find all fourth roots of \[-8-8\sqrt{3i}\] i being a complex number, and show them on an argand diagram

- Ray10

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- anonymous

this is bad :O
are you sure it isn't
\[\large -8-8\sqrt3i\]?

- Ray10

:P my bad!!
yes it is \[-8-8\sqrt{3}j\]

- Ray10

thanks for that! But yes can you help me please?

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## More answers

- anonymous

now it's a j? why has the world gone coconuts?

- anonymous

but more on the fun part later, now we get to work ^.^

- Ray10

it's an i or j, blame my lecturer :P he said use one or the other :P
yes lets get to work! :D

- anonymous

first you need it in polar form.
like this:
\[\Large r\left[\cos(\theta) + \color{red}i\sin(\theta)\right]\]

- anonymous

and lol.. grown-ups -.-

- anonymous

hey did you leave? :(

- Ray10

no I didnt leave!! :P I'm trying to convert it to polar :P
yes ah grown-ups >.<

- Ray10

I get \[r=8\sqrt{2}\]

- anonymous

no :P

- Ray10

what do you mean? :O

- anonymous

I mean no :P
\[\Large r = \sqrt{(-8)^2 + \left(-8\sqrt3\right)^2}\]

- Ray10

16 :P

- anonymous

yes better. what about \(\theta\)?

- Ray10

\[\frac{ \Pi }{ 3 }\] I can confirm :)

- anonymous

Me too :)
I can confirm that it's WRONG!

- Ray10

:O no way?! :O how so???

- anonymous

Oh, Ray ^.^
\[\Large 16\left[\cos\left(\frac \pi 3\right)+\color{red}i\sin\left(\frac \pi3\right)\right]=8+8\sqrt3\color{red}i\]
So... nope :P

- Ray10

\[\tan \theta =\frac{ y }{ x }\] ?

- anonymous

Yes... but that also gives \[\Large \frac{4\pi}{3}\]

- anonymous

So how do you know which is which? XD

- Ray10

\[\frac{ -\Pi }{ 3 }\] :P

- anonymous

No.

- anonymous

You're guessing :P

- Ray10

okay now I am lost, I worked that one out but I got that wrong, I think there is something to do with the quadrants to work it out correct?

- anonymous

Yep, you're lost all right... lost boy ^.^
Anyway, we don't know what \(\theta\) is but we do know that \[\Large \tan \theta = \frac{-8\sqrt3}{-8}=\sqrt3\]right?

- Ray10

ahhh I love all the references!! :D funnest maths tutor :P
yes we do know that :)

- anonymous

So, from \(\large 0 \) to \(\large 2\pi\) it means \(\theta\) could either be \[\Large \frac{\pi}{3}\qquad or\qquad\frac{4\pi}{3}\]
yeah? ^.^

- Ray10

oh yes! that is correct! :D

- Ray10

is there a way to choose the correct one though?

- anonymous

I ALWAYS find a way ^.^
Well, both real \(\large -8\) and imaginary \(\large -8\sqrt3\) parts are negative, so it must be in Quadrant 3 ^.^
So, between \(\Large \frac \pi 3\) and \(\Large \frac{4\pi}3\), which is in Q3?

- Ray10

the second one of course :D (that isn't a guess :P )

- anonymous

Of course it isn't. But I already said \(\Large \frac \pi 3\) was wrong XD

- Ray10

does that make it \[16[\cos (\frac{ 4\Pi }{ 3})+i \times \sin (\frac{ 4\Pi }{ 3 })]\] ?

- Ray10

you did say that was wrong xD

- anonymous

yes moving on ^.^
\[\Large -8-8\sqrt3\color{red}i=16 \ \text{cis}\left(\frac{4\pi}{3}\right) \]

- Ray10

ah yes I see that now :)

- anonymous

Ever heard of this rule?
\[\Large \left[r \ \text{cis}(\theta)\right]^{\color{blue}p}=r^{\color{blue}p} \ \text{cis}(\color{blue}p\theta)\]

- Ray10

actually that does look rather familiar :) do I take it that, p is the amount of roots? :S

- anonymous

not exactly.
Anyway, to get the fourth root, it's just raising to fraction power... whatever that means :P
\[\LARGE\sqrt[4]{-8-8\sqrt3\color{red}i}=\left[16 \ \text{cis}\left(\frac{4\pi}{3}\right)\right]^{\frac14}\]

- anonymous

Now I want you to work THIS
\[\LARGE\sqrt[4]{-8-8\sqrt3\color{red}i}=\color{blue}{\left[16 \ \text{cis}\left(\frac{4\pi}{3}\right)\right]^{\frac14}}\] using this rule:\[\Large \left[r \ \text{cis}(\theta)\right]^{\color{blue}p}=r^{\color{blue}p} \ \text{cis}(\color{blue}p\theta)\]

- Ray10

hmmm I can see now how you got that, I just can't seem to apply it :/

- anonymous

Sheesh, do I have to everything? :/
\[\Large \sqrt[4]{-8-8\sqrt3\color{red}i}=\color{blue}{\left[16 \ \text{cis}\left(\frac{4\pi}{3}\right)\right]^{\frac14}}=16^{\color{blue}{\frac14}} \ \text{cis}\left(\color{blue}{\frac14}\cdot\frac{4\pi}{3}\right)\]

- Ray10

sorry :/ I'm rather new to complex numbers :/

- anonymous

You don't see me giving up :P
Now come on, work out that last bit and simplify.

