Question

Trig question? Please help! I will give a medal!

Answer 1

Find the cube roots of 27(cos 279° + i sin 279°).

Answer 2

I don't know where to start. Please help.

Answer 3

Well, you know that when raising a complex number in polar form to an exponent, you raise its modulus (absolute value) by said exponent, and then you multiply the argument (the angle in question) by the power?

Answer 4

Are you familiar with DeMoivre’s Theorem? :)

Answer 5

Well, I learned about it in this lesson but I really don't understand it.

Answer 6

This is a practice question for the test.

Answer 7

I can see what r, cos theta, and i sin theta are in my formula, but where is the p?

Answer 8

You're taking the cube root, which is the same as raising to an exponent (p) of 1/3
So take p = 1/3

Answer 9

Scratch my formula, it's wrong, stand by...

Answer 10

This is the correct one
\[\huge [r(\cos \theta + i \sin \theta)]^{p}=r^{p}\cos \ p \theta + i \sin \ p \theta\]

Answer 11

Okay, so:

Answer 12

Got it mixed up... sorry
Now, you were saying...?
Yeah, take p = 1/3

Answer 13

\[r ^(1/3) \cos 1/3(279) + i \sin 1/3(279)\]

Answer 14

Is that right?

Answer 15

by the way, from now on, I'll be shortening (for simplicity)
\[\huge \cos \theta + i \sin \theta = cis \theta\]

Answer 16

2b1ask1

Answer 17

@terenzreignz Was that correct?

Answer 18

It was, but your problem was to look for the cube ROOTS right?
There are more than one :)
But your answer was one of them, though please do evaluate 279/3, to make it prettier :)

Answer 19

Haha, okay. I will.

How do I find the other one?

Answer 20

Other TWO, in fact. I'll explain why, along the way.
Now, De Moivre's theorem is all good and useful, but it's not very straightforward for fractional exponents. For instance
\[\cos \theta = \cos \theta + 360^{o}\]
right?
Same goes for sin.

Answer 21

Okay.

Answer 22

Well then, it only follows that
\[r \ cis \ \theta = r \ cis \ \theta+360^{o}\]agree?

Answer 23

Yes, that makes sense.

Answer 24

Now apply substitute and apply de Moivre
(in other words, raise the modulus r to the exponent and divide the angle by the exponent)

Answer 25

\[9 (-\sin(\pi/60) + i \cos(\pi/60))\]

Answer 26

(Don't worry, after this is done, if you don't realise it right away, I'll tell you how to quickly do this)

Answer 27

Is that right?

Answer 28

Are you sure?
Your problem was
27 cis 279
right?
And you already got the solution for that, which is
3 cis 93

now try
27 cis (279+360)

Answer 29

Wait, was it wrong?

27 cis (279 + 360)
27 cis (639)

Answer 30

No, it wasn't, lol
I only said it's only ONE of the solutions, and there are three. Here's the second one
27 cis 639
seems about correct, now take the cube root of this using de Moivre. (the same way you did with
27 cis 279)

Answer 31

Stuck?

Answer 32

Yeah.

Answer 33

Okay, remember de Moivre
\[\huge [r(\cos \theta + i \sin \theta)]^{p}=r^{p}\cos \ p \theta + i \sin \ p \theta\]
and this time, take theta = 639 instead
and p is still 1/3

Answer 34

Okay.\[27/3 (\cos (639/3) + i \sin (639/3))\]

Answer 35

27 is your modulus, remember? You don't multiply it to the exponent, you raise it to that exponent!
What's 27 raised to 1/3?

Answer 36

3?

Answer 37

In other words, 27 is your r value in that de Moivre formula

Answer 38

That's right. So therefore, what's another cube root for your expression?

Answer 39

3 (cos(213) + i sin(213)) ??

Answer 40

That's right :)
So now we have two cube roots
3 cis 93
3 cis 213

Any idea as to how to get the third?

Answer 41

Hmm, add another 360?

Answer 42

I love your intuition already :D
You're correct :)
So add another 360 to 639
And repeat the process
Go ahead now...

Answer 43

Yay!

27^(1/3) (cos(999/3) + i sin(999/3))
3(cos(333) + i sin(333))

Answer 44

That's correct :)
Now, for the sake of information, what happens if you do it again?
Add another 360 to 999 and then repeat the process...

Answer 45

27^(1/3) (cos(1359/3) + i sin(1359/3))
3(-sin(pi/60) + i cos(pi/60))

Oh! So it there can only be three solutions before it repeats itself.

Answer 46

That's right :)
You're catching on pretty quick :D
Now, the quicker way to do this, you want it? :)

Answer 47

THANK YOU SO MUCH! You are a genius. You have really helped me understand it. Oh yeah, if there is a quicker way, sure!

Answer 48

And lol, stop switching between radians and degrees, it's making me dizzy :P
Anyway
Let's pretend we don't know the answer yet, and we're given
3 cis 279
Just do the de Moivre thing once

Answer 49

sorry, we were given
27 cis 279

My bad :)

Answer 50

lol. It's good practice!
27^(1/3) (cos(279/3) + i sin(279/3))

Answer 51

Which is
3 cis 93
Now, that long bit, it can be summarised by......

Answer 52

Just keep adding 360/n to the angle
until you have n solutions.

Answer 53

In our case, we were taking the cube roots, right? So you just keep adding 360/3 to the angle, until you have three solutions :)

Answer 54

You didn't check, did you? :P
You started with
3 cis 93

and did you notice you just kept adding 120 to the angle? :)

Answer 55

Wow! That is so much easier! I will be sure to use that next time. Thank you so much for everything!

Answer 56

Works for radians too
except, of course, replace 360 with 2pi
and just keep adding
2pi/n to the angle
when taking the nth root :)

Answer 57

...until you have exactly n solutions.

Answer 58

I get it now. Thankyou. Have a great night!