Question

Heulp!

Answer 1

Everything equals 1 :|

Answer 2

@zepdrix

Answer 3

hold up i need beverage +_+

Answer 4

haha ok

Answer 5

Oh you have to choose from the 4 graphs?

Answer 6

Yep.

Answer 7

and then do the other equation things

Answer 8

r = 1
tan(theta) = 1 (what is this in radians?)

Answer 9

r=1?

Answer 10

\(\Large\rm r=\sqrt{1^2+1^2}\)

Answer 11

tan(theta)=1 gives theta=pi/4

Answer 12

...my turn to go get a drink :|

Answer 13

lol

Answer 14

How do you get that theta = pi/4 I am still confused how to find that out o.0

Answer 15

Ok time for my trick.
Maybe you won't like it.

Answer 16

xD let's give it a try :)

Answer 17

\[\Large\rm (1+\mathcal i)^{1/3}\]I don't like this radius nonsense.
I just like to try and turn my real and imaginary parts into these "special numbers" that we're used to seeing from trick.

I notice the they're THE SAME, 1 on each, so we want some sqrt(2)/2's.
That's where sine and cosine are the same.

So I'll factor a sqrt(2) out of each term:\[\Large\rm \sqrt{2}^{1/3}\left[\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\mathcal i\right]^{1/3}\]And then I say, oh yah, this is pi/4.\[\Large\rm \sqrt{2}^{1/3}\left[cis\left(\frac{\pi}{4}+2k \pi\right)\right]^{1/3}\]

Answer 18

Too fancy? :U
Confusing??

Answer 19

these special numbers that we're used to seeing from trig*
typo

Answer 20

*blink...blink. That may make sense in the morning ;P it looks cool! but It's a little sketchy haha

Answer 21

Ok back to Earth then.

So we're having trouble understanding why \(\Large\rm \tan(\theta)=1\) gives us \(\Large\rm \theta=\frac{\pi}{4}\)...?

Answer 22

Sowry. I'll read it again in the morning I promise! :)
yeah

Answer 23

oh bed time? :O ok fine fine

Answer 24

no no I meant that fancy thing.
I'm good

Answer 25

oh lol :)

Answer 26

You really have to put some of your special angles to memory.
It would just take too long to go into detail about why tan(x)=1 gives x=pi/4.

tan(x)=sin(x)/cos(x)=1
So you're looking on your unit circle where sin(x) and cos(x) are the same, because they have to divide and give you 1.

Answer 27

\[\Large\rm r=\sqrt2\]\[\Large\rm \theta=\frac{\pi}{4}\]Ya? :o

Answer 28

o. hrmm yes I see. That makes sense :)

Answer 29

ok so back to the question now that we have a formula thing.
w subscript 0 would equal \[w _{0}\sqrt[3]{2}cis(\frac{ \pi }{ 12 }) yeah?\]

Answer 30

ooh shoot it didn't ask for the third root

Answer 31

Out front we should have: \(\Large\rm (\sqrt{2})^{1/3}=2^{1/6}\)

Answer 32

woah what?

Answer 33

You have to take the cube root of your radius, ya?
\(\Large\rm r=\sqrt{2}\).

\(\Large\rm r^{1/3}=(\sqrt{2})^{1/3}\)

Answer 34

for what do I have to take the cube root?

Answer 35

oh gosh hah

Answer 36

\[w _{k}=\sqrt[n]{r}cis \frac{ \theta+2kpi }{ n } \]
where k = 0,1,2

Answer 37

so that's the formula I have..

Answer 38

but there's no n in our case, yeah?
oh. I see what you're saying now.

Answer 39

Bah, I dunno.. stop using formulas >:U

Answer 40

oki.

