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

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?

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, $$ie^{-2t}(\cos(2t)+i\sin(2t)) = ie^{-2t}\cos(2t) + i^2e^{-2t}\sin(2t) = \underbrace{-e^{-2t}\sin(2t)}_{\text{real part}}+i\cdot\underbrace{e^{-2t}\cos(2t)}_{\text{imaginary part}}.$$
• But on the RHS the $i$ is gone. Why is that? – Boris Grunwald Nov 29 '18 at 9:14
• @BorisGrunwald Recall that $i^2 = -1$. – MisterRiemann Nov 29 '18 at 9:15