Solving the following equation with imaginary roots

Question

Okay so firstly I would like to put the question in front of you.

Solve: $^8C_0x^8 + ^8C_2x^6 + ^8C_4x^4 + ^8C_6x^2 + 1 = 0 $

Now it is evident that the equation will only have complex roots.

I wrote the binomial expansions for $ (1+x) ^8 $ and $(1-x)^8$ and added them. After doing some simplification, I am left with this equation:

$$ (1+x)^8 + (1-x)^8 = 0 $$

I cannot proceed further.

Now I believe that I would need to use the De Moivre theorem to solve the equation, but I am not sure how to apply it in this case . Am I doing it right or is there a different way of solving it?

score 1 · Accepted Answer · answered Feb 05 '18 at 17:39

1

$$\left(\dfrac{1+x}{1-x}\right)^8=-1=e^{i\pi(2n+1)}$$ where $n$ is any integer

$\dfrac{1+x}{1-x}=e^{i\pi(2n+1)/8}$ where $0\le n\le7$

Now use Componendo and Dividendo and How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?

answered Feb 05 '18 at 17:39

lab bhattacharjee

274,582

@ManishKundu, $$e^{ix}+e^{-ix}=2\cos x,$$ $$e^{ix}-e^{-ix}=2i\sin x,$$ – lab bhattacharjee Feb 05 '18 at 17:53
Ah, fine. So that would further simplify the result? – Manish Kundu Feb 05 '18 at 17:55

Solving the following equation with imaginary roots

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