# Need help with roots of unity. (1 Viewer)

#### Run hard@thehsc

##### Active Member
How would you do part 5 of this question? I have done questions like these before, but somehow can't find the method I used before. Thanks!

#### ExtremelyBoredUser

##### what
View attachment 34434
I did part 2 but I need help to confirm weather my method was correct... can someone show how this problem can be approached??
$\bg_white z^5 - 1 = 0$ has roots $\bg_white e^{-\frac{4\pi}{5}}, e^{-\frac{2\pi}{5}}, e^{\frac{4\pi}{5}}, e^{-\frac{2\pi}{5}}, 1$

let $\bg_white e^{\frac{2\pi}{5}$ be w, $\bg_white e^{-\frac{2\pi}{5}$ be w conjugate and j for the other root etc.

$\bg_white z^5 - 1 = (z - 1)( z - w)(z-w_{2})(z-j)(z-j_2)$
dividing z^5 -1 with z -1 gets the RHS

$\bg_white z^4 + z^3 + z^2 + 1 = ( z - w)(z-w_{2})(z-j)(z-j_2)$
$\bg_white z^4 + z^3 + z^2 + 1 = (z^2 - z(w+w_{2}) + ww_{2})(z^2 - z(j+j_{2}) + jj_{2})$

$\bg_white w + w_{2} = 2Re(w) = 2(\cos\left(w)\right) = 2\cos\left(\frac{2\pi}{5}\right)$
$\bg_white w \times w_{2} = |w|^2 = \cos^2({\frac{2\pi}{5}) + \sin^2({\frac{2\pi}{5}) =1$
same with the j root.

and then you get the result as required.

edit: don't know how to do conjugates so if anyone reading could show how to latex it i would appreciate
edit 2: fixed a typo

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#### Vipul_K

##### New Member
$\bg_white z^5 - 1 = 0$ has roots $\bg_white e^{\frac{2\pi}{5}}, e^{-\frac{2\pi}{5}}, e^{\frac{4\pi}{5}}, e^{-\frac{2\pi}{5}}, 1$

let $\bg_white e^{\frac{2\pi}{5}$ be w, $\bg_white e^{-\frac{2\pi}{5}$ be w conjugate and j for the other root etc.

$\bg_white z^5 - 1 = (z - 1)( z - w)(z-w_{2})(z-j)(z-j_2)$
dividing z^5 -1 with z -1 gets the RHS

$\bg_white z^4 + z^3 + z^2 + 1 = ( z - w)(z-w_{2})(z-j)(z-j_2)$
$\bg_white z^4 + z^3 + z^2 + 1 = (z^2 - z(w+w_{2}) + ww_{2})(z^2 - z(j+j_{2}) + jj_{2})$

$\bg_white w + w_{2} = 2Re(w) = 2\cos\left(w)\right = 2\cos(\frac{2\pi}{5})$
$\bg_white w \times w_{2} = |w|^2 = \cos^2({\frac{2\pi}{5}) + \sin^2({\frac{2\pi}{5}) =1$
same with the j root.

and then you get the result as required.

edit: don't know how to do conjugates so if anyone reading could show how to latex it i would appreciate
I understand this but can u please expand on how dividing z^5-1 with z-1 gets the rhs? Thanks!

#### ExtremelyBoredUser

##### what
I understand this but can u please expand on how dividing z^5-1 with z-1 gets the rhs? Thanks!
You can prove it through polynomial division. If you were to manually divide both polynomials, you would get that term.

Or

You could factorise z^5 - 1 into two polynomials

$\bg_white (z-1)(z^4+z^3+z^2+1)$

as you can see, all the terms in the polynomial are multiplied by z first, and then the regular polynomial is multiplied by -1 so all the terms between z^5 and -1 would be cancelled out

$\bg_white z^5 + z^4 + z^3 + z^2 + z - z^4 - z^3 - z^2 - z - 1$
which is
$\bg_white z^5 - 1$

so when you divide by z - 1.

$\bg_white \frac{z^5 -1}{z-1} = \frac{(z-1)(z^4+z^3+z^2+1)}{z-1} = (z^4+z^3+z^2+1)$

*z can not equal to 1 btw

edit: i forgot the - z

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#### Vipul_K

##### New Member
You can prove it through polynomial division. If you were to manually divide both polynomials, you would get that term.

