Show $\overline{e^{i\theta}} = e^{-i\theta}$ using trig identities

Question

My book claims we can verify this using "standard trigonometric identities." The notation of the right-hand side is throwing me off. Does that mean the complex conjugate of $e^{i\theta}$ equals its multiplicative inverse, i.e. $\frac{1}{e^{i\theta}}$?

Answer 1

Yes. You have $$ e^{it}=\cos t+i\sin t. $$ And $$ e^{-it}=\frac1{e^{it}}=\frac1{\cos t+i\sin t}=\frac{\cos t-i\sin t}{\cos^2t+\sin^2t}=\cos t-i\sin t=\overline{e^{it}}. $$

Answer 2

Recall that $$ e^{i\theta} = \cos\theta+i\sin\theta, \quad \text{so} \quad \overline{e^{i\theta}} = \cos\theta-i\sin\theta = \cos(-\theta)+i\sin(-\theta) = e^{i(-\theta)} = e^{-i\theta}, $$ since the cosine is even while the sine is odd.