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Math Help - Rotation in complex plane

  1. #1
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    Rotation in complex plane

    If, for complex z and w, |z+w|=|z-w|, show that \arg z and \arg w differ by \pi /2.

    I have an idea how to solve this. Drive out a/b=-y/x from the initial equation (where z=a+bi,w=x+yi), then do something with rotating tangents around 90 degrees blahblah. The problem is I don't really know how to put this down on paper as I get kinda lost when it comes to these funky substitutions into arctan functions. Or maybe there is another way to do this?
    (We didn't do vectors yet, so I should perhaps do without them, but any hint you can drop on dot and cross products and whatnot will be appreciated.) Thanks!
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    Quote Originally Posted by atreyyu View Post
    If, for complex z and w, |z+w|=|z-w|, show that \arg z and \arg w differ by \pi /2.

    I have an idea how to solve this. Drive out a/b=-y/x from the initial equation (where z=a+bi,w=x+yi), then do something with rotating tangents around 90 degrees blahblah. The problem is I don't really know how to put this down on paper as I get kinda lost when it comes to these funky substitutions into arctan functions. Or maybe there is another way to do this?
    (We didn't do vectors yet, so I should perhaps do without them, but any hint you can drop on dot and cross products and whatnot will be appreciated.) Thanks!

    With your symbols, (a+x)^2+(b+y)^2=|z+w|^2=|z-w|^2=(z-x)^2+(b-y)^2\Longrightarrow ax+by=0\Longrightarrow

    \Longrightarrow \frac{b}{a}=-\frac{x}{y}=-\frac{1}{y/x}

    Now all follows from \arctan\omega+\arctan(1/\omega)=\frac{\pi}{2}

    Tonio
    Last edited by tonio; December 12th 2010 at 01:55 PM.
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  3. #3
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    The fact that |z-w|=|z+w| means that z is equally distance from w and –w.
    Thus z is on the perpendicular bisector of the line segment with endpoints w and –w. That line passes through (0,0).
    Argue that the two complex numbers are ‘perpendicular’, (vector wise).
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    Quote Originally Posted by tonio View Post
    Now all follows from \arctan\omega+\arctan(1/\omega)=\frac{\pi}{2}
    I've gotten to that point by still I don't quite get it where it comes from...

    And "to argue [...] vector wise" - I don't really know how to do it!
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    Quote Originally Posted by atreyyu View Post
    And "to argue [...] vector wise" - I don't really know how to do it!
    It is too vague only if you did not bother to draw a diagram. Did you?
    Do you understand the geometry of complex numbers?
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    Quote Originally Posted by atreyyu View Post
    I've gotten to that point by still I don't quite get it where it comes from...

    And "to argue [...] vector wise" - I don't really know how to do it!

    Well, the above equality can be proved either by differentiating the LHS and checking it is zero and then evaluating at x=1 , or

    even more basically by trigonometry: take a straight angle triangle with acute angles \alpha,\beta and opposite

    legs of length a,b , resp. Then \displaystyle{\tan\alpha+\tan\beta=\frac{a}{b}+\fr  ac{b}{a}} and now just

    apply \arctan to both sides and remember that \displaystyle{\alpha+\beta=\frac{\pi}{2}}

    Tonio
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