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Finding the modulus and argument of a complex number

  1. Sep 30, 2014 #1
    Hello,

    I have the point ##-1 + 2i##, for which I am asked to find the modulus and argument. The modulus was simple enough, but I am having difficulty finding the angle. The point is located in the 4th quadrant, and so I need to make certain that I calculate an angle in the range ##(\frac{3 \pi}{2}, 2 \pi )##, if I wish to measure counterclockwise rotations. Here is what I did:

    ##\theta = \arctan(-2) = - \arctan(2)##.

    So, clearly I have a negative angle, which is reasonable. However, I would like to find a positive angle, but I can't seem to wrap my head around the concept of coterminal angles and calculating angles based upon that principle.
     
  2. jcsd
  3. Sep 30, 2014 #2

    Mark44

    Staff: Mentor

    No it isn't.
    Add ##2\pi## radians to get the positive angle.
     
    Last edited: Sep 30, 2014
  4. Sep 30, 2014 #3
    Whoops, you are certainly correct. I meant to type in 1-2i. The principle is the same, however.
     
  5. Sep 30, 2014 #4

    Mark44

    Staff: Mentor

    Arctan(-2) will give you a negative angle. Add ##2\pi## to get a positive angle. For example, if you got ##-\pi/6##, adding ##2\pi## would give you ##11\pi/6## as the positive angle.
     
  6. Sep 30, 2014 #5
    I thought you just said that I should add an angle of ##\pi##.
     
  7. Sep 30, 2014 #6

    Mark44

    Staff: Mentor

    I confused myself. You thought the angle was in the 4th quadrant (it isn't - it's in the 2nd quadrant). Adding π would get your angle to the second quadrant, which isn't what you want. I edited my earlier post.
     
  8. Oct 1, 2014 #7

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    You can also think of the line going through the origin and through the point 1-2i (equiv., (1, -2) in Cartesian ). This may help you visualize how to choose the right angle.
     
  9. Oct 1, 2014 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    For ##0 < \alpha < \pi/2## the angles ##-\alpha## and ##2 \pi - \alpha## both describe the same point in the third quadrant.
     
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