Or

You could factorise z^5 - 1 into two polynomials

$\bg_white (z-1)(z^4+z^3+z^2+1)$

as you can see, all the terms in the polynomial are multiplied by z first, and then the regular polynomial is multiplied by -1 so all the terms between z^5 and -1 would be cancelled out

$\bg_white z^5 + z^4 + z^3 + z^2 + z - z^4 - z^3 - z^2 - 1$
which is
$\bg_white z^5 - 1$

so when you divide by z - 1.

$\bg_white \frac{z^5 -1}{z-1} = \frac{(z-1)(z^4+z^3+z^2+1)}{z-1} = (z^4+z^3+z^2+1)$
Ahhhhh okkk makes sense now thank you!!

#### 5uckerberg

##### Active Member
Ahhhhh okkk makes sense now thank you!!
Note there is a typo there can you spot his mistake.

#### Run hard@thehsc

##### Active Member
Also can someone show me how to do part (5)?

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#### CM_Tutor

##### Moderator
Moderator
edit: don't know how to do conjugates so if anyone reading could show how to latex it i would appreciate
You can use \bar{z} to get $\bg_white \bar{z}$ but I prefer \overline{} so I can get \overline{z} is $\bg_white \overline{z}$ and \overline{a+ib} is $\bg_white \overline{a+ib}$ because \bar{a+ib}, giving $\bg_white \bar{a+ib}$ looks ridiculous.

#### CM_Tutor

##### Moderator
Moderator
I understand this but can u please expand on how dividing z^5-1 with z-1 gets the rhs? Thanks!
An easy way to establish that

$\bg_white \cfrac{z^5 - 1}{z - 1} = 1 + z + z^2 + z^3 + z^4$

is to sum the GP on the RHS.

$\bg_white 1 + z + z^2 + z^3 + z^4$ is a GP of five terms ($\bg_white n = 5$) with first term 1 ($\bg_white a = 1$) and common ratio of $\bg_white z$ ($\bg_white r = z$) which sums to

\bg_white \begin{align*} S_n = \cfrac{a\left(r^n - 1\right)}{r - 1} &= \cfrac{1 \times \left(z^5 - 1\right)}{z - 1} \\ 1 + z + z^2 + z^3 + z^4 &= \cfrac{z^5 - 1}{z - 1} \quad \text{as required} \end{align*}

#### ExtremelyBoredUser

##### what
You can use \bar{z} to get $\bg_white \bar{z}$ but I prefer \overline{} so I can get \overline{z} is $\bg_white \overline{z}$ and \overline{a+ib} is $\bg_white \overline{a+ib}$ because \bar{a+ib}, giving $\bg_white \bar{a+ib}$ looks ridiculous.
$\bg_white \overline{w} + i$

Thanks!

#### ExtremelyBoredUser

##### what
@5uckerberg was it his roots. One his roots were supposed to be e^i(-4pi/5) I believe.... @ExtremelyBoredUser thanks btw!
Yep when I was copying pasting the roots, I must've forgotten to change the 2pi to 4pi. Thanks for that.

#### ExtremelyBoredUser

##### what
How would you do part 5 of this question? I have done questions like these before, but somehow can't find the method I used before. Thanks!
View attachment 34431
Not the solution to (v) but just another method for these type of questions;

Here's what I did.

$\bg_white \sin(A)\sin(B) = \frac{1}{2} \times [\cos(A-B) - \cos(A+B)]$
$\bg_white \sin\left(\frac{\pi}{5}\right)\sin\left(\frac{2\pi}{5}\right) = \frac{1}{2} \times [\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{3\pi}{5}\right)]$
$\bg_white \sin\left(\frac{\pi}{5}\right)\sin\left(\frac{2\pi}{5}\right) = \frac{1}{2} \times [-\cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{2\pi}{5}\right)]$

$\bg_white \cos\left(\frac{2\pi}{5}\right) = - \frac{1}{2} - \cos\left(\frac{4\pi}{5}\right)$

Sub this in

$\bg_white \sin\left(\frac{\pi}{5}\right)\sin\left(\frac{2\pi}{5}\right) = \frac{1}{2} \times [ - \frac{1}{2} - 2\cos\left(\frac{4\pi}{5}\right)$
$\bg_white \sin\left(\frac{\pi}{5}\right)\sin\left(\frac{2\pi}{5}\right) = - \frac{1}{4} - \cos\left(\frac{4\pi}{5}\right)$

from (iv), the exact value is for $\bg_white \cos\left(\frac{4\pi}{5}\right)$ is $\bg_white -\frac{\sqrt{5} + 1}{4}$

Therefore;
$\bg_white \sin\left(\frac{\pi}{5}\right)\sin\left(\frac{2\pi}{5}\right) = - \frac{1}{4} + \frac{\sqrt{5} + 1}{4}$
$\bg_white \sin\left(\frac{\pi}{5}\right)\sin\left(\frac{2\pi}{5}\right) = \frac{\sqrt{5} }{4}$

as required.

