From Euler’s relation eiθ = cosθ + isinθ,(a) find the geometric representation of e-iθ, (b) express cosθ in terms of exponentials, (c) express isinθ in terms of exponentials.

Answer:(a) The graphic representation is in the attached figure.(b) [tex]\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}[/tex].(c) [tex]\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}[/tex].Step-by-step explanation:(a) Given a complex number [tex]e^{i\theta}[/tex] we know, from Euler's formula that [tex]e^{i\theta} = \cos(\theta)+i\sin(\theta)[/tex]. So, it is not difficult to notice that [tex]|e^{i\theta}|^2 = \cos^2(\theta)+\sin^2(\theta) =1[/tex]so it is on the unit circumference. Also, notice that the Cartesian representation of the complex number is [tex] (\cos(\theta), \sin(\theta))[/tex]. Now, [tex]e^{-i\theta} = \cos(\theta)+i\sin(-\theta) = \cos(\theta)-i\sin(\theta)[/tex].Notice that [tex]e^{-i\theta}[/tex] has the same modulus that [tex]e^{i\theta}[/tex], so it is on the unit circumference. Beside, its Cartesian representation is [tex](\cos(\theta), -\sin(\theta))[/tex].So, the points [tex] (\cos(\theta), \sin(\theta))[/tex] and [tex] (\cos(\theta), -\sin(\theta))[/tex] are symmetric with respect to the X-axis. All this can be checked in the attached figure.(b) Notice that [tex]e^{i\theta} + e^{-i\theta} = \cos(\theta)+i\sin(\theta) + \cos(\theta)-i\sin(\theta) = 2\cos(\theta)[/tex]Then, [tex]\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}[/tex].(c) Notice that[tex]e^{i\theta} - e^{-i\theta} = \cos(\theta)+i\sin(\theta) - \cos(\theta)+i\sin(\theta) = 2i\sin(\theta)[/tex]Then, [tex]\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}[/tex].