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Math Help - Complex Numbers

  1. #1
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    Complex Numbers

    I got this question in a complex numbers test and I could not figure it out. Any help would be appreciated.

    Question:

    R is a positive number and  z_1 and z_2 are complex numbers. Show that the points which represent respectively the numbers z_1 , z_2, \frac{z_1-iRz_2}{1-iR} form the vertices of a right angled triangle
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  2. #2
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    Quote Originally Posted by noreaction View Post
    I got this question in a complex numbers test and I could not figure it out. Any help would be appreciated.

    Question:

    R is a positive number and  z_1 and z_2 are complex numbers. Show that the points which represent respectively the numbers z_1 , z_2, \frac{z_1-iRz_2}{1-iR} form the vertices of a right angled triangle

    Put z_3=\frac{z_1-iRz_2}{1-iR} then the lengths of the sides of the triangle are |z_1-z_2| , |z_2-z_3| and |z_1-z_3|.

    Now you need to show that these satisfy Pythagoras's theorem, and then
    by the converse of Pythagoras's the points form a right triangle.

    RonL
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  3. #3
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    Thank you. I can't believe I didn't think of that.
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  4. #4
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    Quote Originally Posted by noreaction View Post
    R is a positive number and  z_1 and z_2 are complex numbers. Show that the points which represent respectively the numbers z_1 , z_2, \frac{z_1-iRz_2}{1-iR} form the vertices of a right angled triangle.
    Let z_3 = \frac{z_1-iRz_2}{1-iR}.

    The displacement from z_1 to z_3 is given by z_3-z_1 = \frac{z_1-iRz_2}{1-iR} - z_1 = \frac{iR(z_1-z_2)}{1-iR}.

    The displacement from z_2 to z_3 is given by z_3-z_2 = \frac{z_1-iRz_2}{1-iR} - z_2 = \frac{z_1-z_2}{1-iR}.

    Therefore z_3-z_1 = iR(z_3-z_2). So to get from z_3-z_2 to z_3-z_1 you have to multiply it by R (which changes its length but not its direction) and then multiply it by i (which has the effect of rotating it through a right angle. This shows that the line from z_1 to z_3 is perpendicular to the line from z_2 to z_3.
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