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Complex question

  1. Oct 23, 2011 #1
    1. Consider the limit limz→i(|z| + iArg(iz))


    2. (i) What value does the limit approach as z approaches i along the
    unit circle |z| = 1 in the first quadrant?
    (ii) What value does the limit approach as z approaches i along the
    unit circle |z| = 1 in the second quadrant?



    3. solutions are as follow:
    (i) Note that iz = −y + ix. When z is in the first quadrant, iz is in the second quadrant. Thus, we have
    limz→i(|z| + iArg(iz)) = 1 + i∏
    when z approaches i along the unit circle |z| = 1 in the first quadrant.
    (ii)Similarly, when z approaches i along the unit circle |z| = 1 in the
    second quadrant,
    limz→i(|z| + iArg(iz)) = 1 − i∏

    My question is (i) why is it in second quadrant when z is in first quadrant?
    is it because the real part is now -y and imaginary part is x which -y is the x-axis and x is the y-axis?
    (ii)Why Arg(iz)=∏, when it is in second quadrant?

    Thanks!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 24, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Any z on the unit circle can be represented as [itex]z= e^{i\theta}[/itex] so [itex]iz= e^{i\pi/2}e^{i\theta}= e^{i(\theta+ \pi/2}[/itex]. Multiplying by i "rotates" the number 90 degrees. When z is in the first quadrant, iz is in the second quadrant. When z is in the second quadrant iz is in the third quadrant.



     
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