Question

Find all fourth roots of \[-8-8\sqrt{3}i\] ?
I come to r=16, \[\theta \] = \[\frac{ \Pi }{ 3}\] , then get lost

Answer 1

That is one big pi.
Use \pi instead of \Pi lol

Answer 2

haha ah thank you! I shall :)

Answer 3

\[\frac{ \pi }{ 3 }\]

Answer 4

Isn't that better? :)

Answer 5

yes it certainly is thanks :) didn't know about that. Do you think you can help me with this question? :)

Answer 6

Can I indeed...

Answer 7

True enough, the modulus r is 16, but tell me how you acquired \(\Large \frac \pi 3\) for your argument.

Answer 8

My guess is that you naïvely took \[\Large \tan^{-1}\left(\frac{-8\sqrt3}{-8}\right)\]

Answer 9

\[\tan^{-1} (\frac{ -8\sqrt{3} }{ -8 }) = \frac{ \pi }{ 3 }\]

Answer 10

the other way I looked at it was, without the negative in front of the \[-8\sqrt{3}\], I thought that was wrong

Answer 11

Unacceptable. :)
You do realise that \(\large \tan^{-1}\) is not the most reliable of functions.

Your logic is far more dependable in these matters.

Answer 12

Now, you know that the tangent of the argument is \(\large \sqrt3\)

Answer 13

A positive, yet your real and imaginary parts are negative, implying naturally that the angle must be in the third quadrant of the Argand Diagram or complex plane.

Answer 14

hmm alright, I can understand where you are coming from :)
yes I see the logic behind it being \[\sqrt{3}\]

Answer 15

The tangent of the angle IS \(\Large \sqrt 3\) and this comes from the elementary division
\[\Large \frac{-8\sqrt3}{-8}=\frac{\cancel{-8}\sqrt3}{\cancel{-8}}\]

Answer 16

therefore \[\theta \] = \[\sqrt{3}\] :)

Answer 17

Most definitely not.
\[\Large \color{red}{\tan}\theta = \sqrt3\]
Do not confuse these things.

Answer 18

oh right because we are still at the point \[\tan \theta = \frac{ -8\sqrt{3} }{ -8 } = \sqrt{3}\]

Answer 19

yes. now do you see a problem?

Answer 20

hmmm sort of, now we must solve for \[\theta\] ?

Answer 21

but in the third quadrant ?

Answer 22

It would appear this is not the first time you've asked this question.

Answer 23

yes it's not the first time, I seem to get lost at this point often :/

Answer 24

Strange...
So what should be \(\large \theta\)?

Answer 25

\[\theta\] would equal \[\frac{ \pi }{ 3}\]

Answer 26

but would be required in the third quadrant right?

Answer 27

I'm almost compelled to mimic Pan and say "I already implied that \(\Large \frac \pi 3\) is wrong, but something tells me you're about to arrive at the correct angle anyway...

Answer 28

so therefore be \[\frac{ 4 \pi }{ 3 }\] :D

Answer 29

You're not just reading that previous thread, are you?
You DO understand why that's the correct angle, right?

Answer 30

because we are looking at the third quadrant, therefore we are required to find the equal such angle in the third quadrant, being that of \[\pi + \frac{ \pi }{ 3 }\]

Answer 31

no I'm not reading that

Answer 32

Not the explanation I was looking for, but in the assumption that you know what you're doing, then okay, that's good enough.

Answer 33

what would be the explanation you would say? :)

Answer 34

Well, now, there you have it, you have the modulus r=16 and the argument \(\large \theta = \frac{4\pi}3\)

Answer 35

The explanation I'd say is that we're looking for the angle whose tangent is \(\large \sqrt3\) and lies in the third quadrant.

Answer 36

Yes I see how that explains it neatly.
so now we have
\[r=16\]
\[\theta = \frac{ 4\pi }{ 3 }\]
and
\[16[cis(\frac{ 4\pi }{ 3 })]\]

Answer 37

yes indeed. Now find the fourth roots

Answer 38

that is where I have trouble with, Pan mentioned to add \[\frac{ \pi }{ 2 }\] but I thought we are meant to add \[2\pi \]

Answer 39

You add \(2\pi\) to the original angle but and THEN you multiply the new angle by \(\Large \frac14\) to get fourth roots... in effect, you're just adding \(\Large \frac{2\pi}{4}\) to the FOURTH ROOT angle, or \(\Large \frac\pi2\)

Answer 40

That might have been a little tough to digest.
You either add \(2\pi\) here:
\[\Large \left\{16\left[\text{cis}\left(\frac{ 4\pi }{ 3 }+\color{red}{2\pi}\right)\right]\right\}^\frac{1}{4}\]

OR you add \(\Large \frac\pi 2\) here:
\[\Large \left.2\left[\text{cis}\left(\frac{ \pi }{ 3 }+\color{red}{\frac\pi2}\right)\right]\right.\]

Answer 41

They mean the same thing.

Answer 42

1. \[\frac{ 4\pi }{ 3 } +2\pi\] = \[\frac{ 10\pi }{ 3 }\]
2. \[\frac{ 10\pi }{ 3 } \times \frac{ 1 }{ 4 } = \frac{ 5\pi }{ 6 }\]
3. \[\frac{ 5\pi }{ 6 } \times ?\]

Answer 43

Pan showed me how to do this, but waking up today, I don't get the part where we find each root after the first one :S

Answer 44

wait...

