Roots of -8: Solving with Exponential Form

[31415]

Homework Statement

I'm supposed to find the 6 roots of -8 + 0i

we are told the usual method of for these problems is to put the complex number into it's exponential form like so

z = |z|exp( i(θ+2πk)) where k: [0 to n-1]

then put it to the relevant power 'n'

z^1/n = |z|^1/nexp(( i(θ+2πk)/n)

so I proceed by taking θ to be pi...

z^1/6 = 8^1/6exp(( i(π+2πk)/6)

thinking the question is easy...

then in the solution given for the question the answer is 8^1/6exp(( i(-π+2πk)/6)

I don't understand where this minus has come from - is the solution wrong or am I missing something?

(picture of question and solution attached)

 

[phyzguy]

Both expressions are the same 6 numbers, it's just that the value of k in one set is shifted by 1 with respect to the other set.

 

[Dustinsfl]

Homework Statement

I'm supposed to find the 6 roots of -8 + 0i

we are told the usual method of for these problems is to put the complex number into it's exponential form like so

z = |z|exp( i(θ+2πk)) where k: [0 to n-1]

then put it to the relevant power 'n'

z^1/n = |z|^1/nexp(( i(θ+2πk)/n)

so I proceed by taking θ to be pi...

z^1/6 = 8^1/6exp(( i(π+2πk)/6)

thinking the question is easy...

then in the solution given for the question the answer is 8^1/6exp(( i(-π+2πk)/6)

I don't understand where this minus has come from - is the solution wrong or am I missing something?

You could always use
$$
w_k = r^{1/n}\left[\cos\frac{\theta + 2\pi k}{n}+i\sin\frac{\theta + 2\pi k}{n}\right]
$$
where $n$ is the number of roots in your case 6, $\theta$ is the angle, and $k = 0,1,..,n-1$

Then $r = 8$ and $\theta = \pi$ where the principle argument is define as $\theta\in (-\pi,\pi]$
So
$$
w_k = 8^{1/6}\left[\cos\frac{\pi + 2\pi k}{6}+i\sin\frac{\pi + 2\pi k}{6}\right] = 8^{1/6}\exp\left[i\frac{\pi + 2\pi k}{6}\right]
$$

So how could $\pi$ be negative? Simply define the the principle argument to be $\theta\in [-\pi,\pi)$.

And sometimes it will be defined be excluding $\pi$ all together like $\theta\in (-\pi,\pi)$

 

[jedishrfu]

what minus sign? I see the -8+0i which you've handled by using pi for theta.

your answer looks right as it is.

 

[31415]

ah i see when I take the principle arguments of each root then the answers agree - thanks all. Still don't understand why he decided to do it like that but I guess it doesn't matter :)

 

[SammyS]

You're supposed to find the 6 sixth roots of -8 .

 

[31415]

You're supposed to find the 6 sixth roots of -8 .

yeah I know that - ... e^i*pi = -1

8 e^i*pi = -8

e^2*pi*k =1 for k: 0,1,2,3...

 

[jedishrfu]

and when you plot them in the complex plane it will look like a six evenly spaced slices of a pizza pi with a centerline about the imaginary axis.

 

[Joffan]

And of course you can write $\sqrt{2}$ in place of $8^{1/6}$ in your answer.