[SOLVED] Geometric Interpretation

**topsquark** · January 22nd 2008, 04:25 PM

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:
$|z_1 + z_2| ^2 + |z_1 - z^2|^2 = 2(|z_1|^2 + |z_2|^2)$
and interpret it geometrically.

(where $z_1$ and $z_2$ are complex numbers.)

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

-Dan

**ThePerfectHacker** · January 22nd 2008, 07:04 PM

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

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

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

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

This seems to be a Needham type of problem.

**Opalg** · January 23rd 2008, 12:42 AM

The points $0,\; z_1,\; z_2,\; z_1 + z_2$ 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.

**topsquark** · January 23rd 2008, 09:26 AM

Originally Posted by Opalg

The points $0,\; z_1,\; z_2,\; z_1 + z_2$ 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.

Well, that explains it. If I have ever heard of this theorem I've completely forgotten it. Thanks!

-Dan

**topsquark** · January 23rd 2008, 09:29 AM

Originally Posted by ThePerfectHacker

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

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

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

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

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, $-z_2$ 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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