Proof: $|\alpha + \beta|^2+|\alpha - \beta|^2=2(|\alpha|^2+| \beta|^2)$

Question

Proof: $|\alpha + \beta|^2+|\alpha - \beta|^2=2(|\alpha|^2+| \beta|^2)$

In this proof, my teacher uses the following step:

Say $\alpha=a+ib$

$|\alpha|^2=\alpha^2=(a+ib)^2=\sqrt{a^2+b^2}^2=a^2+b^2$

I should calculate:

$|\alpha|^2=\alpha^2=(a+ib)^2=a^2+2iab+i^2b^2=a^2+2iab-b^2$

Where am I wrong?

An other example is: $(2+2i)^2$

My solution:

$(2+2i)^2=2(1+i)^2=2(1+2i-1)=2(2i)=4i$ but the solution is $8i$

Answer 1

Your conclusion would be correct if the identity $|z|^2=z^2$ holds in the complex numbers, but it doesn't.

The teacher should have written $$ |\alpha|^2=\alpha\bar{\alpha}=(a+ib)(a-ib)=a^2-i^2b^2=a^2+b^2 $$

Anyway, you don't need that for proving the statement, provided you use the correct definition of modulus and the fact that $\overline{\alpha\pm\beta}=\bar{\alpha}\pm\bar{\beta}$: $$ |\alpha+\beta|^2+|\alpha-\beta|^2= (\alpha+\beta)(\bar{\alpha}+\bar{\beta}) + (\alpha-\beta)(\bar{\alpha}-\bar{\beta}) $$ and then just do the algebra.

Answer 2

Hint

Just use the fact that $$|z|^2=z\bar z.$$

Answer 3

1)$|\alpha +\beta|^2=$

$(\alpha +\beta)(\overline{\alpha +\beta})=$

$(\alpha +\beta)(\overline{\alpha} +\overline{\beta}) =$

$|\alpha|^2 +|\beta|^2 +\alpha\overline{\beta}+\beta\overline {\alpha}.$

2)$ |\alpha -\beta|^2$:

Replace $\beta$ by $-\beta$ in expression 1).

Add 1)+2):The cross terms cancel and you get the desired result.

Note:

1)$\overline{ \gamma}$ : complex conjugate of $\gamma$.

2)$\overline{-\beta} =-\overline{\beta}.$