how to simplify this equation

Question

I know that:

$$\frac{e^{i\theta}+e^{-i\theta}}{2}=\cos(\theta)$$

but I wonder is there way to simplify and rewrite equation above without the imaginary part? :

$$\frac{e^{a+i\theta}+e^{-a-i\theta}}{2}$$

Answer 1

We have $e^{i\theta} = \cos(\theta) + i \sin(\theta)$. This gives us that $$\dfrac{e^{a+i\theta} + e^{-a-i\theta}}2 = \dfrac{e^a(\cos(\theta)+i\sin(\theta)) + e^{-a}(\cos(\theta)-i\sin(\theta))}2 = \dfrac{\cos(\theta)(e^a+e^{-a}) + i \sin(\theta)(e^a-e^{-a})}2$$

Answer 2

$$ e^{a+i\theta} + e^{-a-i\theta} = e^{i(\theta-ia)} + e^{-i(\theta-ia)} = 2\cos{(\theta-ia)}. $$ Or you can use hyperbolic functions to write it as $\cosh{(a+i\theta)}$. These are really the same thing.