Question

Given f(x) = x^6 + 1
(a) Find all six complex roots f(x) = 0
(b) Find the angle between successive roots

It's fine if you can't get (b), but (a) is what I really need now.

Answer 1

@freckles

Answer 2

@mayankdevnani

Answer 3

@ganeshie8

Answer 4

The roots will be all the sixth roots of -1. You have
\[\begin{align*}
x^6+1&=0\\
x^6&=-1\\
x&=(-1)^{1/6}\\
x&=\large\left(e^{i\pi}\right)^{1/6}\\
x&=\Large e^{i\left(\frac{\pi}{6}+\frac{2\pi k}{6}\right)}
\end{align*}\]
where \(k=0,1,...,5\).

Answer 5

we never learned e ^i pi what is that?

Answer 6

It's a part of Euler's formula,
\[\large e^{i\theta}=\cos\theta+i\sin\theta\]

Answer 7

ohh in our class we use z

Answer 8

where did u get 30 degrees from

Answer 9

pi/6

Answer 10

wait oh sorry "z" in our class is =r(cos theta + isin theta)

Answer 11

The polar form of \(-1\) is \(e^{i\pi}\), or \(\cos\pi+i\sin\pi=-1+0i=-1\).

Raising this to the 1/6 power, DeMoivre's theorem says that
\[\large\left(re^{i\theta}\right)^n=r^ne^{in\theta}\]
or
\[\large\left(r(\cos\theta+i\sin\theta)\right)^n=r^n(\cos n\theta+i\sin n\theta)\]
What this means is that
\[\large \left(e^{i\pi}\right)^{1/6}=e^{i\pi/6}\]
where the angle is in radians, or 30 degrees (relative to the positive real axis).

Answer 12

can you use arctan(b/a) to find theta?
Because that is the only way I learned how to do it

Answer 13

Have you not learned DeMoivre's theorem?

Answer 14

\[z ^{n}=[r(\cos \theta + isin \theta)]^{n}=r^n(\cos n \theta + i \sin n \theta)\]

Answer 15

@SithsAndGiggles

Answer 16

thats how we learned it

Answer 17

Okay, well \(\theta\) is the argument/angle of the original number (-1 in this case with angle \(\pi\), or 180 deg). When you raise the polar form to 1/6, you divide whatever this angle is by 6 (so that you have \(\dfrac{\pi}{6}\), or 30 deg).

You can use the inverse tangent approach to find the angle of -1, but it won't give you the angle of the principal sixth root of -1 unless you divide by 6.

Answer 18

so the n is 1/6 and when you multiply that by theta in the formula you get 30?

Answer 19

Yes

Answer 20

then how would you find all 6 roots? +2kpi?

Answer 21

\(+\dfrac{2k\pi}{n}\), yes

Answer 22

This should give you an idea of how to do part (b). Each angle between successive roots is the same.

Answer 23

so when
k=1, is the theta 5 degrees?

Answer 24

No, for \(k=0\) you get your principal root, which has the angle
\[\frac{\pi}{6}+\frac{2\times0 \pi}{6}\]
For \(k=1\), you get
\[\frac{\pi}{6}+\frac{2\times1 \pi}{6}=\frac{3\pi}{6}=\frac{\pi}{2}\]
and so on.

Answer 25

do you know (b)? because if you don't that is completely fine

Answer 26

any ideas?

Answer 27

What's the angle between the first and second roots?

Answer 28

60

Answer 29

There you go.

Answer 30

thats it?

Answer 31

Yes, the agnle between successive roots is the same for all successive roots.

Consider the Argand diagram: