I am asked to write the number $$z=e^{2\pi i/5}=\mathrm{cis}\left(2\pi/5\right)$$ using only numbers from $\mathbb{Q}$, arithmetic operations and the square root operation.
I thought about using $i=\sqrt{-1}$, but I'm not sure if I can do that, because $-i$ also satisfies the equation.
Taylor expansion almost works, but I will have to use $\pi\notin\mathbb{Q}$: $$e^{\frac{2\pi}{5}i}=\sum_{n=0}^{\infty}\frac{\left(\frac{2\pi}{5}\sqrt{-1}\right){}^{n}}{n!}$$
Any help or direction is appreciated!
On
You could write $$z=\cos(2/5\pi)+i\sin(2/5\pi)=\frac{\sqrt{5}-1}{4}+i\sqrt{1-\left(\frac{\sqrt{5}-1}{4}\right)^2{}}.$$
Then use continued fractions for the square root of $\sqrt{5}$, if square roots are not allowed.
Note: $2/5\pi$ is equivalent to an angle of $72°$ and it is always possible to express sine and cosine of multiples of $3°$ as expressions containing roots and rational numbers.
Let $\theta = 2\pi/5$. We have $z^5 = 1$ and $z \neq 1$, so $z^4 + z^3 + z^2 + z + 1 = 0$.
Since $z^4 = \frac1{z} = z^{-1}$, $z^3 = z^{-2}$, we can rewrite:
$$z+ z^{-1} + z^2 + z^{-2} + 1 =0$$
Now $z+z^{-1} = 2\cos\theta$ and $z^2 + z^{-2} = 2 \cos 2\theta = 2(2 \cos^2 \theta -1)$. Putting $r = \cos \theta$, we have:
$$2 r + 2(2r^2 -1) + 1 =0$$
Then $4r^2 +2r - 1 =0$ and you can find $r$ by the quadratic formula. Of course $\sin \theta = \sqrt{1 - r^2}$. So you are done (write $i = \sqrt{-1}$).