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Math Help - complex no's<<<hard!

  1. #1
    Member i_zz_y_ill's Avatar
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    complex no's<<<hard!

    Q;{I'm stuck on part (iii) }
    Given w=1-e^(jx)cosx 0<x<pie/2
    (i) express e^jkx and e^-jkx in the form a+jb
    (ii) show w=-je^(jx)sinx
    (iii)find |W| and argw, hence write down the modulus and argument of each of the two square roots of w.

    k so, bare with me.. (ii) is just an expansion of w in original Q giving on the way w=-jsinx(cosx+jsinx) FROM here i would conclude that |w| is
    |-jsinx| = |sinx| = sinx (correct) ,, my teacher said do,|-j|.|e^(jx)|.|sinx| {more complicated i think} so that's ok.

    Looking at w=-jsinx(cosx+jsinx) i would therefore think that argument is just x,,,,,,,as this is wrong ,presumably the modulus of sinx has some interplay into the argument 'i dunno'. So is it reasonable to conclude everytime a modulus is not a whole no?????? i must write arg(-j)arg(e^jx)arg(sinx), plot each one and get x-pie/2,,,,i presume so,, don't know y,, n e way not to worry,,

    |(w)^o.5| = (sinx)^0.5,,,so argument will be using de moivre (x-pie/2)/2,,,,,,,,,,,,,,and the other one is{don't know where came from}
    ((2x-pie)/4)+pie i think it has something to do with the limits but i can't see how!

    If n e one could take the time to help me on part (iii) that would be most appreciated thnx
    Last edited by i_zz_y_ill; April 3rd 2008 at 07:04 AM. Reason: grammer
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  2. #2
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    Looking at w=-jsinx(cosx+jsinx) i would therefore think that argument is just x,,,,,,,as this is wrong ,presumably the modulus of sinx has some interplay into the argument 'i dunno'. So is it reasonable to conclude everytime a modulus is not a whole no?????? i must write arg(-j)arg(e^jx)arg(sinx), plot each one and get x-pie/2,,,,i presume so,, don't know y,, n e way not to worry,,
    \arg{(w)}=\arg{(-\mathrm{j})}\color{red}+\color{black}\arg{(\mathrm  {e}^{\mathrm{j}x})}\color{red}+\color{black}\arg{(  \sin{x})}=-\,\frac{\pi}{2}+x+0

    Quote Originally Posted by i_zz_y_ill View Post
    |(w)^o.5| = (sinx)^0.5,,,so argument will be using de moivre (x-pie/2)/2,,,,,,,,,,,,,,and the other one is{don't know where came from}
    ((2x-pie)/4)+pie i think it has something to do with the limits but i can't see how!
    The second root is obtained from the first root by rotating it through \pi radians anticlockwise about the origin.
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  3. #3
    Member i_zz_y_ill's Avatar
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    i think i understand

    thanks, when it says the two square roots is this refering to
    +(sinx)^0.5(cosx(-pie/2)+jsin(-pie/2))^0.5

    and -(sinx)^0.5(cosx(-pie/2+x)+jsin(-pie/2+x))^0.5

    the would the minus sign on the second root indicate a anticlockwise pie rotation??? just to clarify that would be gr8???


    also am i right in thinking thn whenever the complex no. is -jsinx(cosx+jsinx) the argument can't be simply x seems bit odd n e genius out there?
    Last edited by i_zz_y_ill; April 3rd 2008 at 09:09 AM. Reason: also
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  4. #4
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    thanks, when it says the two square roots is this refering to
    +(sinx)^0.5(cosx(-pie/2)+jsin(-pie/2))^0.5

    and -(sinx)^0.5(cosx(-pie/2+x)+jsin(-pie/2+x))^0.5

    the would the minus sign on the second root indicate a anticlockwise pie rotation??? just to clarify that would be gr8???
    Those are not the square roots of w.

    Quote Originally Posted by i_zz_y_ill View Post
    also am i right in thinking thn whenever the complex no. is -jsinx(cosx+jsinx) the argument can't be simply x seems bit odd
    Thatís because -\mathrm{j}\sin{x} is not a real number.
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  5. #5
    Member i_zz_y_ill's Avatar
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    r the square roots?

    sorry are the square roots (-jsinx)^0.5(cos(2x-pie)+jsin(2x-pie))^0.5


    and -(-jsinx)^0.5(cos(2x-pie)+jsin(2x-pie))^0.5

    and the modules there sinx for both and second one indicates anticlockwise pei???? if im wrong pls tell me,so confused
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  6. #6
    Member i_zz_y_ill's Avatar
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    arrrrgh complex roots,,lol i was thinking something else

    thankx dude!
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