Absolute value of a complex number proof

Question

Ok, so I have the following proof. Let $z$ and $w$ be complex numbers. Prove $\lvert z+w \rvert ^2 + \lvert z-w \rvert^2 = 2[\lvert z \rvert^2 + \vert w \rvert^2]$. Using $\vert z \rvert^2=z\bar{z}$, I have said

$\lvert z+w \rvert ^2 + \lvert z-w \rvert^2 = 2(\lvert z \rvert^2) + \vert w \rvert^2)$

$(z+w)\bar{(z+w)}+(z-w)(\bar{(z-w)}$

$(z+w)(z-w)+(z-w)(z+w)$

$z^2 - w^2 + z^2 - w^2$

$z^2 + z^2 + -w^2 + -w^2$

$2(z^2) + 2(-w^2)$

$2[z^2 + (-w)^2]$

$2[\lvert z \rvert^2 + \vert w \rvert^2]$

I guess my question is, is my logic correct? I am not so skilled with proof writing and though I feel that this is correct, I have been known to be wrong in the past. Just need some insight. Thank you.

Answer 1

ok, I have taken your comments into consideration and think I am ready to state my new answer.

$\lvert z+w\rvert^2 + \lvert z-w \rvert^2 = 2[\lvert z \rvert^2 + \lvert w \rvert^2$

$(z+w)(\bar{z}+\bar{w})+(z-w)(\bar{z}-\bar{w})$

$z\bar{z}+z\bar{w}+w\bar{z}+w\bar{w}+z\bar{z}-z\bar{w}-w\bar{z}+w\bar{w}$

$\lvert z \rvert^2 + \lvert z \rvert^2 + \lvert w \rvert^2 + \lvert w \rvert^2$

$2\lvert z \rvert^2 + 2\lvert w \rvert^2$

$2[\lvert z \rvert^2 + \lvert w \rvert^2]$

How is this?

Answer 2

Your approach to the problem is good, but you've made some algebraic errors.

Keep in mind that for complex numbers $z$ and $w$, $\overline{z+w} \neq z - w$ - you're probably confusing it with $\overline{z} = \overline{a+bi} = a-bi$.

Also, in general $(-w^2) \neq (-w)^2 = w^2$, and $z^2 \neq |z|^2$ so your last few lines are not valid, but this harldy matters since the main problem stems from the point mentioned in the previous paragraph.

Hope that helps and let me know if you have any other question!