Is Euler's Formula Enough to Prove Trigonometry in Complex Analysis?

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DEMJ
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Homework Statement


Show that

[tex]\overline{e^{i\theta}} = e^{-i\theta}[/tex]


Homework Equations





The Attempt at a Solution



So I what's going through my mind is that the problem above is pretty much the same as saying [tex]\bar{z} = z^{-1}[/tex]

Then to prove it is all I need to say is that since [tex]\overline{e^{i\theta}} = (cos\theta - isin\theta)[/tex] and [tex]e^{-i\theta} = (cos\theta - isin\theta)[/tex]

so then they are equal. Is this sufficient or am I totally under thinking it?
 
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equal is equal. That's a silly problem. maybe show that bar above [tex]cos \theta + i sin \theta[/tex] just to be on the safe side.
 
ignore my last post, do [tex]e^{-i\theta} = cos( -\theta) + i sin (-\theta)[/tex] and take it from there.

The reason being that you want to apply any factors in the exponent to [tex]\theta[/tex] rather than to i.
 
my opinion is to use trigonometry to prove exp functions and try to use reverse; i.e exp functions to prove trigonometry in complex analysis. that's my suggestions.
esp. Euler formula