Math  /  Algebra

QuestionExpress the following in the form z=x+yjz=x+y j a) z=3(cos(5θ)+jsin(5θ))3z=\frac{3}{(\cos (5 \theta)+j \sin (5 \theta))^{3}}

Studdy Solution
Convert the expression to rectangular form z=x+yj z = x + yj using Euler's formula:
ej15θ=cos(15θ)jsin(15θ) e^{-j15\theta} = \cos(15\theta) - j\sin(15\theta)
Thus, the expression becomes:
z=3(cos(15θ)jsin(15θ)) z = 3(\cos(15\theta) - j\sin(15\theta))
Expanding this, we have:
z=3cos(15θ)3jsin(15θ) z = 3\cos(15\theta) - 3j\sin(15\theta)
This is in the form z=x+yj z = x + yj where:
x=3cos(15θ) x = 3\cos(15\theta) y=3sin(15θ) y = -3\sin(15\theta)
The expression in the form z=x+yj z = x + yj is:
z=3cos(15θ)3jsin(15θ) z = 3\cos(15\theta) - 3j\sin(15\theta)

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