Does the imaginary part of a complex exponential function include the sign before it? I am deriving some equations, but encountered a problem.
We know that $$e^{i\theta}=\cos{\theta}+i\sin{\theta}$$
, where $\cos{\theta}$ is the real part, and $\sin{\theta}$ is the imaginary part.
However, for $$e^{-i\theta}=\cos{\theta}-i\sin{\theta}$$
What is the imaginary part?
Is it $\sin{\theta}$? Or $-\sin{\theta}$?

Hint #1: Subsitute $-i \theta$ in the identity $e^{i \theta} = \cos \theta + i \sin \theta$ and note that $\cos (-\theta) = \cos (\theta)$ and $\sin (-\theta) = -\sin (\theta)$.
Hint #2 (using the series expansion for $e^x$)
Since $$e^{i\theta} = i^0 + i^1+\frac {i^2}{2!}+\frac {i^3}{3!}+\frac {i^4}{4!}+\frac {i^5}{5!}+\dots = 1 + i +\frac {1}{2!}+\frac {i}{3!}+\frac {1}{4!}+\frac {i}{5!} +\dots$$ we can separate the real and imaginary parts to get
$$(1 +\frac {1}{2!}+\frac {1}{4!} \dots) = \cos \theta$$ and $$ i(1 +\frac {1}{3!}+\frac {1}{5!} \dots) = i \sin \theta$$ thus $e^{i \theta} = \cos \theta + i \sin \theta$.
Doing the same for $$e^{-i\theta} = i^0 + i^{-1}+\frac {i^{-2}}{2!}+\frac {i^{-3}}{3!}+\frac {i^{-4}}{4!}+\frac {i^{-5}}{5!}\dots = 1 - i +\frac {1}{2!}-\frac {i}{3!}+\frac {1}{4!}-\frac {i}{5!} \dots$$ what do you notice when you separate the real and imaginary parts?

 $$(1 +\frac {1}{2!}+\frac {1}{4!} \dots) = \cos \theta$$ and $$- i(1 +\frac {1}{3!}+\frac {1}{5!} \dots) = -i \sin \theta$$ thus $$e^{-i \theta} = \cos \theta - i \sin \theta$$

