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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?

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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, $$ ie^{-2t}(\cos(2t)+i\sin(2t)) = ie^{-2t}\cos(2t) + i^2e^{-2t}\sin(2t) = \underbrace{-e^{-2t}\sin(2t)}_{\text{real part}}+i\cdot\underbrace{e^{-2t}\cos(2t)}_{\text{imaginary part}}. $$

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  • $\begingroup$ But on the RHS the $i$ is gone. Why is that? $\endgroup$ – Boris Grunwald Nov 29 '18 at 9:14
  • $\begingroup$ @BorisGrunwald Recall that $i^2 = -1$. $\endgroup$ – MisterRiemann Nov 29 '18 at 9:15
  • $\begingroup$ Thanks, I get it now :) $\endgroup$ – Boris Grunwald Nov 29 '18 at 9:17

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