# Maths Help Roots of Complex Numbers (1 Viewer)

#### Lith_30

##### Active Member
Q10 part d and Q11 part d, I am kinda confused as to what to do.

#### Lith_30

##### Active Member
What textbook is this, may I ask?
Cambridge Extension 2 Maths 2020

#### 5uckerberg

##### Well-Known Member
Cambridge Extension 2 Maths 2020
From there we can see that find the values of x in $\bg_white x^{2}+x+2$ which come from $\bg_white \frac{-1\pm{i\sqrt{7}}}{2}$ and here we are going to note that we will test this on $\bg_white \alpha$ because from the question we can immediately identify that the value $\bg_white -\sin(\frac{\pi}{7})+sin(\frac{2\pi}{7})+sin(\frac{3\pi}{7}) > 0$ as such it becomes $\bg_white -sin(\frac{\pi}{7})+sin(\frac{2\pi}{7})+sin(\frac{3\pi}{7})=\frac{\sqrt{7}}{2}$

#### ExtremelyBoredUser

##### Bored Uni Student
how do you do 10c?
from the polynomial in (ii), divide all the terms by w^4 and use the fact $\bg_white z^n + z^{-n} = 2cos(n\theta)$

Here's the working out just in case;

$\bg_white 1 + w + w^2 + w^3 + w^4 + w^5 + w^6 + w^7 + w^8 = 0$
dividing by the middle term.
$\bg_white \frac{1}{w^4} + \frac{1}{w^3}+ \frac{1}{w^2} + \frac{1}{w} + 1 + w + w^2 + w^3 + w^4 = 0$
$\bg_white 2*[cos(4\theta) + cos(3\theta) + cos(2\theta) + cos(\theta)] = -1$
$\bg_white cos\left(\frac{8\pi}{9}\right) + cos\left(\frac{2\pi}{3}\right) + cos\left(\frac{4\pi}{9}\right) + cos\left(\frac{2\pi}{9}\right) = -\frac{1}{2}$
$\bg_white -cos(\frac{\pi}{9}) + cos(\frac{4\pi}{9}) + cos(\frac{2\pi}{9}) = 0$
$\bg_white cos(\frac{4\pi}{9}) + cos(\frac{2\pi}{9}) = cos(\frac{\pi}{9})$
as required

$\bg_white \left(\frac{2\pi}{9}\right)$

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#### Alistruggles

##### Member
from the polynomial in (ii), divide all the terms by w^4 and use the fact $\bg_white z^n + z^{-n} = 2cos(n\theta)$

Here's the working out just in case;

$\bg_white 1 + w + w^2 + w^3 + w^4 + w^5 + w^6 + w^7 + w^8 = 0$
dividing by the middle term.
$\bg_white \frac{1}{w^4} + \frac{1}{w^3}+ \frac{1}{w^2} + \frac{1}{w} + 1 + w + w^2 + w^3 + w^4 = 0$
$\bg_white 2[cos(4\theta) + cos(3\theta) + cos(2\theta) + cos(\theta) = -1$
$\bg_white cos(\frac{8\pi}{9}) + cos(\frac{2\pi}{3}) + cos(\frac{4\pi}{9}) + cos(\frac{2\pi}{9}) = -\frac{1}{2}$
$\bg_white -cos(\frac{\pi}{9}) + cos(\frac{4\pi}{9}) + cos(\frac{2\pi}{9}) = 0$
$\bg_white cos(\frac{4\pi}{9}) + cos(\frac{2\pi}{9}) = cos(\frac{\pi}{9})$
as required
neat thanks a bunch!

#### Lith_30

##### Active Member
from the polynomial in (ii), divide all the terms by w^4 and use the fact $\bg_white z^n + z^{-n} = 2cos(n\theta)$

Here's the working out just in case;

$\bg_white 1 + w + w^2 + w^3 + w^4 + w^5 + w^6 + w^7 + w^8 = 0$
dividing by the middle term.
$\bg_white \frac{1}{w^4} + \frac{1}{w^3}+ \frac{1}{w^2} + \frac{1}{w} + 1 + w + w^2 + w^3 + w^4 = 0$
$\bg_white 2[cos(4\theta) + cos(3\theta) + cos(2\theta) + cos(\theta) = -1$
$\bg_white cos(\frac{8\pi}{9}) + cos(\frac{2\pi}{3}) + cos(\frac{4\pi}{9}) + cos(\frac{2\pi}{9}) = -\frac{1}{2}$
$\bg_white -cos(\frac{\pi}{9}) + cos(\frac{4\pi}{9}) + cos(\frac{2\pi}{9}) = 0$
$\bg_white cos(\frac{4\pi}{9}) + cos(\frac{2\pi}{9}) = cos(\frac{\pi}{9})$
as required
This is more like a cosmetic tip but if you want the brackets to cover the entire fraction use "\left" and "\right", for example "\left(\frac{2\pi}{9}\right)" results in $\bg_white \left(\frac{2\pi}{9}\right)$. So yeah, use it if you want to.

#### ExtremelyBoredUser

##### Bored Uni Student
This is more like a cosmetic tip but if you want the brackets to cover the entire fraction use "\left" and "\right", for example "\left(\frac{2\pi}{9}\right)" results in $\bg_white \left(\frac{2\pi}{9}\right)$. So yeah, use it if you want to.
Thanks!