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The phase of a complex number

  1. Feb 19, 2012 #1
    in cartesian form, a+ ib you can find the phase by doing arctan(b/a).. my question concerns the phase of a purely imaginary number. during a lecture my professor said that the phase of i*2pi= pi/2, he rationalized this by saying that the number lies on the y-axis so the angle between the real axis and the imaginary axis is pi/2. but if you do arctan(2pi/0) you will get an error.. how is he right?
     
  2. jcsd
  3. Feb 19, 2012 #2

    pwsnafu

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    Complex numbers don't have a "phase" they have an "argument" which is defined case by case: if z = x + iy we define [itex]\phi = Arg(z)[/itex] as
    arctan(y/x) when x > 0
    arctan(y/x)+π when x < 0 and y ≥ 0
    arctan(y/x)-π when x < 0 and y < 0
    π/2 when x = 0 and y > 0
    -π/2 when x = 0 and y < 0
    indeterminate when x = 0 and y = 0.
     
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