# Math Help - Help with complex numbers

1. ## Help with complex numbers

Show that $
\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.
$
\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
$
\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 $|a+b|^2=\left| a \right|^2 + 2\left| a \right|\left| b \right|+|b|^2$ which is wrong : if $b=-a$ the LHS equals 0 and the RHS equals $4|b|^2$. I don't think there is a rule that tells us how to expand $|a+b|^2$.

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

\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 ?