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#### Lith_30

##### Member
I think this might be the solution to part v

We know that $\bg_white z^2-2z\cos\left(\frac{2\pi}{5}\right)+1=\left(z-\text{cis}\left(\frac{2\pi}{5}\right)\right)\left(z-\text{cis}\left(\frac{-2\pi}{5}\right)\right)$

$\bg_white \\\therefore \text{cis}^2\left(\frac{2\pi}{5}\right)-2\text{cis}\left(\frac{2\pi}{5}\right)\cos\left(\frac{2\pi}{5}\right)+1=0\\\\ \text{cis}\left(\frac{4\pi}{5}\right)-2\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)\cos\left(\frac{2\pi}{5}\right)+1=0\\\\ \cos\left(\frac{4\pi}{5}\right)+i\sin\left(\frac{4\pi}{5}\right)-2\cos^2\left(\frac{2\pi}{5}\right)+2i\sin\left(\frac{2\pi}{5}\right)\cos\left(\frac{2\pi}{5}\right)+1=0\\\\ \cos\left(\frac{4\pi}{5}\right)-2\cos^2\left(\frac{2\pi}{5}\right)+1+i\left(\sin\left(\frac{4\pi}{5}\right)+2\sin\left(\frac{2\pi}{5}\right)\cos\left(\frac{2\pi}{5}\right)\right)=0\\\\\text{Equate real component}\\\\ \cos\left(\frac{4\pi}{5}\right)-2\cos^2\left(\frac{2\pi}{5}\right)+1=0\\\\ \cos\left(\frac{4\pi}{5}\right)-2\left(1-\sin^2\left(\frac{2\pi}{5}\right)\right)+1=0\\\\ 2\sin^2\left(\frac{2\pi}{5}\right)=1-\cos\left(\frac{4\pi}{5}\right)\\\\ \sin^2\left(\frac{2\pi}{5}\right)=\frac{1-\cos\left(\frac{4\pi}{5}\right)}{2}$

We can do a similar thing with $\bg_white z^2-2z\cos\left(\frac{4\pi}{5}\right)+1=\left(z-\text{cis}\left(\frac{4\pi}{5}\right)\right)\left(z-\text{cis}\left(\frac{-4\pi}{5}\right)\right)$ but we sub in $\bg_white z=\text{cis}\left(\frac{4\pi}{5}\right)$

and the result should be after equating the real components...

$\bg_white \\\cos\left(\frac{-2\pi}{5}\right)-2\cos^2\left(\frac{4\pi}{5}\right)+1=0\\\\ \cos\left(\frac{2\pi}{5}\right)-2\left(1-\sin^2\left(\frac{4\pi}{5}\right)\right)+1=0\\\\ \sin^2\left(\frac{4\pi}{5}\right)=\frac{1-\cos\left(\frac{2\pi}{5}\right)}{2}\\\\ \sin^2\left(\frac{\pi}{5}\right)=\frac{1-\cos\left(\frac{2\pi}{5}\right)}{2}$

therefore by multiplying the two results we get

$\bg_white \\\sin^2\left(\frac{2\pi}{5}\right)\sin^2\left(\frac{\pi}{5}\right)=\left(\frac{1-\cos\left(\frac{4\pi}{5}\right)}{2}\right)\left(\frac{1-\cos\left(\frac{2\pi}{5}\right)}{2}\right)\\\\\sin^2\left(\frac{2\pi}{5}\right)\sin^2\left(\frac{\pi}{5}\right)=\frac{1}{4}\left(1-\cos\left(\frac{2\pi}{5}\right)-\cos\left(\frac{4\pi}{5}\right)+\cos\left(\frac{2\pi}{5}\right)\cos\left(\frac{4\pi}{5}\right)\right)\\\\\text{using identities from part iii}\\\\ \sin^2\left(\frac{2\pi}{5}\right)\sin^2\left(\frac{\pi}{5}\right)=\frac{1}{4}\left(1-\left(-\frac{1}{2}\right)\frac{1}{4}\right)\\\\ \sin^2\left(\frac{2\pi}{5}\right)\sin^2\left(\frac{\pi}{5}\right)=\frac{1}{4}\cdot\frac{5}{4}\\\\ \sin\left(\frac{2\pi}{5}\right)\sin\left(\frac{\pi}{5}\right)=\pm\frac{\sqrt{5}}{4}\\\\ \text{since} \ \sin\left(\frac{2\pi}{5}\right),\sin\left(\frac{\pi}{5}\right) > 0 \\\\\therefore \sin\left(\frac{2\pi}{5}\right)\sin\left(\frac{\pi}{5}\right)=\frac{\sqrt{5}}{4}$

