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Homework Help: Argument of complex expr.

  1. Jul 26, 2007 #1
    The problem statement, all variables and given/known data
    find the argument of-:
    [tex]Z=\frac{\iota-\sqrt{3}}{\iota+\sqrt{3}} [/tex]


    Method 1

    [tex]Z={2e^{\frac{- \iota \pi}{3}} \mbox{ divided by } {2e^{\frac{\iota \pi}{3}} [/tex]

    [tex]Z=e^{\frac{-2 \iota \pi}{3}}[/tex]

    Therefor arg(Z)=[tex] \ -\frac{2\pi}{3} [/tex]

    Method 2

    [tex]Z=\frac{4}{1+2\sqrt{3}\iota-3} [/tex]

    [tex]Z=\frac{2}{-1+\sqrt{3}\iota} [/tex]

    [tex] \mbox{arg}(Z)=0+\frac{\pi}{3} = \frac{\pi}{3} [/tex]

    What am i getting wrong ?
     
    Last edited: Jul 26, 2007
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  3. Jul 26, 2007 #2

    George Jones

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    Is the middle term in the denominator of the right side of the first line of Method 2 right?
     
  4. Jul 26, 2007 #3

    CompuChip

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    And where did you get the last step,
    [tex]\arg\left( \frac{2}{-1 + i \sqrt{3}} \right) [/tex]
    anyway?

    Here's another one: try multiplying top and bottom by the conjugate of the denominator, [itex]i - \sqrt{3}[/itex]. Then write the result as [itex]r e^{i \phi}[/itex] and read off [itex]\phi[/itex].
    This method always works with fractions.
     
  5. Jul 26, 2007 #4

    Dick

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    In your first method - note 2*exp(-i*pi/3)=1-i*sqrt(3). That's not the numerator that you want. Nor is the other one the denominator. How are you getting these? (PS put i's in the exponentials).
     
  6. Jul 26, 2007 #5
    Thats [tex] 2\times 1\times \iota , \mbox{the 2ab term in }(a+b)^2 [/tex]

    I've rationalized, by multiplying numerator and denominator with i+sqrt(3) , and then divided by 2 ; although the standard practice is to remove i from denominator as you suggested.

    Yeah sry about those missing i's..
    I've expressed both as Re^(i.theta) form, (theta=arctan(b/a))
    Sorry I am unclear as to what you mean by "not the numerator that you want
    ". Could you plz explain ? Thx.

    PS: I have a feeling that method 2 has a bug, but i cant find it :tongue:
     
  7. Jul 26, 2007 #6

    George Jones

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    Did you forget a minus sign?

    You need to factor out the r before finding the argument.
     
  8. Jul 26, 2007 #7

    Dick

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    I meant that 2*exp(-i*pi/3)=1-i*sqrt(3) which is what I said. Maybe I didn't explicitly point out that the numerator of your fraction is i-sqrt(3), which is NOT the same thing.
     
  9. Jul 26, 2007 #8
    oh sorry , i mixed up the questions...i get it
    Thank you and sorry about this :blushing:
     
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