Ray10
  • Ray10
find all fourth roots of \[-8-8\sqrt{3i}\] i being a complex number, and show them on an argand diagram
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
this is bad :O are you sure it isn't \[\large -8-8\sqrt3i\]?
Ray10
  • Ray10
:P my bad!! yes it is \[-8-8\sqrt{3}j\]
Ray10
  • Ray10
thanks for that! But yes can you help me please?

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anonymous
  • anonymous
now it's a j? why has the world gone coconuts?
anonymous
  • anonymous
but more on the fun part later, now we get to work ^.^
Ray10
  • 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
  • anonymous
first you need it in polar form. like this: \[\Large r\left[\cos(\theta) + \color{red}i\sin(\theta)\right]\]
anonymous
  • anonymous
and lol.. grown-ups -.-
anonymous
  • anonymous
hey did you leave? :(
Ray10
  • Ray10
no I didnt leave!! :P I'm trying to convert it to polar :P yes ah grown-ups >.<
Ray10
  • Ray10
I get \[r=8\sqrt{2}\]
anonymous
  • anonymous
no :P
Ray10
  • Ray10
what do you mean? :O
anonymous
  • anonymous
I mean no :P \[\Large r = \sqrt{(-8)^2 + \left(-8\sqrt3\right)^2}\]
Ray10
  • Ray10
16 :P
anonymous
  • anonymous
yes better. what about \(\theta\)?
Ray10
  • Ray10
\[\frac{ \Pi }{ 3 }\] I can confirm :)
anonymous
  • anonymous
Me too :) I can confirm that it's WRONG!
Ray10
  • Ray10
:O no way?! :O how so???
anonymous
  • 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
  • Ray10
\[\tan \theta =\frac{ y }{ x }\] ?
anonymous
  • anonymous
Yes... but that also gives \[\Large \frac{4\pi}{3}\]
anonymous
  • anonymous
So how do you know which is which? XD
Ray10
  • Ray10
\[\frac{ -\Pi }{ 3 }\] :P
anonymous
  • anonymous
No.
anonymous
  • anonymous
You're guessing :P
Ray10
  • 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
  • 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
  • Ray10
ahhh I love all the references!! :D funnest maths tutor :P yes we do know that :)
anonymous
  • 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
  • Ray10
oh yes! that is correct! :D
Ray10
  • Ray10
is there a way to choose the correct one though?
anonymous
  • 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
  • Ray10
the second one of course :D (that isn't a guess :P )
anonymous
  • anonymous
Of course it isn't. But I already said \(\Large \frac \pi 3\) was wrong XD
Ray10
  • Ray10
does that make it \[16[\cos (\frac{ 4\Pi }{ 3})+i \times \sin (\frac{ 4\Pi }{ 3 })]\] ?
Ray10
  • Ray10
you did say that was wrong xD
anonymous
  • anonymous
yes moving on ^.^ \[\Large -8-8\sqrt3\color{red}i=16 \ \text{cis}\left(\frac{4\pi}{3}\right) \]
Ray10
  • Ray10
ah yes I see that now :)
anonymous
  • 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
  • Ray10
actually that does look rather familiar :) do I take it that, p is the amount of roots? :S
anonymous
  • 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
  • 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
  • Ray10
hmmm I can see now how you got that, I just can't seem to apply it :/
anonymous
  • 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
  • Ray10
sorry :/ I'm rather new to complex numbers :/
anonymous
  • anonymous
You don't see me giving up :P Now come on, work out that last bit and simplify.
Ray10
  • Ray10
I get \[1+\sqrt{3} \times i\] :S I'm wrong huh?
anonymous
  • anonymous
As a matter of fact, you're right (for a change >:) )
anonymous
  • anonymous
but that's not where the problems end... as you probably predicted, there are more than one fourth roots ^.^
Ray10
  • 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
  • Ray10
yaaay I'm right for a change :D
anonymous
  • 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
  • anonymous
giving up already @Ray10 ?
Ray10
  • Ray10
I think I am! :P I fought that there crocodile off, unlike Captain Hook xP
Ray10
  • Ray10
I would really appreciate knowing the secret :)
anonymous
  • 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
  • Ray10
therefore giving the value of yet another root?
Ray10
  • Ray10
yes I did :)
anonymous
  • 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
  • Ray10
I seem to get an odd answer.... \[-\sqrt{3} + i\] :S
anonymous
  • anonymous
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
  • anonymous
and it's still the same. Right?
anonymous
  • anonymous
Hey @Ray10 stay with me here! (and how old are you anyway?)
Ray10
  • Ray10
it's like magic!!! :O
anonymous
  • 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
  • Ray10
(I am \[\frac{ 5 }{ 10 } + \frac{ 8 }{ \frac{ 16 }{ 37 } }\] :P ) (how old are you?) \[-1-\sqrt{3}i\]
anonymous
  • anonymous
19?
anonymous
  • anonymous
steep. ^.^
Ray10
  • Ray10
haha yes that is correct :P
anonymous
  • anonymous
but, you're right. you seem to be on a roll :D
anonymous
  • anonymous
You are ready for the secret ^.^
Ray10
  • Ray10
yaaaaay!!! :D proved my worth :D
anonymous
  • anonymous
Which, if you were actually listening (something I don't do) you'd probably have figured out by now :D
Ray10
  • Ray10
add \[2\Pi \] ?
