Need help with roots of unity. (1 Viewer)

Run hard@thehsc · Dec 5, 2021

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 · Dec 5, 2021

Vipul_K said:
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??

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

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

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

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

$w + w_{2} = 2Re(w) = 2(\cos\left(w)\right) = 2\cos\left(\frac{2\pi}{5}\right)$
$w \times w_{2} = |w|^2 = \cos^2({\frac{2\pi}{5}) + \sin^2({\frac{2\pi}{5}) =1$
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

Vipul_K · Dec 5, 2021

ExtremelyBoredUser said:
$z^5 - 1 = 0$ has roots $e^{\frac{2\pi}{5}}, e^{-\frac{2\pi}{5}}, e^{\frac{4\pi}{5}}, e^{-\frac{2\pi}{5}}, 1$

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

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

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

$w + w_{2} = 2Re(w) = 2\cos\left(w)\right = 2\cos(\frac{2\pi}{5})$
$w \times w_{2} = |w|^2 = \cos^2({\frac{2\pi}{5}) + \sin^2({\frac{2\pi}{5}) =1$
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 · Dec 5, 2021

Vipul_K said:
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

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

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

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

so when you divide by z - 1.

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

*z can not equal to 1 btw

edit: i forgot the - z

Vipul_K · Dec 5, 2021

ExtremelyBoredUser said:
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

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

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

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

so when you divide by z - 1.

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

Ahhhhh okkk makes sense now thank you!!

5uckerberg · Dec 5, 2021

Vipul_K said:
Ahhhhh okkk makes sense now thank you!!

Note there is a typo there can you spot his mistake.

Run hard@thehsc · Dec 5, 2021

@5uckerberg was it his roots. One his roots were supposed to be e^i(-4pi/5) I believe.... @ExtremelyBoredUser thanks btw!

Run hard@thehsc · Dec 5, 2021

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

CM_Tutor · Dec 5, 2021

ExtremelyBoredUser said:
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 $\bar{z}$ but I prefer \overline{} so I can get \overline{z} is $\overline{z}$ and \overline{a+ib} is $\overline{a+ib}$ because \bar{a+ib}, giving $\bar{a+ib}$ looks ridiculous.

CM_Tutor · Dec 5, 2021

Vipul_K said:
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

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

is to sum the GP on the RHS.

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

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

ExtremelyBoredUser · Dec 5, 2021

CM_Tutor said:
You can use \bar{z} to get $\bar{z}$ but I prefer \overline{} so I can get \overline{z} is $\overline{z}$ and \overline{a+ib} is $\overline{a+ib}$ because \bar{a+ib}, giving $\bar{a+ib}$ looks ridiculous.

$\overline{w} + i$

Thanks!

ExtremelyBoredUser · Dec 5, 2021

Run hard@thehsc said:
@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.

5uckerberg · Dec 5, 2021

Run hard@thehsc said:
@5uckerberg was it his roots. One his roots were supposed to be e^i(-4pi/5) I believe.... @ExtremelyBoredUser thanks btw!

Well he has changed it.

ExtremelyBoredUser · Dec 7, 2021

Run hard@thehsc said:
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.

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

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

Sub this in

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

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

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

as required.

Lith_30 · Dec 7, 2021

I think this might be the solution to part v

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

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

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

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

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

therefore by multiplying the two results we get

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

CM_Tutor · Dec 7, 2021

If part (iv) has established that

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

then applying the result

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

shows that

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

from which we can deduce that

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

Now, we seek

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

CM_Tutor · Dec 7, 2021

The two results that you have derived:

Lith_30 said:
$\sin^2\left(\frac{2\pi}{5}\right)=\frac{1-\cos\left(\frac{4\pi}{5}\right)}{2}$

and

Lith_30 said:
$\sin^2\left(\frac{\pi}{5}\right)=\frac{1-\cos\left(\frac{2\pi}{5}\right)}{2}$

Can be found without complex numbers:

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

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

Lith_30 · Dec 8, 2021

CM_Tutor said:
The two results that you have derived:

and

Can be found without complex numbers:

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

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

Yeah that's true, I was trying to use part ii cause the question asked to.

CM_Tutor · Dec 9, 2021

Lith_30 said:
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 · Dec 9, 2021

CM_Tutor said:
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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