For $z_1, z_2 \in \mathbb C$ prove that $|z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2)$, and interpret this result geometrically.

Question

For $z_1, z_2 \in \mathbb C$ prove that $|z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2)$, and interpret this result geometrically.

We have \begin{align*} |z_1+z_2|^2+|z_1-z_2|^2&=(z_1+z_2)(\overline{z_1+z_2})+(z_1-z_2)(\overline{z_1-z_2})\\ &=(z_1+z_2)(\overline{z_1}+\overline{z_2})+(z_1-z_2)(\overline{z_1}-\overline{z_2})\\ &=z_1\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_2}+\overline{z_1}z_2+z_1\overline{z_1}+z_2\overline{z_2}-z_1\overline{z_2}-\overline{z_1}z_2\\ &=2z_1\overline{z_1}+2z_2\overline{z_2}\\ &=2|z_1|^2+2|z_2|^2\\ &=2(|z_1|^2+|z_2|^2). \end{align*}

However, I am having trouble "interpreting this result geometrically"?

What does this mean?

Answer 1

in components: $$ (x_1+x_2)^2 +(y_1+y_2)^2 + (x_1-x_2)^2 +(y_1-y_2)^2 = 2(x_1^2 + y_1^2 + x_2^2 + y_2^2) $$ if a parallelogram has sides $a,b$ making an angle of $\theta$ then the cosine rule tells us that if the two diagonals have lengths $d_1, d_2$ then $$ d_1^2 = a^2 +b^2 - 2ab\cos\, \theta \\ d_2^2 = a^2 +b^2 + 2ab\cos\, \theta $$ now add the two

Answer 2

User Al Jebr has proven the parallelogram law:

https://en.wikipedia.org/wiki/Parallelogram_law (Robert Z).

Consider the complex plane:

Draw the vectors $\vec z_1$, $\vec z_2$, not collinear from the origin.

Let $\vec OA:= \vec z_1$; $\vec OC := \vec z_2$.

Complete the parallelogram $OABC$ by drawing a parallel through $A$, parallel to $\vec z_2$, and a parallel through C, parallel to $\vec z_1$.

These $2$ parallels intersect at point $B$.

$OABC$ is a parallelogram.

$|\vec z_1+\vec z_2| = \overline{OB}=$

length of diagonal $OB$.

$|\vec z_1- \vec z_2| =\overline{OC}=$

length of diagonal $OC$.

$|\vec z_1| = \overline{OA} =$ length of side $OA$.

$|\vec z_2| = \overline{OC} =$ length of side $OC$.

Parallelogram law:

$[\overline{OB}]^2 +[\overline{AC}]^2 = $

$2(\overline{OA})^2 +2(\overline{OC})^2$.