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A level FP2 question I got stuck on. Please help!

Show that $$(1+e^{ix})(1+e^{-ix}) = 2 + 2\cos x$$

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closed as off-topic by Semiclassical, Stefan Mesken, Henrik, Davide Giraudo, SBareS Dec 6 '16 at 0:04

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    $\begingroup$ Just multiply it out and use Euler's formula. $\endgroup$ – copper.hat Dec 5 '16 at 19:49
  • $\begingroup$ well e^i + 1 = 0, so i can kinda see where the two comes from, but i dont know about the 2cosx $\endgroup$ – Alex Robinson Dec 5 '16 at 19:49
  • $\begingroup$ @copper.hat Please can you show me. I'm not too sure on how to use Euler's formula sorry. $\endgroup$ – Edward Burfield Dec 5 '16 at 19:50
  • $\begingroup$ What do you mean? Euler's formula just says $e^{it} = \cos t +i \sin t$. Multiply the above and expand the remaining exponential terms with this formula and that will give the result. $\endgroup$ – copper.hat Dec 5 '16 at 19:51
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$(1+e^{ix})(1+e^{-ix})=1+e^{-ix}+e^{ix}+1$

$=2+e^{-ix}+e^{ix}$ $=2+2\cfrac{e^{-ix}+e^{ix}}{2}$ $=2+2\cos(x)$

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  • $\begingroup$ Thank you very much that's perfect :) $\endgroup$ – Edward Burfield Dec 5 '16 at 19:52
  • $\begingroup$ 2 minutes sorry wont let me $\endgroup$ – Edward Burfield Dec 5 '16 at 20:00
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The Euler formula states that : $$e^{ix}=cos(x)+isin(x)$$ And : $$e^{-ix}=cos(x)-isin(x)$$ Also remember that : $$sin^2(x)+cos^2(x)=1$$

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$$z \bar{z}=||z||^2$$

If $z=1+e^{ix}$ then $\bar{z}=1+e^{-ix}$

$$(1+e^{ix})(1+e^{-ix})=||1+e^{ix}||^2=(1+\cos{x})^2+\sin^2{x}=2+2\cos{x}$$

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