5
$\begingroup$

For context, here is the entire question:

15. (a) Find the fifth roots of unity in exponential form.
$\hspace{14.5pt}$(b) Let $\alpha$ be the complex fifth root of unity with the smallest positive argument, and suppose that $u = \alpha + \alpha^4$ and $v = \alpha^2 + \alpha^3$,
$\hspace{30pt}$(i) Find the values of $u + v$ and $u - v$.

Particularly, I am struggling with the second part of (b)(i) where the answer is $\sqrt{5}$ for $u-v$. However, I cannot fathom how this answer is obtained without using a calculator.

What I have done so far: $u-v = \alpha + \alpha^4 - \alpha^2 - \alpha^3$, then through substitution of $\alpha = e^{i(2\pi/5)}$, I eventually obtained $u - v = 2\cos(2\pi/5) - 2\cos(4\pi/5)$.

How do I go further from here to calculate √5, or is it a distinct approach to the question entirely?

Any help greatly appreciated!

$\endgroup$
7
  • $\begingroup$ It wouldn't take much of this to clean that up a little. $\endgroup$
    – John Douma
    Commented Jul 11, 2022 at 1:25
  • 4
    $\begingroup$ $\sqrt 5$ is not a fifth root of unity. $\endgroup$
    – user317176
    Commented Jul 11, 2022 at 1:28
  • 4
    $\begingroup$ @DougM I don't think this was ever claimed in the question. $\endgroup$ Commented Jul 11, 2022 at 1:34
  • $\begingroup$ I'm lost, I mean if you look at $z^2-5$ ok. You need a second root of unity (sort of). But what then? You still need $\sqrt5$. $\endgroup$ Commented Jul 11, 2022 at 1:35
  • $\begingroup$ @JohnDouma Thanks for that link, I was unsure how to use MathJax :) $\endgroup$
    – qwerty
    Commented Jul 11, 2022 at 1:52

3 Answers 3

2
$\begingroup$

Hint: $\,0 = \alpha^5 -1 = (\alpha-1)(\alpha^4+\alpha^3+\alpha^2+\alpha+1) \implies \alpha^4+\alpha^3+\alpha^2+\alpha+1 = 0\,$. Then:

  • $u + v = \alpha + \alpha^2 + \alpha^3 + \alpha^4 = -1$

  • $uv = \alpha^3+\alpha^4+\alpha^6+\alpha^7 = \alpha^3+\alpha^4+\alpha+\alpha^2 = -1$

Knowing their sum and product, it follows that $\,u,v\,$ are the roots of the quadratic $\,t^2 + t - 1 = 0\,$, so $\,u,v = \frac{-1 \pm \sqrt{5}}{2}\,$, then $\,|u-v| = \sqrt{5}\,$. What's left to prove is that $\,u \gt v\,$, so $\,u-v = \sqrt{5}\,$.

$\endgroup$
4
  • $\begingroup$ P.S. The "what's left to prove" part is actually covered in OP's attempt described in the question, which can be paraphrased as $\,\text{Re}(\alpha) \gt 0 \gt \text{Re}(\alpha^2)\,$. $\endgroup$
    – dxiv
    Commented Jul 12, 2022 at 0:15
  • $\begingroup$ Thanks for this explanation! Just checking, I'm still a bit confused about how I should exactly prove that 𝑢 > 𝑣 ? $\endgroup$
    – qwerty
    Commented Jul 12, 2022 at 0:40
  • 1
    $\begingroup$ @tktk You found that $u - v = 2\cos(2\pi/5) - 2\cos(4\pi/5)$. Since $0 \lt 2\pi/5 \lt \pi/2 \lt 4\pi/5 \lt \pi$ it follows that $2 \cos(2\pi/5) \gt 0$ and $2 \cos(4\pi/5) \lt 0$ so their difference is positive. But their difference is $u-v$, so $u-v \gt 0$ which is equivalent to $u \gt v$. $\endgroup$
    – dxiv
    Commented Jul 12, 2022 at 1:00
  • 1
    $\begingroup$ That's makes sense, thanks a lot! $\endgroup$
    – qwerty
    Commented Jul 12, 2022 at 1:10
2
$\begingroup$

$z^5 - 1 = 0\\ (z-1)(z^4+z^3+z^2 + z+1)=0$

$z^4+z^3+z^2 +z+ 1$ might not be immediately obvious how to factor, but it is a symmetric polynomial so we can do this.

