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Thread: [SOLVED] Geometric Interpretation

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    [SOLVED] Geometric Interpretation

    I suppose this might fall under the heading of Pre-Calc, but since I got it out of a book on Complex Analysis I thought I'd put it here.

    The problem is this:
    Prove the identity:
    $\displaystyle |z_1 + z_2| ^2 + |z_1 - z^2|^2 = 2(|z_1|^2 + |z_2|^2)$
    and interpret it geometrically.
    (where $\displaystyle z_1$ and $\displaystyle z_2$ are complex numbers.)

    I was able to easily prove the identity. But what is the geometric interpretation of it?

    -Dan
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    It seems to be the geometric interpretation of the law of cosine.

    1)Let $\displaystyle z_1$ be the vector seen below.
    2)Let $\displaystyle z_2$ be the other vector.
    3)Then $\displaystyle z_1+z_2$ is the vector sum.
    4)And $\displaystyle -z_2$ is the reflection of the vector through the x-axis.
    5)Then $\displaystyle z_1 - z_2$ is the sum of those two vectors.

    Let $\displaystyle a$ be the length of $\displaystyle z_1+z_2$ and let $\displaystyle b$ be the length of $\displaystyle z_1-z_2$ and let $\displaystyle c$ be the length of the line joining the tips of $\displaystyle z_1+z_2$ and $\displaystyle z_1-z_2$.

    Then,
    $\displaystyle c^2 = a^2 + b^2 - 2ab\cos \theta$
    Now $\displaystyle a = |z_1+z_2|$ and $\displaystyle b=|z_1-z_2|$ and $\displaystyle c = |(z_1+z_2) - (z_1 - z_2)| = |2z_2|$.
    This means,
    $\displaystyle 2|z_2| = |z_1+z_2|^2 + |z_1-z_2|^2 - 2 |z_1^2 - z_2^2|\cos \theta$.
    If we can show that $\displaystyle |z_1^2 - z_2^2|\cos \theta = 2|z_1|$ then, $\displaystyle 2(|z_1|^2+|z_2|^2) = |z_1+z_2|^2 + |z_1-z_2|^2$.

    This seems to be a Needham type of problem.
    Attached Thumbnails Attached Thumbnails [SOLVED] Geometric Interpretation-picture1.jpg  
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    The points $\displaystyle 0,\; z_1,\; z_2,\; z_1 + z_2$ form a parallelogram with sides of length |z_1| and |z_2|, and diagonals of length |z_1+z_2| and |z_1–z_2|. The geometrical interpretation is the theorem which says that the sum of the squares of the diagonals of a parallelogram is twice the sum of the squares of the sides.
    Last edited by Opalg; Jan 23rd 2008 at 01:59 AM. Reason: Clarity
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Opalg View Post
    The points $\displaystyle 0,\; z_1,\; z_2,\; z_1 + z_2$ form a parallelogram with sides of length |z_1| and |z_2|, and diagonals of length |z_1+z_2| and |z_1z_2|. The geometrical interpretation is the theorem which says that the sum of the squares of the diagonals of a parallelogram is twice the sum of the squares of the sides.
    Well, that explains it. If I have ever heard of this theorem I've completely forgotten it. Thanks!

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    It seems to be the geometric interpretation of the law of cosine.

    1)Let $\displaystyle z_1$ be the vector seen below.
    2)Let $\displaystyle z_2$ be the other vector.
    3)Then $\displaystyle z_1+z_2$ is the vector sum.
    4)And $\displaystyle -z_2$ is the reflection of the vector through the x-axis.
    5)Then $\displaystyle z_1 - z_2$ is the sum of those two vectors.

    Let $\displaystyle a$ be the length of $\displaystyle z_1+z_2$ and let $\displaystyle b$ be the length of $\displaystyle z_1-z_2$ and let $\displaystyle c$ be the length of the line joining the tips of $\displaystyle z_1+z_2$ and $\displaystyle z_1-z_2$.

    Then,
    $\displaystyle c^2 = a^2 + b^2 - 2ab\cos \theta$
    Now $\displaystyle a = |z_1+z_2|$ and $\displaystyle b=|z_1-z_2|$ and $\displaystyle c = |(z_1+z_2) - (z_1 - z_2)| = |2z_2|$.
    This means,
    $\displaystyle 2|z_2| = |z_1+z_2|^2 + |z_1-z_2|^2 - 2 |z_1^2 - z_2^2|\cos \theta$.
    If we can show that $\displaystyle |z_1^2 - z_2^2|\cos \theta = 2|z_1|$ then, $\displaystyle 2(|z_1|^2+|z_2|^2) = |z_1+z_2|^2 + |z_1-z_2|^2$.

    This seems to be a Needham type of problem.
    I had considered Law of Cosines, so I'll have to look at it again. However I'd like to point out that in line 4, $\displaystyle -z_2$ is not a reflection over the x-axis, but an inversion through the origin. But the construction could work anyway.

    By the way, "Needham" problem? I googled it but mainly came up with only Needham, MA.

    -Dan
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