- Ray10

I get \[1+\sqrt{3} \times i\] :S
I'm wrong huh?

- anonymous

As a matter of fact, you're right (for a change >:) )

- anonymous

but that's not where the problems end... as you probably predicted, there are more than one fourth roots ^.^

- Ray10

hmm I see, so that was a step to obtain one fourth root hey, there must be more to the formula to find the rest :P

- Ray10

yaaay I'm right for a change :D

- anonymous

Of course.
I also happen to know the secret of the roots :3
It's an all-powerful way to quickly find the other fourth (or any nth) roots of a complex number after finding the first...
I can tell you, but are you worthy? ^.^

- anonymous

giving up already @Ray10 ?

- Ray10

I think I am! :P I fought that there crocodile off, unlike Captain Hook xP

- Ray10

I would really appreciate knowing the secret :)

- anonymous

I'll decide if you're worthy :P
Did you notice that we can add \(\large 2\pi\) to the angle and its cis wouldn't change one bit? :D
\[\Large 16 \ \text{cis}\left(\frac{4\pi}{3}\right) = \ 16 \ \text{cis}\left(\frac{4\pi}{3}+2\pi\right)\]

- Ray10

therefore giving the value of yet another root?

- Ray10

yes I did :)

- anonymous

That's right ^.^
So work out
\[\Large \left[16 \ \text{cis}\left(\frac{4\pi}{3}+2\pi\right)\right]^\frac14 \]
And I'll see if you're worthy :D

- Ray10

I seem to get an odd answer.... \[-\sqrt{3} + i\] :S

- anonymous

- anonymous

And I'd probably sound like what you'd call a broken record, but we can add another \(\large 2\pi\) to the angle
\[\Large16 \ \text{cis}\left(\frac{4\pi}{3}+2\pi\right)=16 \ \text{cis}\left(\frac{4\pi}{3}+4\pi\right)\]

- anonymous

and it's still the same. Right?

- anonymous

Hey @Ray10 stay with me here!
(and how old are you anyway?)

- Ray10

it's like magic!!! :O

- anonymous

So let's get a new fourth root.
\[\Large \left[16 \ \text{cis}\left(\frac{4\pi}{3}+4\pi\right)\right]^\frac14 \]

- Ray10

(I am \[\frac{ 5 }{ 10 } + \frac{ 8 }{ \frac{ 16 }{ 37 } }\] :P ) (how old are you?)
\[-1-\sqrt{3}i\]

- anonymous

19?

- anonymous

steep. ^.^

- Ray10

haha yes that is correct :P

- anonymous

but, you're right.
you seem to be on a roll :D

- anonymous

You are ready for the secret ^.^

- Ray10

yaaaaay!!! :D proved my worth :D

- anonymous

Which, if you were actually listening (something I don't do) you'd probably have figured out by now :D

- Ray10

add \[2\Pi \] ?

- anonymous

Close. Still wrong :P
First, figure out one fourth root.
\[\Large 2 \ \text{cis}\left(\frac{4\pi}3\right)\]

- anonymous

And then keep adding \(\LARGE \frac{2\pi}{\color{red}4}=\frac\pi2\) to the angle until you have four roots ^.^

- Ray10

so in this case, where we did \[+2\Pi and +4\Pi \] , do they count as roots?

- anonymous

whoopsie, the fourth root was
\[\Large 2 \ \text{cis}\left(\frac\pi 3\right)\]sorry ^.^

- Ray10

oh so \[\frac{ \Pi }{ 3}\] is one?

- anonymous

So, to list them down, the four fourth roots are \[\Large 2 \ \text{cis}\left(\frac{\pi}3\right)\]\[\Large 2 \ \text{cis}\left(\frac{\pi}3+\frac\pi2\right)=\Large 2 \ \text{cis}\left(\frac{5\pi}6\right)\]\[\Large 2 \ \text{cis}\left(\frac{\pi}3+\pi\right)=\Large 2 \ \text{cis}\left(\frac{4\pi}3\right)\]\[\Large 2 \ \text{cis}\left(\frac{\pi}3+\frac{3\pi}2\right)=\Large 2 \ \text{cis}\left(\frac{11 \pi}6\right)\]

- anonymous

Notice we just kept adding \[\Large \frac{2\pi}4=\frac\pi2\]to the angle ^.^

- Ray10

ah I see, now that makes sense!! :D
but do I keep it in cis form or put it into like; \[1+\sqrt{3}i\] form?
because I need to show it on an Argand Diagram :/

- anonymous

It looks like they're readily convertible to a+bi form ^.^
And I don't know what an Argand Diagram is.

