# Differentiating Complex Exponentials

Hello 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:

$$x = 2e^{-t}\sin(3t)$$.

Express this in terms of a single complex exponential.

Hence evaluate the particle's velocity, $$v = \frac{dy}{dx}$$

## Homework Equations

$$e^{i \theta} = cos(\theta) + isin(\theta)$$

Chain Rule: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

## The Attempt at a Solution

$$x = 2e^{-t}sin(3t)$$

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

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

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

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

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

$$v = \frac{dx}{dt}$$

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

$$= 2e^{-t} (-cos(3t) - 3sin(3t))$$

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:

$$v = 2e^{-t}(3cos(3t) - sin(3t))$$

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

Thanks.

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tiny-tim
Homework Helper
Hello Matty R! You've taken the real part …

just take the other bits instead! (and chuck away the i)

Hiya tim. Nice to see you again. That makes perfect sense. I've managed to get the same answer as the solutions now, and I was wondering if you'd be so kind as to confirm what I did.

For the real part:

$$2e^{-t} \left[(-1 \times cos(3t)) + (3i \times isin(3t) \right]$$

For the imaginary part:

$$2e^{-t} \left[(3i \times cos(3t)) + (-1 \times isin(3t) \right]$$

Then discard the "$$i$$"s

tiny-tim
Yup! Brilliant. Thank you very much. 