Answer 41

We start with this:\[\Large\rm z=1+\mathcal{i}\]We determined that we can write it in this form:\[\Large\rm z=r\left[cis \theta\right]\]With our r=sqrt(2) and theta=pi/4.\[\Large\rm z=\sqrt2\left[cis \frac{\pi}{4}\right]\]And yes, we should allow for rotations, so we'll throw a 2k pi into the mix,\[\Large\rm z=\sqrt2\left[cis\left(\frac{\pi}{4}+2k \pi\right)\right]\]Then we're taking the cube root of our complex number, giving us,\[\Large\rm z=(\sqrt2)^{1/3}\left[cis\left(\frac{\pi}{4}+2k \pi\right)\right]^{1/3}\]Apply De'Moivre's Theorem let's us bring the 1/3 inside,\[\Large\rm z=(\sqrt2)^{1/3}\left[cis\left(\frac{\pi}{12}+\frac{2k \pi}{3}\right)\right]\]

Answer 42

Those last two lines should be \(\Large\rm z^{1/3}\) on the left side, my bad.

Answer 43

If you want to use your formula, that's ok.
Just make sure you plug in your radius correctly.\[\Large w _{k}=\sqrt[n]{\color{orangered}{r}}~cis\left(\frac{ \theta+2k \pi }{ n }\right)\]\[\Large w _{k}=\sqrt[n]{\color{orangered}{\sqrt{2}}}~cis\left(\frac{ \theta+2k \pi }{ n }\right)\]

Answer 44

You were simply plugging a 2 in for your r before.

Answer 45

your way is simplier. thanks

Answer 46

simplier :3 lol

Answer 47

final would be
\[2^{\frac{ 1 }{ 6 }}(\cos \frac{ \pi }{ 12 }+isin \frac{ \pi }{ 12 })\]

Answer 48

for w0, yes.

Answer 49

haa, yay!
the next ones would be
5pi/12 and 13pi/12?

Answer 50

is that something you keep in y our tool chest? simpliers? kinda like pliers but simpler? 0_o

Answer 51

lol xD

Answer 52

Hmm, no.
I think the next one is 9pi/12.

You're plugging in k=1, and then you need a common denominator between your fractions, ya?

Answer 53

Oh stop it. It should be a word!

Answer 54

hrmm let me check my work :o

Answer 55

oh yeah, sorry. is the 13 right?

Answer 56

13pi/12? No I don't think so.

Answer 57

\[\Large\rm \frac{\pi}{12}+\frac{2k \pi}{3}=\frac{\pi}{12}+\frac{8k \pi}{12}\]

Answer 58

oh asdgpahspdgi i keep doing pi/3 instead of pi/4

Answer 59

Ah you silly billy :O

Answer 60

16pi/12?

Answer 61

no -_- come onnnnnnn

Answer 62

waaaahh

Answer 63

\[\Large\rm \frac{\pi}{12}+\frac{8\cdot0 \pi}{12}\]
\[\Large\rm \frac{\pi}{12}+\frac{8\cdot 1\pi}{12}\]
\[\Large\rm \frac{\pi}{12}+\frac{8\cdot 2\pi}{12}\]

Answer 64

Those are your three angles, ya?

Answer 65

So the last one is pi/12.......... . .. . .
PLUS 16pi/12.

Answer 66

Yeahh.
siiigh yeah, i see.

Answer 67

DONE. Except for picking the graph.

Answer 68

Third one? (on the left in the second pic)

Answer 69

Third option?
I don't think so.. Look at the first dot,
it looks like it's located somewhere around pi/4.

Our first angle should way way smaller than that, ya?

Answer 70

numberu 1?

Answer 71

mmmmmm ya i would go with #1.
that seems to make the most sense to me :O

Answer 72

:))

Answer 73

*happy dance.
Thanks for putting up with my ridiculous answers at times. yay for green check marks.

Answer 74

always happy to see those :)

red x's are so stupid, *grumble grumble*

Answer 75

Preach it.

Answer 76

Okies, I am retiring to my bed :P

Answer 77

Night, sleep happy whenever you go to bed ^_^

Answer 78

~\c: