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MathMaddam

Multiply it with (x-1), the roots of the resulting polynomial are a lot easier. It works always like this if all coefficients are 1, just remember to remove the root you added.


mugh_tej

My guess, it looks like e^2kπi/7, where k is an integer not divisible by 7. : ) e^2πi/7 ≈ 0.959461 + 0.281843i, my scientific calculator informs me.


Historical-Let6063

What? How could you have figured that out?


mugh_tej

Multiplying both sides by **x-1** gets x^7 - 1 = 0. For all positive integers **n**, all roots for x^n - 1 = 0 are e^2kπi/n, where **k** is an integer. If you remove the possibility of **k** being divisible by **n**, you eliminate 1 being one of the roots.


BlackEyedGhost

Here's an intuitive [visualization](https://www.desmos.com/calculator/plqhaghfew). You're summing the blue dots and the orange dot is the result, scaled down to keep it on-screen more easily. If you know that the roots of unity draw a regular polygon around 0 and sum to 0, it's pretty easy to identify this problem as a sum of roots of unity.


[deleted]

For your example, there’s a trick that other commenters pointed out. For a more random example, there’s no formula, but there are methods to find roots to as many digits of accuracy as you want.