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Find the argument of a complex number

  1. Sep 13, 2008 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Find the argument of [tex](-2+i)(1+2i)[/tex].
    2. The attempt at a solution [tex]z=(-2+i)(1+2i)=-4-3i[/tex]. I've read on the Internet that I can write [tex]z=\rho (\cos \theta +i\sin \theta)[/tex] where [tex]\rho[/tex] is the modulo of [tex]z[/tex]. I've calculated the modulo of [tex]z[/tex] which is [tex]5[/tex].
    So I have that [tex]-4=5\cos \theta \Leftrightarrow \theta = \arccos \left( -\frac{4}{5} \right)[/tex]. Am I right? Because Mathematica says it's my result but with a negative sign...
    Also, is this the common used way to find the argument of a complex number?
     
  2. jcsd
  3. Sep 13, 2008 #2

    CompuChip

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    The cosine of both arccos(-4/5) and -arccos(-4/5) is -4/5, because cos(x) = cos(-x). You also need to take the other one into account: -3 = 5 sin(theta). Because sin(-x) and sin(x) are different (sin(-x) = -sin(x)) this will determine which one to use.

    In geometric terms: the magnitude of the angle is completely determined by the cosine of an angle, but you also need to specify the sine to say in which quadrant.
     
  4. Sep 13, 2008 #3

    fluidistic

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    Ah yes, you're right! I underestimated the sine part.
    So this is generally the way one finds out the argument of a complex number?
     
  5. Sep 14, 2008 #4

    Gib Z

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    No. Plot a+bi on the complex plane, and make a triangle. It should be easier to see that way.
     
  6. Sep 14, 2008 #5

    HallsofIvy

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    Most commonly given is: arg(x+ iy)= arctan(y/x).
     
  7. Sep 14, 2008 #6

    fluidistic

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    Ah ok, thanks to both. Nice to know.
     
  8. Jan 16, 2009 #7
    Arg(-pi)= arctan(-pi)??
     
  9. Jan 16, 2009 #8

    CompuChip

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    Where did you get that from?
    Note that HallsOfIvy's argument doesn't entirely work here, because it only gives you the right answer modulo pi.
    But you don't even need that, you can see it right away: -pi is on the negative real axis.
     
  10. Jan 16, 2009 #9
    sorry i am trying to learn this concept ...but don't have real good source !...
    so if you could help me would be great!
    please and thank you !
     
  11. Jan 17, 2009 #10

    CompuChip

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    Do you know how to visualise a complex number on a (x, y) - plane?
     
  12. Jan 17, 2009 #11

    jambaugh

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    Only if x>0. Else = arctan(y/x) +/- pi.
    [tex] \theta = arctan(y/x) \pm \pi[/tex]

    Anytime we use inverse trig we must account for the fact that the trig functions are not invertible over their entire domain. We must first restrict the domain and this restricts the range of the inverse trig functions. We must use different restrictions for different trig functions so in this case it helps to think of the tangent of the angle as the slope of the line (through the origin) at that angle. The ambiguity then is on which side of the origin lies a point on the line with given slope=tan(theta).

    The convention for tangent is to work in the right half-plane, [itex] -\pi/2 \le arctan(m) \le \pi/2[/itex] (so that the domain is connected) thus we only get correct results with [itex] x\ge 0 [/itex]. Again remembering we are working via tan with slopes we simply rotate the line [itex] 180^o = \pi[/itex] modulo the full circle to get the other two quadrants.

    For general inverse trig issues one picks a connected part of the unit circle which spans the definition of the value trig function... e.g.
    [itex] -\pi/2 \le \theta \le \pi/2[/itex] for tangent and sine (slope and y-coordinate)
    and
    [itex] 0\le \theta \le \pi[/itex] for cotangent and cosine (1/slope and x-coordinate)

    This is all analogous to the [itex]\pm[/itex] ambiguity we put in a square root because the principle root is defined for the restricted domain [itex] x \ge 0[/itex] of the square function.
    [itex] x^2 = a \quad \Longleftrightarrow \quad x = \pm\sqrt{a}[/itex]
     
  13. Sep 19, 2010 #12
    I would just like to confirm, is the answer to the 1st part -2.4980915 radians ?
     
    Last edited: Sep 19, 2010
  14. Sep 19, 2010 #13

    Mentallic

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    Yes, but you should keep it in exact form, [tex]tan^{-1}\left(\frac{3}{4}\right)-\pi[/tex]
     
  15. Sep 19, 2010 #14
    My examination board prefers it as a decimal. :)
     
  16. Sep 19, 2010 #15

    Mentallic

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    Then if they don't require you to use a million decimal places, just pop out your protractor and start measuring away on an accurate diagram :wink:
     
  17. Sep 20, 2010 #16
    I cant tell if you're being sarcastic.... but never mind. :rolleyes:

    If you must know, I decided to write my answer to 'a million' decimal places to ensure that it truly was the correct answer and not some fluke on my part. ;)

    Anyways, peace.
     
  18. Sep 20, 2010 #17

    Mentallic

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    The actual reason why examiners ask for you to leave such questions in exact form is because some students would spend their time trying to draw an accurate diagram and finding the angle with a protractor. Exact answers eliminate this possibility.
     
  19. Sep 20, 2010 #18
    Are you aware of the Edexcel GCE A Level examinations ?
     
  20. Sep 20, 2010 #19

    Mentallic

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    Nope I'm from Australia.
     
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