How exactly do you take the complex conjugate of a term with an imaginary unit in the exponent? (Example below.)

Question

mendicant_bias · Answer

If I was trying to take the complex conjugate of an expression like$$e^{\alpha + i \beta}$$How would I get to the result? I see countless examples of the complex conjugate taken on the web where there is simply a single term in the exponent, like $$e^{i \beta}$$where the complex conjugate is $$e^{-i \beta}$$
Which makes sense enough, but how would I arrive at the result for the first exponent? I'm really hesitant to just mirror the rules applied to a normal complex number like taking the complex conjugate of a+bi, *what rules do I follow* by to get to the result when it's a more complex term than the second? 

@agent0smith

mendicant_bias · Answer

(maybe?) figured it out here: http://www.math.pitt.edu/~sparling/23012/23012complex1/node13.html Let's see.

mendicant_bias · Answer

Oh, whoops. Wrong section.

agent0smith · Answer

Well $$\large e^{a+ib} = e^a*e^{ib}$$ so I'd think you just change the sign on the i term to get the conjugate.

kainui · Answer

There are a few ways you could do this depending on what sorta thing you want. Personally I can see a proof and be dissatisfied so I'd rather feel convinced in a way that makes it common sense to me.

So, that being said, one semi lame way that might convince you is:

$$(e^{a+bi})^* = \left(\sum_{n=0}^\infty \frac{(a+bi)^n}{n!} \right)^* =\sum_{n=0}^\infty \frac{(a-bi)^n}{n!}  = e^{a-bi}$$

But maybe something else is better idk how you feel about this one haha.

anonymous · Answer

Why not just use Euler's formula? (Maybe it's listed in the link above, I didn't check.)

If you're okay with $e^{it}=\cos t+i\sin t$, then
$$\begin{align*}\overline{e^{x+iy}}&=\overline{e^x}\overline{e^{iy}}\[1ex]&=e^x\overline{(\cos y+i\sin y)}\[1ex]&=e^x\left(\overline{\cos y}+\overline{i\sin y}\right)\[1ex]&=e^x(\cos y-i\sin y)\[1ex]&=e^xe^{-iy}\[1ex]&=e^{x-iy}\end{align*}$$