$z^2(z^2 + \frac 1{z^2} + z+ \frac {1}{z} + 1)$
let $s = z+ \frac 1z$
$s^2 = z^2 + 2 + \frac {1}{z^2}$
$s^2 + s - 1 = 0$
$s = \frac {-1\pm\sqrt{5}}{2}$

There is a shortcut from here to the end, and the next few lines can be bypassed.

$z + \frac 1z = \frac {-1\pm\sqrt{5}}{2} $
$z^2 + \frac {1 \pm \sqrt 5}{2}z + 1 = 0$
$z = \frac {-1\pm\sqrt{5}}{4} \pm \sqrt{\frac {5 \pm \sqrt 5}{8}}i$

or $z = \cos(\frac {2n\pi}{5})+i\sin(\frac {2n\pi}{5})$

The root with the smallest argument is $a = \frac {-1 + \sqrt{5}}{4} + \sqrt{\frac {5 + \sqrt 5}{8}}i = \cos(\frac {2\pi}{5})+i\sin(\frac {2\pi}{5}) = e^{\frac {2\pi}5 i} $

$u = a+a^4 = e^{\frac {2\pi}5 i}+ e^{\frac {8\pi}5 i} = e^{\frac {2\pi}5 i}+e^{\frac {-2\pi}5 i} = 2\text{ Re} (a) = \frac{-1+\sqrt 5}{2}$

$v= a^2+a^3 = e^{\frac {4\pi}5 i}+ e^{\frac {6\pi}5 i} = e^{\frac {4\pi}5 i}+e^{\frac {-4\pi}5 i} = 2\text{ Re} (a^2) = \frac {-1-\sqrt 5}{2}$

Since we only needed the real part of $a$ we could save ourselves a trip through the quadratic formula to find the imaginary part of $a$

$u+v = -1$
$u-v = \sqrt 5$

$\endgroup$
2
  • $\begingroup$ Thank you! Could you clarify with me please how you obtained the subsequent line after letting 𝑠=𝑧+1/𝑧 ? $\endgroup$
    – qwerty
    Commented Jul 11, 2022 at 2:22
  • $\begingroup$ $s^2 = (z + \frac 1z)^2 = z^2 + 2 z\frac 1z + \frac {1}{z^2} = z^2+2+ \frac 1z^2$ Then rearranging $z^2+\frac 1z^2 + z + \frac 1z + 1 = z^2+2+\frac 1z^2 + z + \frac 1z + 1-2= s^2+s-1$ $\endgroup$
    – user317176
    Commented Jul 11, 2022 at 2:26
1
$\begingroup$

The cosine (or sine) of a rational multiple of $\pi$ is always algebraic. Often a simple way to express such a number as a polynomial root is using the Chebyshev polynomials.

Here the second and third Chebyshev polynomials of the first kind give the identities

$$ \cos(2x) = 2 \cos^2 x - 1 $$ $$ \cos(3x) = 4 \cos^3 x - 3 \cos x $$

If $x = \frac{2\pi}{5}$, then we know $\cos(2x) = \cos \frac{4\pi}{5} = \cos \frac{6\pi}{5} = \cos(3x)$, so $t = \cos \frac{2\pi}{5}$ is a solution to

$$ 2 t^2 - 1 = 4 t^3 - 3 t $$ $$ 4 t^3 - 2 t^2 - 3t + 1 = 0 $$

$t=1$ is one obvious root, so the cubic is factored:

$$ (t-1)(4t^2 + 2t - 1) = 0 $$ $$ 4(t-1)\left(t-\frac{-1+\sqrt{5}}{4}\right)\left(t-\frac{-1-\sqrt{5}}{4}\right) = 0 $$

Since $0 < \cos \frac{2\pi}{5} < 1$, we must have

$$ \cos \frac{2\pi}{5} = \frac{-1+\sqrt{5}}{4} $$

Then the rest is just plugging in and computations.

$$ \cos \frac{4\pi}{5} = 2 \cos^2 \frac{2\pi}{5} - 1 = \frac{-1-\sqrt{5}}{4} $$

(This makes sense since $x=\frac{4\pi}{5}$ is another value where $\cos(2x)=\cos(3x)$.)

$$ 2\cos \frac{2\pi}{5} - 2\cos \frac{4\pi}{5} = \sqrt{5} $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .