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Complex numbers (arg)

  1. Aug 25, 2010 #1
    1. The problem statement, all variables and given/known data

    I need help on a little review please. z=2-i What is arg(iz)

    2. Relevant equations

    Well iz= 1+2i

    3. The attempt at a solution

    I think this should end up being arg(2i/1) But this doesn't seem to make sense because I am wanting to find an angle here right? I am somewhat confused. This is supposed to be an inverse tangent type problem, correct?
  2. jcsd
  3. Aug 25, 2010 #2
    I am thinking now that this should be something like 3pi/8 but I can't explain my logic....
  4. Aug 26, 2010 #3


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    Homework Helper

    It's simple trigonometry if you take a look at a complex number on the argand diagram. The argument of the complex number is the angle between the vector and the x-axis. The vector is just a line between the origin and the complex number.

    So, [tex]arg(a+ib)=tan^{-1}(b/a)[/tex]
  5. Aug 26, 2010 #4
    Ok. Doesn't that mean that it would be tan-1(2) which would be 1.107?
  6. Aug 26, 2010 #5


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    Science Advisor

    No, it doesn't! The arg of a+ bi is the angle the line from 0 to a+bi, thought of as points in the complex plane (a+ bi-> (a, b)) makes with the positive real axis. Specifically it is, as [itex]tan^{-1}(b/a)[/itex] as Mentallic says.

    But your number is z= 2- i. a= 2 and b= -1, NOT a= 1 and b= 2 as you seem to think.
  7. Aug 26, 2010 #6
    Yeah but the question is what is What is arg(iz) so doesn't that mean that I mult i times 2-i to give me 1+2i?

    I think what you just said would be for arg(z). correct?
  8. Aug 26, 2010 #7


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    Sorry, my fault, I didn't read the problem carefully enough. If z= 2- i then iz= 2i+ 1 so, yes, the arg of 1+ 2i is, in fact, [itex]tan^{-1}(2)[/itex].
  9. Aug 26, 2010 #8
    Ok. Thanks!
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