Thread: prove of equation with complex numbers

1. prove of equation with complex numbers

Here is the equation to prove:
$\displaystyle |a-b|^2=|a|^2-2 re \hat{a} b+|b|^2$
I tried with something like this:
$\displaystyle |z|^2=z \hat{z}$
$\displaystyle z+\hat{z}=2rez$
but i still can not get the right result.

a,b are different complex numbers.

2. Originally Posted by Vermax
Here is the equation to prove:
$\displaystyle |a-b|^2=|a|^2-2 re \hat{a} b+|b|^2$
I tried with something like this:
$\displaystyle |z|^2=z \hat{z}$
$\displaystyle z+\hat{z}=2rez$
but i still can not get the right result.

a,b are different complex numbers.
Multiply out $\displaystyle (a - b)(\hat a - \hat b)$ then note that $\displaystyle \hat a b = \hat {a \hat b}$.

3. Thank you, now almost everything is clear for me except one thing. How do you know that:
$\displaystyle |a-b|^2=|a|^2-2 re \hat{a} b+|b|^2=(a - b)(\hat a - \hat b)$?

I am sure you, as an expert, presented me it correctly, but i am just curious where it comes from. How do you know these identities? (they are not the common ones i think...)

4. Originally Posted by Vermax
Thank you, now almost everything is clear for me except one thing. How do you know that:
$\displaystyle |a-b|^2=|a|^2-2 re \hat{a} b+|b|^2=(a - b)(\hat a - \hat b)$?

I am sure you, as an expert, presented me it correctly, but i am just curious where it comes from. How do you know these identities? (they are not the common ones i think...)
I'm going to use $\displaystyle \overline z$ for conjugate, it's easier to use (see below).

You already have that $\displaystyle |z|^2 = z \overline z$.

So $\displaystyle |a - b|^2 = (a-b) \overline{(a-b)}$.

But you also have that $\displaystyle \overline{(a-b)} = \overline a - \overline b$ (it's a standard result, but simple to prove).

So $\displaystyle |a - b|^2 = (a-b) (\overline a - \overline b)$.

Multiply this out, you get

$\displaystyle a \overline a + b \overline b - a \overline b - \overline a b$.

But as $\displaystyle \overline a b = \overline {a \overline b}$ (which you should be able to prove, it's very straightforward), this works out as

$\displaystyle a \overline a + b \overline b - (\overline a b + \overline {a \overline b})$.

The result follows from $\displaystyle z + \overline z = 2 Re z$.

5. Thank you for that post. I did not think that complex numbers can be that complicated
I proved these two identities that I was supposed to, but still I think I should work on beeing more "fluent" in it again