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Complex Numbers: Equation involving the argument operator.

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Question:
    Sr5oxE3.png

    2. Relevant equations
    Any relevant to complex numbers.

    3. The attempt at a solution

    Given,

    Arg([itex]\frac{z}{w}[/itex])= Arg(z)-Arg(w)

    z=x+yi
    z1 = -1-2i
    z2 = 2+3i

    Arg(z-z1)=Arg(z2-z1)

    LHS:
    Arg(x+yi+1+2i)
    Arg((x+1) + i(y+2))

    tan([itex]\theta[/itex])=[itex]\frac{y+2}{x+1}[/itex]

    RHS
    Arg(2+3i+1+2i)
    Arg(3+5i)

    tan([itex]\theta[/itex])=[itex]\frac{5}{3}[/itex]

    RHS = LHS

    [itex]\frac{y+2}{x+1}[/itex] = [itex]\frac{5}{3}[/itex]

    Simplified
    y = [itex]\frac{5}{3}[/itex]x -[itex]\frac{1}{3}[/itex]

    Is this right?
     
  2. jcsd
  3. Apr 3, 2013 #2
    Yes it is correct.
     
  4. Apr 3, 2013 #3
    Cool beans,

    Is there a simple way to solve these though?

    A shortcut method?


    Wait, I found it...

    It's the line created from the points z1= -1-2i and z2 = 2+3i

    Gradient = [itex]\frac{3 - (-2)}{2 - (-1)}[/itex] = [itex]\frac{5}{3}[/itex]

    y = [itex]\frac{5}{3}[/itex]x + b

    (-2) = [itex]\frac{5}{3}[/itex](-1) + b

    -6 = -5 + 3b
    -1 = 3b
    b = -[itex]\frac{1}{3}[/itex]
     
    Last edited: Apr 3, 2013
  5. Apr 3, 2013 #4
    Yes. That's how I solved it. Your original method is still fine though. If you have loads of practice,, you'll be able to use both the methods equally efficiently.
     
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