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Solutions to four complex numbers showing how arg(z) is obtained

  1. Jun 22, 2004 #1
    Hi...

    my notes show the solutions to four complex numbers showing how arg(z) is obtained...they also show an argand diagram showing theta...there's a couple of things i don't understand so i was hoping that someone could shed some light...thank you...

    (i) z = 1 + i
    (ii) z = -1 + i
    (iii) z = -sqrt(3) - i
    (iv) z = 1 - i

    for (i), arg(z) = arctan(1) = pi/4...ane the argand diagram shows theta being projected anti-clockwise within the first quadrant...

    for (ii), arg(z) = arctan(-1) = pi - pi/4 = 3pi/4...the argand diagram shows theta moving into the second quadrant anti-clockwise...

    for (iii), arg(z) = arctan[1/(sqrt3)] = -5pi/6...the argand diagram shows theta in the third quadrant and is projected clockwise from the x-axis...

    for (iv), arg(z) = arctan (-1) = -pi/4...the argand diagram shows theta in the fourth quadrant and it is projected clockwise from the x-axis...

    okay, here's the bits i'm having trouble with...

    1. in part (i) the complex number is in the first quadrant...arg(z) = pi/4 and i guess that's straightforward...

    for part (ii), the complex number is in the second quadrant...so to find arctan (-1)...do we take the principal angle which is pi/4 and then because it's in the second quadrant, which is pi - (the related angle), do we find arctan(-1) by calculating pi-pi/4...???...

    2. for part(iii), i can see that the complex number is in the third quadrant...and i know that tan there is negative and and the angle there is found by pi + (the related angle)...i know also that arctan[1/(sqrt3)] = pi/6...i know also that arctan is restricted between -pi and pi...

    so how was arg(z) found to be -5pi/6...???...

    why is theta shown to be projected anticlockwise this time...???...

    3. basically the same concerns as question 2...i can see that the complex number is in the fourth quadrant...in the fourth quadrant, theta is 2pi - (the related angle)...i know that arctan(1) is pi/4...

    so how was arctan(-1) = -pi/4...???...

    theta is shown projected anticlockwise from the x-axis...why, when the first two questions show theta projected clockwise...???...
     
  2. jcsd
  3. Jun 22, 2004 #2

    AKG

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    There is something called the CAST rule. Bascially, going counter-clockwise, label quadrants, IV, I, II, and III with C, A, S, and T, respectively. The letter tells you which function gives a positive value. In the C quadrant, only Cosine is positive, in the A quadrant, All functions give positive values, and you can guess the rest. Another way to think about it is this: cosine, for example, is adjacent/hypoteneuse. The hypoteneuse is always positive, but the adjacent changes depending on what quadrant you're in. If you think of the arm or line segment swinging around the origin, and the angle theta being the angle between the arm and the polar axis, or the positive x-axis, then consider that arm to the hypoteneuse. Let's say the arm goes from (0,0) to (1,1) = 1 + i. Now, you know the angle is pi/4. The x value is +1, and the hypoteneuse is sqrt{2}. Now, if you imagine the arm going into the II quadrant, i.e. the S quadrant, then the x-values are negative, so the adjacent part of the triangle is negative, and so you have negative cosine values. However, the y-values are still positive, so your sine values are still positive. Since tangent depends on opposite and adjacent (not hypoteneuse), it's positive when the x and y values are both positive (1st quadrant, A) or both negative (3rd quadrant, T).

    Now, I'm not sure, but I don't think it matters if the angle is shown clockwise or anti-clockwise. For all intents and purposes, [itex]\dots \ -13\pi /4,\ -5\pi /4,\ 3\pi /4,\ 11\pi /4,\ \dots[/itex] are all the same. They are all coterminal.

    Now, you want to know how to find arctan values, and I assume you mean without a calculator. In most cases, this is only possible if you are dealing with a special triangle. The following special triangles are right angled triangles:

    side lengths 1, hypoteneuse [itex]\sqrt{2}[/itex], angles [itex]\pi /4[/itex].
    side length 1 opposite angle [itex]\pi /6[/itex]; side length [itex]\sqrt{3}[/itex] opposite angle [itex]\pi /3[/itex]; hypoteneuse 2.

    Draw these triangles out, and these would allow you to be able to find the arc tan values that you need.
     
  4. Jun 22, 2004 #3
    For number 3, see my image. Consider the easier problem of calculating the angle v. We see that |a| = sqrt(3), and |b| = 1. Thus tan(v) = a/b = sqrt(3)/1, which implies v = pi/3 (since we can see that 0 < v < pi/2). The relationship between v and arg(z) is: v + arg(z) = 270 degrees = 270pi/180 radians = 3pi/2. So arg(z) = 3pi/2 - v = 3pi/2 - pi/3 = 7pi/6, which is "the same angle" as -5pi/6. I'm not sure why the choose to make it anticlockwise though... tan is not negative in the third quadrant btw.
     

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  5. Jun 22, 2004 #4

    AKG

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    I'm not sure what your question is.
    I never said it was.
     
  6. Jun 22, 2004 #5
    Did you confuse me with naav (who I was addressing) or something? I'm quite flattered that you think that I was able to read your post, paint the image, and compose my reply in a mere minute, though ;)

    Probably because I didn't ask one. I responded to this (which is in the first post):

    ...

    And I never said you said it was. I responded to this (see the first post):

     
    Last edited: Jun 22, 2004
  7. Jun 22, 2004 #6
    thank you very much...i'm still trying to get my head round this...

    sorry, for question 3, i meant to say that theta was projected clockwise...

    1. okay how do we know that 7pi/6 is the same as -5pi/6...???...

