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Complex number question

  1. Apr 18, 2010 #1

    [tex] P = 4e^{-j\frac{\pi}{3}} [/tex]

    [tex] Q = 4-3j [/tex]

    [tex] R = 2e^{j\frac{\pi}{2}} [/tex]

    [tex] S = 5 [/tex]

    note: I'm using j to be a complex number, it's equivalent to i in mathematics


    A: express p q r s in both cartesian (a+ib) and polar (re^itheta) forms

    B: sketch p q r and s on the complex plane

    I'm not too certain as to how P and R work,
    I know from the taylor series that r(cos(theta)+isin(theta)) = re^itheta

    P is already in polar form, [tex] P = 4e^{-j\frac{\pi}{3}} [/tex]
    so the Cartesian form,

    (a+jb) = [tex] 4 cos\frac{\pi}{3} = a [/tex] and [tex] 4 sin \frac{\pi}{3} = b [/tex]

    so it is 2+j(3.464) or [tex] 2+j\sqrt{12} [/tex]
    SO SHOULD I PUT, [tex] 2-j\sqrt{12} [/tex] ????????????????
    [tex] Q = 4-3j [/tex]
    Q is already in cartesian form,

    |z| q = 5, = root 4^2 + -(3^2)
    [tex] \theta = cos^{-1} \frac{4}{5} = 0.643 rad [/tex]

    so the polar form is , 5cis(0.643) = 5(cos(0.643) + jsin(0.643)) = 5e^(j0.643)

    [tex] R = 2e^{j\frac{pi}{2}} [/tex]
    R is already in polar form,

    2 = |z|

    2 cos(pi/2) = 0,
    2 sin(pi/2) = 2,
    this means if i wear to imagine the vector R, it would be going straight up,

    so in cartesian form, the equivalent equation of [tex] R = 2e^{j\frac{pi}{2}} [/tex] is j2

    I'm assuming S is already in cartesian form, since it is 5+j0

    |z| =5
    the angle it makes with the x axis, is 0,
    so i think the equation for S in polar form is, [tex] 5e^{j * 0 } [/tex]
    Last edited: Apr 18, 2010
  2. jcsd
  3. Apr 18, 2010 #2
    Determine numberical answers for each of the following:
    either in cartesian form or in polar form with the angle in degrees


    1) (PQ)^1/2
    2) (R/P)^1/3

    Last edited: Apr 18, 2010
  4. Apr 18, 2010 #3
    i hate macs,

    I calculate pq to be,

    [tex]20e^{-j(\frac{\pi}{3} - 0.643)}[/tex]

    how do i square root it?

    [tex] \sqrt{20e^{-j(\frac{\pi}{3} - 0.643)}} [/tex]
  5. Apr 19, 2010 #4
    =>P is already in Exponential form. To convert to Cartesian:

    Euler's Identity:
    [PLAIN]https://dl.dropbox.com/u/4645835/MATH/EulersID.gif [Broken]

    [PLAIN]https://dl.dropbox.com/u/4645835/MATH/Pcart.gif [Broken]

    =>Q is already in Cartesian; so convert to Polar:
    [PLAIN]https://dl.dropbox.com/u/4645835/MATH/Qpol.gif [Broken]

    =>Convert Q to Exponential:
    [PLAIN]https://dl.dropbox.com/u/4645835/MATH/Qexp.gif [Broken]

    I think these are right and I hope they help.
    Last edited by a moderator: May 4, 2017
  6. Apr 19, 2010 #5


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    No, [itex]\theta= -\frac{\pi/3}[itex], not [itex]\frac{\pi}{3}[/itex].

  7. Apr 19, 2010 #6
    It is possible for an expression to fit into more than one category or into none of the categories. I'm not sure how this works anymore..as I did this a long time ago. But I think S satisfies exponential and cartesian.
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