#### CM_Tutor

##### Moderator
Moderator
If part (iv) has established that

$\bg_white \cos{\cfrac{4\pi}{5}} = -\cfrac{1 + \sqrt{5}}{4}$

then applying the result

$\bg_white \cos{(\pi - \theta)} = -\cos\theta$

shows that

$\bg_white \cos{\cfrac{\pi}{5}} = -\cos{\left(\pi - \cfrac{\pi}{5}\right)} = -\cos{\cfrac{4\pi}{5}} = \cfrac{1 + \sqrt{5}}{4}$

from which we can deduce that

$\bg_white \sin^2{\cfrac{\pi}{5}} = 1 - \cos^2{\cfrac{\pi}{5}} = 1 - \left(\cfrac{1 + \sqrt{5}}{4}\right)^2 = 1 - \cfrac{1 + 2\sqrt{5} + 5}{16} = \cfrac{5 - \sqrt{5}}{8}$

Now, we seek

\bg_white \begin{align*} \sin{\cfrac{\pi}{5}}\sin{\cfrac{2\pi}{5}} &= \sin{\cfrac{\pi}{5}} \times 2\sin{\cfrac{\pi}{5}}\cos{\cfrac{\pi}{5}} \\ &= 2\sin^2{\cfrac{\pi}{5}}\cos{\cfrac{\pi}{5}} \\ &= 2 \times \cfrac{5 - \sqrt{5}}{8} \times \cfrac{1 + \sqrt{5}}{4} \\ &= \cfrac{\left(5 - \sqrt{5}\right)\left(1 + \sqrt{5}\right)}{4^2} \\ &= \cfrac{5 - \sqrt{5} + 5\sqrt{5} - 5}{4^2} \\ &= \cfrac{4\sqrt{5}}{4^2} \\ &= \cfrac{\sqrt{5}}{4} \end{align*}

#### CM_Tutor

##### Moderator
Moderator
The two results that you have derived:

$\bg_white \sin^2\left(\frac{2\pi}{5}\right)=\frac{1-\cos\left(\frac{4\pi}{5}\right)}{2}$
and
$\bg_white \sin^2\left(\frac{\pi}{5}\right)=\frac{1-\cos\left(\frac{2\pi}{5}\right)}{2}$
Can be found without complex numbers:

$\bg_white \cos{2\theta} = 1 - 2\sin^2\theta \quad \implies \quad \sin^2\theta = \cfrac{1 - \cos{2\theta}}{2}$

and then set first $\bg_white \theta = \cfrac{2\pi}{5}$ and then $\bg_white \theta = \cfrac{\pi}{5}$

#### Lith_30

##### Member
The two results that you have derived:

and

Can be found without complex numbers:

$\bg_white \cos{2\theta} = 1 - 2\sin^2\theta \quad \implies \quad \sin^2\theta = \cfrac{1 - \cos{2\theta}}{2}$

and then set first $\bg_white \theta = \cfrac{2\pi}{5}$ and then $\bg_white \theta = \cfrac{\pi}{5}$
Yeah that's true, I was trying to use part ii cause the question asked to.

#### CM_Tutor

##### Moderator
Moderator
Yeah that's true, I was trying to use part ii cause the question asked to.
I understand... I was wondering whether it meant to use part (iv), or about noting that part(iv) was derived from (ii), and so using it is building from part (ii)

#### Lith_30

##### Member
I understand... I was wondering whether it meant to use part (iv), or about noting that part(iv) was derived from (ii), and so using it is building from part (ii)
I think the question would have explicitly stated to use part (iv) if they meant for you to use it. But it I guess the question was kinda building up to that point through all the previous parts, so your way would have been correct too and in timed conditions my method would take way to long for 2 marks.
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