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Complex numbers and modulus

  1. Nov 15, 2005 #1
    I have z=-(1/2)-(sqrt3/2)i
    r=|z|

    is this right?

    r=cos*2Pi/3+i*sin*Pi/3 = 1 + sqrt3/2*i

    Now I have to find Z^2004, how do I do that?
     
  2. jcsd
  3. Nov 15, 2005 #2

    dextercioby

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    Nope. The modulus of a complex number has to be real.

    Daniel.
     
  4. Nov 15, 2005 #3

    dextercioby

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    Show that the trigonometric form of your "z" is

    [tex] z=\cos 210\mbox{deg} \ +i\sin 210\mbox{deg} [/tex]

    Daniel.
     
  5. Nov 15, 2005 #4

    benorin

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    no

    No. If [tex]z=a+bi[/tex] where a and b are real, then [tex]\left| z \right| = \sqrt{a^2+b^2} \geq 0[/tex] for any complex number z. Notably, [tex]\left| z \right| [/tex] is always a positive real number, and hence your answer for r cannot be correct.

    If [tex]z = -\frac{1}{2}-\sqrt{\frac{3}{2}}i[/tex], then [tex]r=\left| z \right| = \sqrt{\left( -\frac{1}{2} \right) ^2 + \left( -\sqrt{\frac{3}{2}} \right) ^2} = \sqrt{ \frac{1}{4}+ \frac{3}{2}} =\frac{1}{2}\sqrt{7}[/tex]
     
  6. Nov 15, 2005 #5

    HallsofIvy

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    [tex]r= \sqrt{\left(\frac{1}{2}\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2}[/tex]
    which, as Dextercioby and denorin point out, is a real number.
    [tex]tan(\theta)= \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}= \sqrt{3}[/tex]
    Since both real and imaginary parts of negative (the negatives disappear in the fraction) the angle is in the 3rd quadrant.
    (Thanks, Dextercioby. One of these days, I really need to learn to count!)

    Once you know r and [itex]\theta[/itex],
    [tex](r(cos(\theta)+ i sin(\theta))^n= r^n(cos(n\theta)+ i sin(n\theta))[/tex]
     
    Last edited: Nov 15, 2005
  7. Nov 15, 2005 #6

    dextercioby

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    You mean the third quadrant, right...?

    Daniel.
     
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