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Complex numbers: Argument

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    When finding the argument of a complex number using tan(\theta) = y/x (where z = x + iy), sometimes I do not get the correct answer. I assume this is because tangent is only defined for -pi/2 to pi/2 (and from this it is periodic).

    So is it a good idea always to find the argument of a complex number using cosine and sine? Or am I missing something here?

    Thanks in advance.

    Regards,
    Niles.
     
  2. jcsd
  3. Nov 10, 2008 #2
    if you use cos and sine, you should still get the same theta if you use tan. Think of it as a triangle, where z is the hypotenuse, x is the base, and y is the height.
     
  4. Nov 10, 2008 #3

    berkeman

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    As in the figure at the top of this page:

    http://en.wikipedia.org/wiki/Complex_plane

    .
     
  5. Nov 10, 2008 #4

    HallsofIvy

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    There is nothing wrong with using tangent (more correctly arctangent) as long as you keep track of the signs of x and y. For example if you find that y/x= 1, then arctan(1) could be [itex]\pi/4[/itex] or [itex]5\pi/4[/itex]. If you know that x and y are positive then you know the argument is [itex]\pi/4[/itex], if negative, [itex]5\pi/4[/itex].
     
  6. Nov 11, 2008 #5
    First, thanks to all for replying.

    But I can avoid all of this "confusion" by using sine or cosine? I mean, what if only x is negative and y is positive. Then using arctangent I have to add pi/2 to get the correct result.
     
  7. Nov 11, 2008 #6

    HallsofIvy

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    What do you mean by "using" sine and cosine? Yes, [itex]tan(\theta)= y/x[/itex] can be interpreted as [itex]sin(\theta)= y/r[/itex] and [itex]cos(\theta)= x/r[/itex]. If you only find [itex]arcsin(y/r)[/itex] where y is positive, you still have two possible answers [itex]\theta[/itex] between 0 and [itex]\pi/2[/itex] and [itex]\pi/2- \theta[/itex]. You can, of course, decide which you want by looking at [itex]arccos(x/r)[/itex] but it seems to me simpler to just find [itex]\arctan(y/x)[/itex] and look at the signs of x and y. That way you don't have to calculate [itex]r= \sqrt{x^2+ y^2}[/itex].
     
  8. Nov 11, 2008 #7
    Perhaps getting to know the handy-dandy unit circle will help you avoid this confusion? That way you'll get a sense on what happens when either X or Y is positive or negative.
     
  9. Nov 11, 2008 #8
    So what I can conclude from your answers is that there is no way using arctan, arccos or arcsin to find the "correct" angle all the time, but it depends on the values of x and y?
     
  10. Nov 11, 2008 #9
    no, you can find the angle using arctan, arccos, arcsin if you know x,y, and z.
    are you familiar with the term: SOHCAHTOA?

    just remember that the answer can range from 0 to 2pi, because that depends on the signs of x and y.
     
  11. Nov 11, 2008 #10

    HallsofIvy

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    You cannot determine it by using any ONE of those without taking the signs of x and y separately into account.

    I think you have misunderstood the question.
     
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