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I'm currently using past papers to revise for January exams, and I've found a bit of a problem with something I thought I was okay with.

## Homework Statement

The position at time t of a particle undergoing damped oscillations is given by:

[tex]x = 2e^{-t}\sin(3t)[/tex].

Express this in terms of a single complex exponential.

Hence evaluate the particle's velocity, [tex]v = \frac{dy}{dx}[/tex]

## Homework Equations

[tex]e^{i \theta} = cos(\theta) + isin(\theta)[/tex]

Chain Rule: [tex]\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}[/tex]

## The Attempt at a Solution

[tex]x = 2e^{-t}sin(3t)[/tex]

[tex]sin(3t) = \text{Im} \left[e^{i3t} \right][/tex]

[tex]x = \text{Im} \left[2e^{-t}\cdot e^{i3t} \right][/tex]

[tex]= 2 \text{Im} \left[e^{-t} \cdot e^{i3t} \right][/tex]

[tex]= 2 \text{Im} \left[e^{-t + i3t} \right][/tex]

[tex]= 2 \text{Im} \left[ e^{(-1 + 3i)t} \right][/tex]

[tex]v = \frac{dx}{dt}[/tex]

[tex]= 2 \text{Im} \left[(-1 + 3i) e^{-t} (cos(3t) + isin(3t)) \right][/tex]

[tex]= 2e^{-t} (-cos(3t) - 3sin(3t))[/tex]

Thats the answer I get by following the only example I have of this from my lecture notes. However, that example was concerned with the real part of the complex exponential.

The baseline solutions to the paper give the following answer:

[tex]v = 2e^{-t}(3cos(3t) - sin(3t))[/tex]

Does anyone know what I've done wrong, or missed?

Thanks.