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

[DEMJ]

Homework Statement

Show that

\overline{e^{i\theta}} = e^{-i\theta}

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 \bar{z} = z^{-1}

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

so then they are equal. Is this sufficient or am I totally under thinking it?

 

[aPhilosopher]

equal is equal. That's a silly problem. maybe show that bar above cos \theta + i sin \theta just to be on the safe side.

 

[rock.freak667]

Well that is how I would do it.

 

[aPhilosopher]

ignore my last post, do e^{-i\theta} = cos( -\theta) + i sin (-\theta) and take it from there.

The reason being that you want to apply any factors in the exponent to \theta rather than to i.

 

[NJunJie]

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