Maths Help Roots of Complex Numbers (1 Viewer)

Lith_30 · Nov 2, 2021

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

tickboom · Nov 2, 2021

This might help you for 10 d):

Lith_30 · Nov 2, 2021

tickboom said:
This might help you for 10 d):

Thankyou!

ExtremelyBoredUser · Nov 2, 2021

Lith_30 said:
Q10 part d and Q11 part d, I am kinda confused as to what to do.
View attachment 33055

What textbook is this, may I ask?

Lith_30 · Nov 2, 2021

ExtremelyBoredUser said:
What textbook is this, may I ask?

Cambridge Extension 2 Maths 2020

5uckerberg · Nov 2, 2021

Lith_30 said:
Cambridge Extension 2 Maths 2020

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

Alistruggles · Nov 13, 2021

Lith_30 said:
Q10 part d and Q11 part d, I am kinda confused as to what to do.
View attachment 33055

how do you do 10c?

ExtremelyBoredUser · Nov 13, 2021

Alistruggles said:
how do you do 10c?

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

Here's the working out just in case;

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

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

Alistruggles · Nov 14, 2021

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

Here's the working out just in case;

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

neat thanks a bunch!

Lith_30 · Nov 14, 2021

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

Here's the working out just in case;

$1 + w + w^2 + w^3 + w^4 + w^5 + w^6 + w^7 + w^8 = 0$
dividing by the middle term.
$\frac{1}{w^4} + \frac{1}{w^3}+ \frac{1}{w^2} + \frac{1}{w} + 1 + w + w^2 + w^3 + w^4 = 0$
$2[cos(4\theta) + cos(3\theta) + cos(2\theta) + cos(\theta) = -1$
$cos(\frac{8\pi}{9}) + cos(\frac{2\pi}{3}) + cos(\frac{4\pi}{9}) + cos(\frac{2\pi}{9}) = -\frac{1}{2}$
$-cos(\frac{\pi}{9}) + cos(\frac{4\pi}{9}) + cos(\frac{2\pi}{9}) = 0$
$cos(\frac{4\pi}{9}) + cos(\frac{2\pi}{9}) = cos(\frac{\pi}{9})$
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 $\left(\frac{2\pi}{9}\right)$ . So yeah, use it if you want to.

ExtremelyBoredUser · Nov 14, 2021

Lith_30 said:
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 $\left(\frac{2\pi}{9}\right)$ . So yeah, use it if you want to.

Thanks!

Maths Help Roots of Complex Numbers (1 Viewer)

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