# Thread: [SOLVED] Roots in Polar Form

1. ## [SOLVED] Roots in Polar Form

Hi

If I am calculating roots in polar form how should I be displaying them?

eg. The seventh roots of -128 are:

$z_{0} = \langle2,\pi/7\rangle$
$z_{1} = \langle2,3\pi/7\rangle$
$z_{2} = \langle2,5\pi/7\rangle$
$z_{3} = \langle2,\pi\rangle = -2$
$z_{4} = \langle2,9\pi/7\rangle$
$z_{5} = \langle2,11\pi/7\rangle$
$z_{6} = \langle2,13\pi/7\rangle$

Do I need to be displaying them like this?

$z_{0} = \langle2,\pi/7\rangle = 2(cos(\pi/7)+isin(\pi/7))$

etc, etc...

Or like in the first example

The reason I ask is because i'm asked to show how they make 3 complex conjugate pairs which I can see are

$z_{0}$ and $z_{6}$
$z_{1}$ and $z_{5}$
$z_{2}$ and $z_{4}$

Thanks

2. Originally Posted by Ian1779
eg. The seventh roots of -128 are:
$z_{0} = \langle2,\pi/7\rangle$
$z_{1} = \langle2,3\pi/7\rangle$
$z_{2} = \langle2,5\pi/7\rangle$
$z_{3} = \langle2,\pi\rangle = -2$
$z_{4} = \langle2,9\pi/7\rangle$
$z_{5} = \langle2,11\pi/7\rangle$
$z_{6} = \langle2,13\pi/7\rangle$

Do I need to be displaying them like this?

$z_{0} = \langle2,\pi/7\rangle = 2(cos(\pi/7)+isin(\pi/7))$
I think that really up to your instructor/textbook

The reason I ask is because i'm asked to show how they make 3 complex conjugate pairs which I can see are
$z_{0}$ and $z_{6}$
$z_{1}$ and $z_{5}$
$z_{2}$ and $z_{4}$
As to the complex conjugate pairs, take note.
$\arg (z) = - \arg \left( {\overline z } \right)$

3. Thanks for the explanation about the arg!!

Another reason I ask is that the follow on question asks me to factorise the polynomial $x^7+128$

I've worked out that one of the roots are $2(cos(\pi/7)+isin(\pi/7))$ and another is $-2$ hence my question about the polar forms as I feel i'm duplicating my work

Thanks

4. Originally Posted by Ian1779
Another reason I ask is that the follow on question asks me to factorise the polynomial $x^7+128$
$x^7+128=(x-z_0)(x-z_1)(x-z_2)(x-z_3)(x-z_4)(x-z_5)(x-z_6)$

5. Thanks