I need help proving $\operatorname{Im}(ie^{-2t}(\cos(2t)+i\sin(2t))=e^{-2t}\cos(2t)$

Question

This is an example from my textbook.

$$\operatorname{Im}(ie^{-2t}(\cos(2t)+i\sin(2t))=e^{-2t}\cos(2t)$$

I don't understand why the imaginary part of this expression equals $e^{-2t}\cos(2t)$

Can anyone clarify this?

Answer 1

Recall that for a complex number $z=a+ib$, where $a,b\in\mathbb{R}$, the number $a$ is said to be the real part and $b$ the imaginary part. In your case, $$ ie^{-2t}(\cos(2t)+i\sin(2t)) = ie^{-2t}\cos(2t) + i^2e^{-2t}\sin(2t) = \underbrace{-e^{-2t}\sin(2t)}_{\text{real part}}+i\cdot\underbrace{e^{-2t}\cos(2t)}_{\text{imaginary part}}. $$