Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

and lol.. grown-ups -.-

Answer 9

hey did you leave? :(

Answer 10

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

Answer 11

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

Answer 12

no :P

Answer 13

what do you mean? :O

Answer 14

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

Answer 15

16 :P

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

So how do you know which is which? XD

Answer 24

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

Answer 25

No.

Answer 26

You're guessing :P

Answer 27

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?

Answer 28

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?

Answer 29

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

Answer 30

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? ^.^

Answer 31

oh yes! that is correct! :D

Answer 32

is there a way to choose the correct one though?

Answer 33

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?

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

you did say that was wrong xD

Answer 38

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

Answer 39

ah yes I see that now :)

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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:\[\Large \left[r \ \text{cis}(\theta)\right]^{\color{blue}p}=r^{\color{blue}p} \ \text{cis}(\color{blue}p\theta)\]

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

yaaay I'm right for a change :D

Answer 53

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? ^.^

Answer 54

giving up already @Ray10 ?

Answer 55

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

Answer 56

I would really appreciate knowing the secret :)

Answer 57

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
\[\Large 16 \ \text{cis}\left(\frac{4\pi}{3}\right) = \ 16 \ \text{cis}\left(\frac{4\pi}{3}+2\pi\right)\]

Answer 58

therefore giving the value of yet another root?

Answer 59

yes I did :)

Answer 60

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

Answer 61

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

Answer 62

<claps>

Answer 63

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

Answer 64

and it's still the same. Right?

Answer 65

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

Answer 66

it's like magic!!! :O

Answer 67

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

Answer 68

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

Answer 69

19?

Answer 70

steep. ^.^

Answer 71

haha yes that is correct :P

Answer 72

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

Answer 73

You are ready for the secret ^.^

Answer 74

yaaaaay!!! :D proved my worth :D

Answer 75

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

Answer 76

add \[2\Pi \] ?

Answer 77

Close. Still wrong :P

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

Answer 78

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

Answer 79

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

Answer 80

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

Answer 81

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

Answer 82

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

Answer 83

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

Answer 84

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 :/

Answer 85

It looks like they're readily convertible to a+bi form ^.^

And I don't know what an Argand Diagram is.

Answer 86

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

Answer 87

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 !!!

Answer 88

Oh, and I'm 14 ^.^

Answer 89

O.o
14????

Answer 90

genius much?!?! :D

Answer 91

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

Answer 92

o.O you are incredibly smart!!!

Answer 93

That and a lot of other things ^.^

haha

Need any more help?

Answer 94

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?

Answer 95

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

Answer 96

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

Answer 97

no promises. ^.^
hit me...