Multiplying imaginary factors of a polynomial

**Glitch** · August 20th 2010, 06:33 AM

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.

**Opalg** · August 20th 2010, 07:12 AM

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.

**Glitch** · August 20th 2010, 07:13 AM

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

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