# Thread: Help with complex numbers

1. ## Help with complex numbers

Show that $\displaystyle \left| {a + b} \right|^2 + \left| {a - b} \right|^2 = 2\left( {\left| a \right|^2 + \left| b \right|^2 } \right)$

where a and b are complex numbers. Interpret this results geometrically.
$\displaystyle \begin{array}{l} \left| {a + b} \right|^2 + \left| {a - b} \right|^2 \\ = \left| a \right|^2 + 2\left| a \right|\left| b \right| + \left| b \right|^2 + \left| a \right|^2 - 2\left| a \right|\left| b \right| + \left| b \right|^2 \\ = 2\left| a \right|^2 + 2\left| b \right|^2 \\ = RHS \\ \end{array}$

I don't know how to interpret this result geometrically though, any help would be appreciated.

2. Hello,
Originally Posted by free_to_fly
$\displaystyle \begin{array}{l} \left| {a + b} \right|^2 + \left| {a - b} \right|^2 \\ = \left| a \right|^2 + 2\left| a \right|\left| b \right| + \left| b \right|^2 + \left| a \right|^2 - 2\left| a \right|\left| b \right| + \left| b \right|^2 \\ \end{array}$
You're saying that $\displaystyle |a+b|^2=\left| a \right|^2 + 2\left| a \right|\left| b \right|+|b|^2$ which is wrong : if $\displaystyle b=-a$ the LHS equals 0 and the RHS equals $\displaystyle 4|b|^2$. I don't think there is a rule that tells us how to expand $\displaystyle |a+b|^2$.

However, I'm sure you know that if $\displaystyle z$ is a complex number then $\displaystyle |z|^2=z\overline{z}$ so

\displaystyle \begin{aligned} \left| {a + b} \right|^2 + \left| {a - b} \right|^2&=(a + b)\overline{(a + b)} + (a - b)\overline{(a-b)}\\ &=\ldots \end{aligned}

I don't know how to interpret this result geometrically though, any help would be appreciated.
Take a look at equation #10 of Parallelogram -- from Wolfram MathWorld. It looks like the one you're asked to show, doesn't it ?