# Expressing a vector in the exponential form

All below

## Relevant Equations:

All below   I managed to expand a general expression from the alternatives that would leave me to the answer, that is:
I will receive the alternatives like above, so i find the equation: C = -sina, P = cosa

So reducing B: R: Reducing D: R: Is this right?

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anuttarasammyak
Gold Member
I prefer simply the way
(b) ##\ \ Re(e^{i(\omega t-\pi/3)}-e^{i\omega t})=Re(e^{i\omega t}(e^{-i\pi/3}-1))=Re(e^{i\omega t} e^{-i2\pi/3})##

(d) ##\ \ Re(e^{i\omega t-i \pi/2}-2e^{i(\omega t - \pi/4 )}+e^{i\omega t})##

Last edited:
• LCSphysicist and etotheipi
vela
Staff Emeritus
Homework Helper
The idea is to use the fact that ##\cos \alpha = {\rm Re}(e^{i\alpha})##, then simplify the complex expression.

One technique you can use is
\begin{align*}
e^{i\theta} + 1 &= e^{i\theta/2}(e^{i\theta/2} + e^{-i\theta/2}) = e^{i\theta/2}[2 \cos (\theta/2)] \\
e^{i\theta} - 1 &= e^{i\theta/2}(e^{i\theta/2} - e^{-i\theta/2}) = e^{i\theta/2}[2i \sin (\theta/2)]
\end{align*} or some variation.

• • LCSphysicist and etotheipi