Expressing a vector in the exponential form

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LCSphysicist
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Homework Statement
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Relevant Equations
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Text reading: Express the following in the form z = Re[Ae^(i(ωt+α))]

Printed math expression (b) z = cos(ωt − π/3) − cos ωt.

Printed equation (d): z = sin ωt − 2 cos(ωt − π/4) + cos ωt.

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:

Equation z = A P cos(w t) + C A sin(w t), with a = tan^-1(-C/P), written in curly braces as a set of expressions.
C = -sina, P = cosa

So reducing B:
Expression: (1/2) times square root of 3 times sin(t w), minus (1/2) times cos(t w).


R:
Handwritten-style equation Z = 1 e^{i(wt+2×π/3)}, a complex exponential expression with a phase term of 2π/3.
Reducing D:
Expression: -√2 sin(t w) + sin(t w) - √2 cos(t w) + cos(t w)


R:
Mathematical expression (2 - √2) multiplied by e raised to i(wt - π/4), written in exponential form.


Is this right?
 
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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})##
 
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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.
 
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