Question

Roots of unity help! Write each expression in the standard form for the complex number a + bi.
1. The complex fifth roots of 5-5√3i.
2. All seventh roots of unity.

Answer 1

Hey :)

Answer 2

\[\large\rm 5-5\sqrt{3}~i\]Factor a 10 out of each number,
(I can explain this step further if this is confusing),\[\large\rm =10\left(\frac{1}{2}-\frac{\sqrt3}{2}i\right)\]Let's convert to polar.
So the real portion corresponds to cosine.
Cosine is 1/2 at pi/3 and 5pi/3.
Sine is negative, so we must be dealing with 5pi/3.\[\large\rm =10\left(\cos\frac{5\pi}{3}+i \sin\frac{5\pi}{3}\right)\]

Then we can look for 5th roots at this point, apply De'Moivre's Theorem and such.
I'll give you a sec to read that,
lemme know if any of it is crazy confusing.

Answer 3

I'm not quite sure why we would factor out the 10, or how to apply De Moivre's Theorem to an equation like this. Other than that, it seems pretty clear so far...

Answer 4

You can do the more traditional route if that was strange.
\(\large\rm r=\sqrt{a^2+b^2}\)

Answer 5

I like that little trick though,
what I was doing was...
I was attempting to turn the a and b values into special trig values.

Answer 6

But we can do it the more normal way,
\(\large\rm r=\sqrt{5^2+(5\sqrt{3})^2}=\sqrt{25+3\cdot25}=\sqrt{100}=10\)

Answer 7

\[\large\rm \theta=\arctan\left(\frac{b}{a}\right)=\arctan\left(\frac{-5\sqrt3}{5}\right)=\arctan\left(-\sqrt3\right)\]

Answer 8

And uhhh yah, 5pi/3 right? :)
Remember your trig stuff?

maybe instead we'll start at -pi/3 (which is co-terminal to 5pi/3).

Answer 9

\[\large\rm =10\left[\cos\left(\frac{5\pi}{3}\right)+i \sin\left(\frac{5\pi}{3}\right)\right]\]We do have a small issue to work out here though.
We're located at some point in the complex plane, on the circle of radius 10.
What happens if we spin a full time around the circle 2pi?
We land right back in the same spot, ya?

Answer 10

\[\large\rm =10\left[\cos\left(-\frac{\pi}{3}\right)+i \sin\left(-\frac{\pi}{3}\right)\right]\]Oh I said -pi/3 didn't I? :d woops.

Answer 11

If we spin a full 2pi, we end up in the same spot,
therefore,\[\rm 10\left[\cos\left(-\frac{\pi}{3}\right)+i \sin\left(-\frac{\pi}{3}\right)\right]\quad=\quad 10\left[\cos\left(-\frac{\pi}{3}+2\pi\right)+i \sin\left(-\frac{\pi}{3}+2\pi\right)\right]\]Do you understand why these are equal based on that explanation? :o

Answer 12

Yes, since 2pi=one rotation or one circle, if you add or subtract 2pi you should end up where you started.

Answer 13

Good :)
And notice that we can add ANY multiple of 2pi for this to happen.
\[\large\rm =10\left[\cos\left(-\frac{\pi}{3}+2k\pi\right)+i \sin\left(-\frac{\pi}{3}+2k\pi\right)\right],\qquad\qquad k\in \mathbb Z\]As long as we're doing full spins, this is true.

Answer 14

Ok , with this insane looking monstrosity, NOW let's look for our 5th roots ^^

Answer 15

\[\rm 10^{1/5}\left[\cos\left(-\frac{\pi}{3}+2k\pi\right)+i \sin\left(-\frac{\pi}{3}+2k\pi\right)\right]^{1/5}\qquad\qquad k=?\]Something is going to happen to our k here.

Answer 16

Since we're looking for `fifth roots`,
we'll let k take on the first 4 positive integer values starting from 0,
after the first 5 integers, the roots will start to repeat,

so we get our 5 unique roots of this complex number from those first 5 values of k
k=0,1,2,3,4

Answer 17

\[\rm 10^{1/5}\left[\cos\left(-\frac{\pi}{3}+2k\pi\right)+i \sin\left(-\frac{\pi}{3}+2k\pi\right)\right]^{1/5}\qquad\qquad k=0,1,2,3,4\]

Answer 18

Do you remember De'Moivre's Theorem maybe?\[\large\rm \left[\cos(\theta)+i \sin(\theta)\right]^n=\cos(n \theta)+i \sin(n \theta)\]Does that look familiar?

Answer 19

Does it mean it works out to cos(1/5 -pi/3)+i sin (1/5 -pi/3)?

Answer 20

you should have some kind of multiplication in there, but yes :)
But keep in mind, you need to `distribute` the 1/5.
The 2k pi is also being multiplied by 1/5!