anonymous
  • anonymous
Close. Still wrong :P First, figure out one fourth root. \[\Large 2 \ \text{cis}\left(\frac{4\pi}3\right)\]
anonymous
  • anonymous
And then keep adding \(\LARGE \frac{2\pi}{\color{red}4}=\frac\pi2\) to the angle until you have four roots ^.^
Ray10
  • Ray10
so in this case, where we did \[+2\Pi and +4\Pi \] , do they count as roots?
anonymous
  • anonymous
whoopsie, the fourth root was \[\Large 2 \ \text{cis}\left(\frac\pi 3\right)\]sorry ^.^
Ray10
  • Ray10
oh so \[\frac{ \Pi }{ 3}\] is one?
anonymous
  • 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
  • anonymous
Notice we just kept adding \[\Large \frac{2\pi}4=\frac\pi2\]to the angle ^.^
Ray10
  • 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
  • anonymous
It looks like they're readily convertible to a+bi form ^.^ And I don't know what an Argand Diagram is.
anonymous
  • anonymous
And now I do. ^.^ In that case, you can either use polar or rectangular, whichever is easier to "argandise" XD
Ray10
  • 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
  • anonymous
Oh, and I'm 14 ^.^
Ray10
  • Ray10
O.o 14????
Ray10
  • Ray10
genius much?!?! :D
anonymous
  • anonymous
14. 7+7 7 x 2 \[\Large \sum_{n=0}^\infty \frac{7}{2^n }\]
Ray10
  • Ray10
o.O you are incredibly smart!!!
anonymous
  • anonymous
That and a lot of other things ^.^ haha Need any more help?
Ray10
  • 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
  • anonymous
Let's see... this \[\Large \frac{\partial}{\partial x}\] comes to mind. What's on yours?
Ray10
  • Ray10
yeah thats about right :) think you can have a look at a question I've done so far?
anonymous
  • anonymous
no promises. ^.^ hit me...
Ray10
  • 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
  • anonymous
hmm
anonymous
  • 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
  • Ray10
x=1 y=1, x-1 y=-1 :)
anonymous
  • anonymous
No.
Ray10
  • Ray10
ah true, one can never read enough!
anonymous
  • anonymous
Oh wait, you made a bloody typo XD
anonymous
  • anonymous
(1,1) and (1,-1) aye?
Ray10
  • Ray10
yeah :P sorry it's 12:50am here and I am close to falling asleep in my glossy keyboard xD
anonymous
  • 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
  • Ray10
12
anonymous
  • anonymous
Well that can't be true. What's \(\Large z_{yy}\)?
Ray10
  • Ray10
6y ?
anonymous
  • anonymous
Now reevaluate your Hessian -.-
Ray10
  • Ray10
zy = \[3y ^{2} -3\] zyy = \[6y\]
anonymous
  • anonymous
Yes, I know that :D I said reevaluate the Hessian...
Ray10
  • Ray10
zx = \[2x-2\] zxx = 2
Ray10
  • Ray10
ohhh right :P
Ray10
  • Ray10
12 and -12 ? :S
anonymous
  • anonymous
HESSIAN! \[\Large z_{xx}z_{yy}-(z_{xy})^2\]
Ray10
  • Ray10
\[2\times 6 - (0)^{2}\] is the Hessian
Ray10
  • Ray10
is that what you are saying? :S
Ray10
  • Ray10
@PeterPan I think my mind is fried :S
anonymous
  • anonymous
So, you say this:
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anonymous
  • anonymous
And then you say this:
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Ray10
  • Ray10
yes
Ray10
  • Ray10
\[2\times 6y - (0)^{2}\]
anonymous
  • anonymous
NOW he gets it -.- Simplify it :P
Ray10
  • Ray10
12y :P
anonymous
  • anonymous
And that, really-old-and-very lost boy, is the Hessian. Now, at (1,1), what is its value?
Ray10
  • Ray10
12...
Ray10
  • Ray10
I like your method of teaching actually :P
anonymous
  • 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
  • Ray10
-12..
anonymous
  • anonymous
Negative Hessian means it's a saddle point. And that concludes this adventure ^.^
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Second_partial_derivative_test
Ray10
  • Ray10
oh gosh O.o why are you sooo smart?!?!
anonymous
  • anonymous
Probably because of all the candy ^.^
Ray10
  • Ray10
I need this candy :P
Ray10
  • Ray10
You're quite bright for your age! :D
anonymous
  • anonymous
It only works when you're a kid :P And you're a grown-up already :(
Ray10
  • Ray10
oh please!! If you knew what I still did, you would differ that answer :P
anonymous
  • anonymous
Aww yea! Grow big and never grow up ^.^
Ray10
  • Ray10
thats me!!! :D Pokemon nerd..just a bit too much ^.^
anonymous
  • anonymous
So... I'll leave you to your musings, then? I'm sure you need to sleep :D
Ray10
  • Ray10
yes thank you ! But once again, thank you very much for all your help and advice! Really appreciate it! :D
Ray10
  • Ray10
I'll talk to you soon!
anonymous
  • anonymous
Keep your window open ^.^
Ray10
  • Ray10
fairy dust?! :D
anonymous
  • anonymous
A happy thought. Now get your rest XD A drowsy adventurer... is a DEAD adventurer >:)
Ray10
  • Ray10
ah hahah soo true! Night @PeterPan! :D
anonymous
  • anonymous
bye ^.^
Ray10
  • Ray10
quick question; why is it 2 in front?
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Ray10
  • Ray10
@PeterPan
Ray10
  • Ray10
all good, I worked it out :)

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