# How to obtain $\sqrt{5}$ (without a calculator) using complex numbers & roots of unity?

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!

• It wouldn't take much of this to clean that up a little. Commented Jul 11, 2022 at 1:25
• $\sqrt 5$ is not a fifth root of unity. Commented Jul 11, 2022 at 1:28
• @DougM I don't think this was ever claimed in the question. Commented Jul 11, 2022 at 1:34
• 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 $\sqrt5$. Commented Jul 11, 2022 at 1:35
• @JohnDouma Thanks for that link, I was unsure how to use MathJax :) Commented Jul 11, 2022 at 1:52

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 $$\,u-v = \sqrt{5}\,$$.

• 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 $\,\text{Re}(\alpha) \gt 0 \gt \text{Re}(\alpha^2)\,$.
– dxiv
Commented Jul 12, 2022 at 0:15
• Thanks for this explanation! Just checking, I'm still a bit confused about how I should exactly prove that 𝑢 > 𝑣 ? Commented Jul 12, 2022 at 0:40
• @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 $u \gt v$.
– dxiv
Commented Jul 12, 2022 at 1:00
• That's makes sense, thanks a lot! Commented Jul 12, 2022 at 1:10

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

$$z^4+z^3+z^2 +z+ 1$$ 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 $$a$$

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

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

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 $$\cos(2x)=\cos(3x)$$.)

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