    2. why the different projections of theta - i mean anti-clockwise for the first two complex numbers and clockwise for the other two...???...
     
  8. Jun 22, 2004 #7

    Hurkyl

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    Try drawing the two angles.


    arg is a multivalued "function"; each input gives lots of outputs.

    Arg is specifically restricted to a particular range.... (-pi, pi] I think. Thus, you have to pick the one output that lies in this range.
     
  9. Jun 22, 2004 #8

    AKG

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    :uhh: Oops! Sorry, you're right, I confused you two.
     
  10. Jun 22, 2004 #9
    thank you...

    i've tried to think about it more...here's what i've come up with...

    1. complex number 1 is in the first quadrant so arctan(1) is pi/4 and pi/4 is between -pi and pi...so we can accept it as arg(z)...

    2. complex number 2 is in the second quadrant...for arctan(-1), we see that the related angle is pi/4 and since we are in the second quadrant tan is negative...so pi - pi/4 = 3pi/4...which is again within our restriction...

    3. complex number 3 is in the third quadrant...arctan[1/(sqrt3)] is pi/6...and since tan is in the third quadrant and tan in the third quadrant is 7pi/6...but because we are restricted to theta being between -pi and pi we then go clockwise...so theta then is 2pi - 7pi/6 = 5pi/6 and SINCE WE WENT CLOCKWISE we say that theta is -5pi/6...is that why it's negative...???...
     
  11. Jun 23, 2004 #10

    HallsofIvy

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    or you could add pi/2- it gives the same thing.
    Knowing that the "base" angle is pi/6 but that the angle is in the third quadrant, you could either add pi: pi+ pi/6= 7pi/6 or subtract pi:
    pi/6- pi= -5pi/6. Since your problem specified (more or less arbitrarily) that the angle must be between pi and -pi, the second is appropriate for this problem. Of course the two differ by 2pi so they are really the same.
     
  12. Jun 24, 2004 #11
    Hi...thank you again...

    1. why does the fact that they differ by 2pi mean that they are they same...isn't the tan function periodic for pi...???...

    2. have i explained the following correctly...for the complex number z = -sqrt3 - i, arg(z) = arctan (1/sqrt3)....

    1st quadrant = pi/6 which is also the same as -11pi/6...

    2nd quadrant = pi - pi/6 = 5pi/6 which is also the same as -7pi/6...

    3rd quadrant = pi + pi/6 = 7pi/6 which is also equal to -5pi/6...

    4th quadrant = 2pi - pi/6 = 11pi/6 which is also equal to -pi/6...

    so then because the complex number is found to be in the 3rd quadrant...arg(z) will either be 7pi/6 or -5pi/6...and then because we are restricted to theta being between -pi and pi we say that arg(z) = -5pi/6...???...

    3. so if you have an angle of pi/6...this is the same as saying -11pi/6...???...
     
  13. Jun 24, 2004 #12

    HallsofIvy

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    "tan" is not relevant. What is relevant is that arg(z) is the angle measured from the positive real axis. if arg(z)= [theta] then arg(z) is also equal to [theta]+ any multiple of 2[pi].
     
  14. Jun 25, 2004 #13
    Hi...thank you...

    i don't understand how to get arg(z) for the following...

    1. z = 2i

    2. z = 3

    3. z = -3i

    4. z = -5

    for part 1, arg(z) = arctan(2/0)...but that's undefined so does that mean it will be at pi/2 which is also undefined..??..

    for part 2, arg(z) = arctan (0/3) = arctan(0)...for this one pi, 0 and pi would all be valid i guess..??..

    for part 3, arg(z) = arctan(-3)...i don't know how to find this one..??.

    for part 4, arg(z) = arctan[0/(-5)] = arctan(0) which again i would guess to be -pi, 0 and pi..??..
     
  15. Jun 26, 2004 #14

    HallsofIvy

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    Draw a picture! z= 2i is on the positive y-axis, at (0, 2) so arg(z)= pi/2.
    z= 3 is on the positive x-axis, at (3,0), so arg(z)= 0. No, pi would NOT be valid!
    0, 2pi, -2pi, in general any even multiple of pi would work (0 + any multiple of 2pi).

    3. z= -3i is on the negative y- axis, at (0, -3) so arg(z)= 3pi/2 or -pi/2. In general,
    3pi/2+ any multiple of 2pi.,

    4. z= -5 is on the negative x-axis. arg(z)= pi or -pi or, more generally, pi+ any multiple of 2pi. (NOT 0!)
     
  16. Jun 26, 2004 #15
    thank you...

    1. why multiples of 2pi...isn't the period of the tan function pi...???...

    2. how come arg(z) = arctan[(img(z))/(re(z))] doesn't work here for the four complex numbers mentioned in my last post...???...
     
  17. Jun 26, 2004 #16

    HallsofIvy

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    One more time: the period of tan(z) is irrelevant. Adding a multiple of 2pi to arg(z) gives you the same point because you are plotting the points on a circle (said another way, the inverse of arg(z) has period 2pi).

    For the same reason that the period of tan(z) is irrelevant. If z= a+ bi, then img(z)/re(z)= b/a and if z= -a-bi, then img(z)/re(z)= -b/-a= b/a. The args differ by pi but you can't distinguish that since the period of tan(z) is pi rather than 2pi. If you insist upon using arg(z)= arctan(img(z)/re(z)), you will have to use a little common sense to distinguish between them.
     
  18. Jun 27, 2004 #17
    Hi...thank you for trying...
     
  19. Jun 27, 2004 #18

    HallsofIvy

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    Oooh! That doesn't sound good!
     
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