- anonymous

And now I do. ^.^
In that case, you can either use polar or rectangular, whichever is easier to "argandise" XD

- Ray10

argandise! xD haha that is a good one!! :D
well you have been a MASSIVE help!!
Fastest and most consistent responses ever!! :D Thank you so much @PeterPan !!!

- anonymous

Oh, and I'm 14 ^.^

- Ray10

O.o
14????

- Ray10

genius much?!?! :D

- anonymous

14.
7+7
7 x 2
\[\Large \sum_{n=0}^\infty \frac{7}{2^n }\]

- Ray10

o.O you are incredibly smart!!!

- anonymous

That and a lot of other things ^.^
haha
Need any more help?

- Ray10

you are a genius in my eyes ^.^
Well actually I need an answer of mine confirmed now that you mention it :P do you know about partial differentiation?

- anonymous

Let's see... this \[\Large \frac{\partial}{\partial x}\] comes to mind.
What's on yours?

- Ray10

yeah thats about right :) think you can have a look at a question I've done so far?

- anonymous

no promises. ^.^
hit me...

- Ray10

all good :) lets see
Given the function \[z=x ^{2}-2x+y ^{3}-3y\]
determine the position of the stationary point and their types (maxima, minima or saddle points)
so far I gett
\[zx\] = \[2x-2\]
\[zy\] = \[3y ^{2}-3\]
\[zxy\] = 0

- anonymous

hmm

- anonymous

One can never read enough :/
Oh well, first find the critical points... points where BOTH \(\large z_x\) and \(\large z_y\) are zero.

- Ray10

x=1 y=1, x-1 y=-1 :)

- anonymous

No.

- Ray10

ah true, one can never read enough!

- anonymous

Oh wait, you made a bloody typo XD

- anonymous

(1,1) and (1,-1) aye?

- Ray10

yeah :P
sorry it's 12:50am here and I am close to falling asleep in my glossy keyboard xD

- anonymous

Not my problem. >:)
Anyway, find the Hessian ^.^
\[\Large \left| \begin{matrix}z_{xx}&z_{xy}\\z_{yx}&z_{yy}\end{matrix}\right|=z_{xx}z_{yy}-(z_{xy})^2\]

- Ray10

12

- anonymous

Well that can't be true. What's \(\Large z_{yy}\)?

- Ray10

6y ?

- anonymous

Now reevaluate your Hessian -.-

- Ray10

zy = \[3y ^{2} -3\]
zyy = \[6y\]

- anonymous

Yes, I know that :D
I said reevaluate the Hessian...

- Ray10

zx = \[2x-2\]
zxx = 2

- Ray10

ohhh right :P

- Ray10

12 and -12 ? :S

- anonymous

HESSIAN!
\[\Large z_{xx}z_{yy}-(z_{xy})^2\]

- Ray10

\[2\times 6 - (0)^{2}\] is the Hessian

- Ray10

is that what you are saying? :S

- Ray10

@PeterPan I think my mind is fried :S

- anonymous

So, you say this:

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- anonymous

And then you say this:

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- Ray10

yes

- Ray10

\[2\times 6y - (0)^{2}\]

- anonymous

NOW he gets it -.-
Simplify it :P

- Ray10

12y :P

- anonymous

And that, really-old-and-very lost boy, is the Hessian.
Now, at (1,1), what is its value?

- Ray10

12...

- Ray10

I like your method of teaching actually :P

- anonymous

That's what she said ^.^
Anyway, positive Hessian, and positive \(\large z_{xx}\) which is 2 and always positive anyway... means it's a local minimum. :D
What about (1,-1)?

- Ray10

-12..

- anonymous

- anonymous

http://en.wikipedia.org/wiki/Second_partial_derivative_test

- Ray10

oh gosh O.o
why are you sooo smart?!?!

- anonymous

Probably because of all the candy ^.^

- Ray10

I need this candy :P

- Ray10

You're quite bright for your age! :D

- anonymous

It only works when you're a kid :P
And you're a grown-up already :(

- Ray10

oh please!! If you knew what I still did, you would differ that answer :P

- anonymous

Aww yea!
Grow big and never grow up
^.^

- Ray10

thats me!!! :D
Pokemon nerd..just a bit too much ^.^

- anonymous

So... I'll leave you to your musings, then? I'm sure you need to sleep :D

- Ray10

yes thank you ! But once again, thank you very much for all your help and advice! Really appreciate it! :D

- Ray10

I'll talk to you soon!

- anonymous

Keep your window open ^.^

- Ray10

fairy dust?! :D

- anonymous

A happy thought.
Now get your rest XD
A drowsy adventurer... is a DEAD adventurer >:)

- Ray10

ah hahah soo true! Night @PeterPan! :D

- anonymous

bye ^.^

- Ray10

quick question; why is it 2 in front?

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- Ray10

@PeterPan

- Ray10

all good, I worked it out :)

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