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    z = -j * e^(jx) * sin(x)

    how do you possibly find the magnitude and argument of that?
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    (Original post by Bealzibub)
    z = -j * e^(jx) * sin(x)

    how do you possibly find the magnitude and argument of that?
    Express e^(jx) as \cos x + j\sin x and then you'll get it in the form z = a+ bj.
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    (Original post by Zacken)
    Express e^(jx) as \cos x + j\sin x and then you'll get it in the form z = a+ bj.
    Thanks got it

    w=-je^{j\Theta}sin(\Theta)\\w=-j(cos(\Theta) + jsin(\Theta))sin\Theta\\w=-j(cos(\Theta)sin(\Theta)+jsin^2(  \Theta))\\w=-jcos(\Theta)sin(\Theta)+sin^2(\T  heta)\\|w| = \sqrt{(sin^2(\Theta))^2 + (cos(\Theta)sin(\Theta))^2}\\|w| = \sqrt{sin^4(\Theta) + cos^2(\Theta)sin^2(\Theta)}\\|w| = \sqrt{sin^2\Theta(sin^2(\Theta)+  cos^2(\Theta))}\\|w| = sin(\Theta)\\
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    (Original post by Bealzibub)
    Thanks got it

    w=-je^{j\Theta}sin(\Theta)\\w=-j(cos(\Theta) + jsin(\Theta))sin\Theta\\w=-j(cos(\Theta)sin(\Theta)+jsin^2(  \Theta))\\w=-jcos(\Theta)sin(\Theta)+sin^2(\T  heta)\\|w| = \sqrt{(sin^2(\Theta))^2 + (cos(\Theta)sin(\Theta))^2}\\|w| = \sqrt{sin^4(\Theta) + cos^2(\Theta)sin^2(\Theta)}\\|w| = \sqrt{sin^2\Theta(sin^2(\Theta)+  cos^2(\Theta))}\\|w| = sin(\Theta)\\
    Perfect!
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    (Original post by Zacken)
    Perfect!
    The question does ask to find arg(w) too but i'm getting that wrong.

    w =sin^2(x)-jcos(x)sin(x)

    tan(x) = O/A = cos(x)sin(x)/sin^2(x)
    tan(x) = cos(x)/sin(x)
    tan(x) = 1/tan(x)
    tan^2(x) = 1
    tan(x) = +- 1
    x = -pi/4 or pi/4

    so x = -pi/4 but text book answer is theta - pi/2
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    (Original post by Bealzibub)
    The question does ask to find arg(w) too but i'm getting that wrong.

    w =sin^2(x)-jcos(x)sin(x)

    tan(x) = O/A = cos(x)sin(x)/sin^2(x)
    tan(x) = cos(x)/sin(x)
    tan(x) = 1/tan(x)
    tan^2(x) = 1
    tan(x) = +- 1
    x = -pi/4 or pi/4

    so x = -pi/4 but text book answer is theta - pi/2
    This is what happens when you use/confuse the same variables. The argument is not x, so why are you writing tan x = f(x).Let's call the argument y, then:

    \tan y = \frac{\cos x}{\sin x} \iff \tan y = \cot x \iff \tan y = \tan(\frac{\pi}{2} - x) \Rightarrow y = \frac{\pi}{2} - x.
 
 
 
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