Let $z = x + iy$ be a fifth root of unity. Show that the quotient y/x can only take on five distinct values, and determine those five values.

Asked Oct 20 '16 at 08:15

Active Oct 20 '16 at 11:25

Viewed 65 times

It's easy to prove that y/x can only take on 5 different values through the use of the exponential form of z. However that doesn't allow us to give a precise value. For example, for k=1, $z = e^{i\frac{2\pi}{5}} $, so $x/y = \tan(\frac{2\pi}{5}) $.

What I think we need to do is find the cartesian form of those numbers and I have no idea how to go about that.

I could take the route mentioned by this question, but that seems like too much work, which will amount to finding tan of all of those angles:

Find exact value of $\cos (\frac{2\pi}{5})$ using complex numbers.

edited Apr 13 '17 at 12:21

Community

asked Oct 20 '16 at 08:15

Quantaliinuxite

1,557

1

Why so many cartesian forms? – Sarvesh Ravichandran Iyer Oct 20 '16 at 08:16
1

I edited my question, haha. I think my brain bugged out – Quantaliinuxite Oct 20 '16 at 08:21
1

The method given here is the best way to proceed. You have to look for nice relations of $\frac {2 \pi}{5}$.One way to start is to note that $\pi - \frac{4\pi }{5} = \frac{2 \pi}{ 5 * 2}$. Now use the half angle formula on one side, and double angle on the other. – Sarvesh Ravichandran Iyer Oct 20 '16 at 08:28
@астонвіллаолофмэллбэрг Is there really no easier way to go about? The reason I ask is because the full problem seems to imply that we find y/x before any tangents... – Quantaliinuxite Oct 20 '16 at 08:29
I actually am not sure, but I would go about it this way, it's more comfortable. – Sarvesh Ravichandran Iyer Oct 20 '16 at 08:40
1

Strictly speaking, you've answered the question when you observe that $y/x =\cot(2k\pi/5)$, $k=1,2,3,4,5$. The question doesn't explicit ask for an algebraic solution(?) – B. Goddard Oct 20 '16 at 11:41

Let $z = x + iy$ be a fifth root of unity. Show that the quotient y/x can only take on five distinct values, and determine those five values.

0 Answers0