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Math Help - Complex Geometry

  1. #1
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    Complex Geometry

    The points A and B in the complex plane correspond to complex nos. z1 and z2 respectively. Both triangle OAP and OBQ are right angled isosceles triangles.

    i) Explain why P correspond to (1+i)z1: Done
    ii) M is midpoint of PQ, what complex no. correspond to M

    Here is my attempt, which conflict with solutions

    Q reqpresents z2(1-i), by rotation.

    Since M is mid point of QP, then QM = 1/2 (p-q)

    =1/2 (z1(1+i)-z2(1-i))
    ...
    =1/2(z1-z2) + i/2 (z1+ z2)

    But QM = M - z2
    z2 + QM = M
    therefore,
    M = 1/2 ((z1+z2) + i(z1 + z2))

    However solutions has 1/2(z1+z2 + i(z1-z2))

    Thanks in advance.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Lukybear View Post
    The points A and B in the complex plane correspond to complex nos. z1 and z2 respectively. Both triangle OAP and OBQ are right angled isosceles triangles.

    i) Explain why P correspond to (1+i)z1: Done
    ii) M is midpoint of PQ, what complex no. correspond to M

    Here is my attempt, which conflict with solutions

    Q reqpresents z2(1-i), by rotation.
    Why does Q not correspond to (1+i)z2, given that you have already accepted that P corresponds to (1+i)z1, accordingt to i)?
    It seems to me that you are getting P by rotating A around 0 by +90 degrees, whereas you are getting Q by rotating B around 0 by -90 degrees. Which is right: rotating by +90 or rotating by -90 degrees?

    Since M is mid point of QP, then QM = 1/2 (p-q)

    =1/2 (z1(1+i)-z2(1-i))
    ...
    =1/2(z1-z2) + i/2 (z1+ z2)

    But QM = M - z2
    z2 + QM = M
    therefore,
    M = 1/2 ((z1+z2) + i(z1 + z2))

    However solutions has 1/2(z1+z2 + i(z1-z2))

    Thanks in advance.
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  3. #3
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    Sorry to classify, P and Q are separate isoceles right triangles, with P in 1st quad and Q in 4th quad. So for this case lets assume my P and Q are correct. What about resulting solution?

    So this is what it would look like. Sorry for confusion.
    Attached Thumbnails Attached Thumbnails Complex Geometry-complex-rotation.jpg  
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Lukybear View Post
    Sorry to classify, P and Q are separate isoceles right triangles, with P in 1st quad and Q in 4th quad. So for this case lets assume my P and Q are correct. What about resulting solution?

    So this is what it would look like. Sorry for confusion.
    The midpoint M is M=\frac{P+Q}{2} (I am treating points like complex numbers for simplicity here).

    Thus you get

    M=\frac{P+Q}{2}=\frac{(1+i)z_1+(1-i)z2}{2}=\frac{(z_1+z_2)+i(z_1-z_2)}{2}

    You can get the same result like this
    M=P+\frac{1}{2}\cdot (Q-P)=(1+i)z_1+\frac{1}{2}\cdot\big((1+i)z_2-(1-i)z_1\big)=\ldots
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  5. #5
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    Thanks for that.
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