Answer 21

\[\rm =10^{1/5}\left[\cos\left(-\frac{\pi}{15}+\frac{2k\pi}{5}\right)+i \sin\left(-\frac{\pi}{15}+\frac{2k\pi}{5}\right)\right]\qquad\qquad k=0,1,2,3,4\]Bringing the 1/5 into the angle gives us something like this ya?

Answer 22

Woah they want us to write the roots in the form a+bi?
We'll probably have to do some approximating then...
these aren't going to be very nice angles.

Answer 23

Uh ohhh you got quiet :U
too confusing? I know that silence!!

Answer 24

After applying De'Moivre's Theorem,
we can proceed to plug in the different k values and look for roots.

k=0 gives us our first root:\[\rm =10^{1/5}\left[\cos\left(-\frac{\pi}{15}\right)+i \sin\left(-\frac{\pi}{15}\right)\right]\]You would need a calculator in order to put it into standard a+bi form.
Maybe I misread the directions though.

Answer 25

I was stuck on how to find the denominator for a bit, but it's just multiplying fractions, so I think I understand now. But to convert it to a+bi, I have no idea..

Answer 26

Ya, the first fraction has a 15,
the second fraction has a 5,
so I guess the second fraction just needs a 3 on top and bottom. :)

Answer 27

Lemme write that out with the common denominator just so it's clear.\[\rm =10^{1/5}\left[\cos\left(-\frac{\pi}{15}+\frac{6k\pi}{15}\right)+i \sin\left(-\frac{\pi}{15}+\frac{6k\pi}{15}\right)\right]\qquad\qquad k=0,1,2,3,4\]Don't try to combine them any further, the k is preventing you from doing that until a plug in a value.

Answer 28

Ah, then plug it all into the equation so you get it in a+bi form. Would this equation also work?
\[10^{1/5}[\cos (\frac{ \pi }{ 3 }+2k \pi)+i \sin (\frac{ \pi}{ 3 }+2k \pi)] ?\]

Answer 29

Hmm no :( That doesn't have the 5th root applied to it.

Answer 30

Just so it's clear, I wanna do another root.
For k=1, you get:\[\rm =10^{1/5}\left[\cos\left(-\frac{\pi}{15}+\frac{6(1)\pi}{15}\right)+i \sin\left(-\frac{\pi}{15}+\frac{6(1)\pi}{15}\right)\right]\]\[\rm =10^{1/5}\left[\cos\left(\frac{5\pi}{15}\right)+i \sin\left(\frac{5\pi}{15}\right)\right]\]\[\rm =10^{1/5}\left[\cos\left(\frac{\pi}{3}\right)+i \sin\left(\frac{\pi}{3}\right)\right]\]Oh this one actually gives us nice clean values :O Interesting.\[\large\rm =10^{1/5}\left[\frac{1}{2}+\frac{\sqrt3}{2}i\right]\]And the final a+bi form would be\[\large\rm =\frac{10^{1/5}}{2}+\frac{10^{1/5}\sqrt3}{2}i\]

Answer 31

Has your teacher mentioned anything about a "branch cut"?
Does that sound familiar?

Answer 32

I've never heard about it before.

Answer 33

It might change what your teacher would consider the "principle" root,
like which interval he/she wants you to stay in.

usually we do our angle from -pi to pi,
at least that's what I'm familiar with.
but maybe your teacher would prefer 0 to 2pi,
in which case we wouldn't want your starting angle to be -pi/3.

Ok ok ignore that though :) just something to keep in mind.

Answer 34

I think as long as it falls within -2pi to 0 to 2pi he won't mind, because he never specified which domain or range anyways, and -pi/3 would still be the right answer. It'd be kinda unfair if he docked points for that, in my opinion.

Answer 35

Wolfram is really helpful for.. at least checking your work:
https://www.wolframalpha.com/input/?i=%285%2B5sqrt%283%29i%29%5E%281%2F5%29

If you scroll down, it lists the 5 roots in decimal form, a+bi.
Don't get into a habit of using it for EVERY problem though :) lol
that will mess up your studying :P

Answer 36

I plugged in k=1 which equals 0.79+1.39i, but the root stated in the link above says it should be 0.79-1.39i, should I be concerned? Sorry for all the questions!

Answer 37

Mmm sec, I better check my work :)

Answer 38

Oh oh because I put the wrong equation into wolfram,
I put 5+5sqrt(3) instead of 5-5sqrt(3),
sorry bout that :(

https://www.wolframalpha.com/input/?i=%285-5sqrt%283%29i%29%5E%281%2F5%29

Answer 39

Ohhhhh, no worries, you've explained this problem perfectly! I'll just plug the rest of the values (0, 2, 3, 4) into the equation, find the decimal in a+bi format, and double check using wolfram. Thank you so much!

Answer 40

Yay team!
Np! ୧ʕ•̀ᴥ•́ʔ୨