1-isqrt(3))^3 into trig form

Answer 21

omg site crashed again :( I lost of the stuff i was writing...
so frustrating...

I dunno what's going on lately, this thing is so unstable :C

Answer 22

:o oh no try writing it somewhere else and copy and paste?

Answer 23

ya maybe I'll try that :c

Answer 24

nah its cool i figured it out xD

Answer 25

ok cool c: sorry