# Thread: Multiplying imaginary factors of a polynomial

1. ## [Solved] Multiplying imaginary factors of a polynomial

The problem:
$(x - e^{\frac{i\pi}{4}})(x - e^{\frac{-i\pi}{4}})$

My attempt:
I used the distributive rule to expand it, and got:

$x^2 - x(e^{\frac{i\pi}{4}} + e^{\frac{-i\pi}{4}}) + 1$

I recognise that $cos(\frac{\pi}{4}) = \frac{1}{2}(e^{\frac{i\pi}{4}} + e^{\frac{-i\pi}{4}})$

So I get $x^2 - x(2cos(\frac{\pi}{4})) + 1$
= $x^2 - \frac{2x}{\sqrt{2}} + 1$
= $x^2 - \sqrt{2}x + 1$

I'm unsure if this is correct, since I don't have a solution. Any assistance would be greatly appreciated.

2. Originally Posted by Glitch
The problem:
$(x - e^{\frac{\pi}{4}})(x - e^{\frac{-\pi}{4}})$

My attempt:
I used the distributive rule to expand it, and got:

$x^2 - x(e^{\frac{\pi}{4}} + e^{\frac{-\pi}{4}}) + 1$

I recognise that $cos(\frac{\pi}{4}) = \frac{1}{2}(e^{\frac{\pi}{4}} + e^{\frac{-\pi}{4}})$

So I get $x^2 - x(2cos(\frac{\pi}{4})) + 1$
= $x^2 - \frac{2x}{\sqrt{2}} + 1$
= $x^2 - \sqrt{2}x + 1$

I'm unsure if this is correct, since I don't have a solution. Any assistance would be greatly appreciated.
$\cos x = \frac12(e^{ix} + e^{-ix})$ (notice the i's in the exponents). So either the problem should read $(x - e^{\frac{\pi i}{4}})(x - e^{\frac{-\pi i}{4}})$, or the solution should have cosh instead of cos. If the problem should have had i's in the exponents then the solution is correct.

3. Whoops, yeah, there's supposed to be 'i'. Thanks.