Answer 45

\[\Large \frac{5\pi}{6}\] is right.

Answer 46

Okay, assuming you found the first root, which happens to be
\[\Large 2\text{cis}\left(\frac{\pi}{3}\right)\]

Answer 47

I get lost after that :P I multiply by \[\frac{ 1 }{ 4 }\]

Answer 48

Okay, relax.... just take a deep breath and ready yourself...

Answer 49

I get lost after that :P I multiply by \[\frac{ 1 }{ 4 }\]

Answer 50

We have one fourth root, right?
\[\Large 2\text{cis}\left(\frac{\pi}{3}\right)\]

Answer 51

I'm going to ask you to do something a little crazy, but it might just ease up your load for now.

Answer 52

yes that is correct :) okay hit me! :D

Answer 53

Forget adding 2pi.
As soon as you found one fourth root, such as this one
\[\Large 2\text{cis}\left(\frac{\pi}{3}\right)\] then just keep adding 2pi/4 (4 in this case because we're looking for fourth roots)

Answer 54

just keep adding 2pi/4 to the angle until you have a total of FOUR fourth roots.
(if we were looking for fifth roots, then you'd have to get one fifth root, and then keep adding 2pi/5 until you get a total of FIVE fifth roots)

Answer 55

So, root number 1:
\[\Large 2\text{cis}\left(\frac{\pi}{3}\right)\]

Root number 2:
\[\Large 2\text{cis}\left(\frac{\pi}{3}+\frac{\pi}{2}\right)\]

Answer 56

O.o, it worked! And I really understood that O.o

Answer 57

I make a point of making sure I'm understood, at least, to the best of my ability.
Don't tell my you understood it... SHOW me.

Answer 58

me*

Answer 59

\[\frac{ \pi }{ 3 }, (\frac{ \pi }{ 3 }+\frac{ \pi }{ 2 })=\frac{ 5\pi }{ 6 },(\frac{ 5\pi }{ 6 }+\frac{ \pi }{ 2 })= \frac{ 4\pi }{ 3 }, (\frac{ 4\pi }{ 3 }+\frac{ \pi }{ 2 })=\frac{ 11\pi }{ 6 } \]

Answer 60

Again, I'm trusting that you didn't get this from simply reading Pan's work :)

Answer 61

I hope I am right this time :S

Answer 62

Note that those, in and of themselves aren't the answers yet.

Answer 63

noo I didn't read his work, you said that I must add \[\frac{ 2\pi }{ 4 }\] to each new angle

Answer 64

But
\[\Large 2\text{cis}\left(\frac{\pi}{3}\right)\]\[\Large 2\text{cis}\left(\frac{5\pi}{6}\right)\]\[\Large 2\text{cis}\left(\frac{4\pi}{3}\right)\]\[\Large 2\text{cis}\left(\frac{11\pi}{6}\right)\]

Answer 65

I'll leave you to work these out, I have confidence in you :)

Answer 66

you said a little while ago, "I'm going to get you to do something crazy" I understand the ease of use, but what other way is standard to be used? :)

Answer 67

Thank you @terenzreignz !!!

Answer 68

Actually, this way is used so often, this may well BE the standard way.

Answer 69

I was just worried that you'll love the shortcut so much (the secret of the roots?) that you'll forget why it works.

Answer 70

oh woah, that is a relief to hear! :)
oh no not at all, I asked again how to do this question because I want to have the confidence to teach this myself :)

Answer 71

You're teaching this?

Answer 72

not teaching as teacher, but say if another colleague of mine is stuck, I want to be able to help them while knowing the foundation behind it :)

Answer 73

Well because while adding 2\pi to the original angle:
\[\Large 16\text{cis}\left(\frac{4\pi}{3}+2\pi\right)\] doesn't change it, it DOES give you a new fourth root (or any nth root for that matter)
\[\Large 2\text{cis}\left[\frac14\left(\frac{4\pi}{3}+2\pi\right)\right]= 2\text{cis}\left[\frac{\pi}{3}+\frac\pi2\right]\]

Answer 74

Understood?

Answer 75

ahh I see, so the way you asked me to do it just then is another way, and I prefer the way you taught me. This way I must multiply by \[\frac{ 1 }{ 4 }\] after adding \[2\pi \] ?

Answer 76

Yes, that's true. That's exactly the same way.

How observant of you :)

Answer 77

Don't you just feel smarter? :)

Answer 78

I sure do!! It's always great knowing why you're doing something than just knowing how :D

Answer 79

I'm sure.
But you were still outplayed by a 14-year-old :P
(whom I taught)

By the way, it's TJ (for Terence Jason) to you ^_^

Answer 80

I know xD Pan is incredibly good! :D I'm wondering how old you are now :O
TJ, very nice to meet you, and you certainly helped me out really well!

Answer 81

I'm 17, as my profile states ^_^
I turned 17 two months ago.

Well, I have to go to school, so... I'll be seeing you, I guess ^_^
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Terence out

Answer 82

talk soon! Have a good time in school ^_^
thanks heaps again!