I was doing a question on complex numbers (see the linked image) and I'm quite unsure why you can say that $|α + β|² = (α + β)(\overline{\alpha} + \overline{\beta})$, and similarly why $|α - β|² = (α - β)(\overline{\alpha} - \overline{\beta})$.
Isn't $|α + β|² = (|α + β|)(|α + β|)$? Why even consider using the conjugates?
Is there a proof for why $|α + β|² = (α + β)(\overline{\alpha} - \overline{\beta})$
For any(!) complex number $z = a+bi$, we have $$ z\bar z = (a+bi)(a-bi) = a^2 - (bi)^2 = a^2 + b^2 = \sqrt{a^2 + b^2}^2 = |z|^2 $$ Hence, for $z = \alpha \pm \beta$: $$ |\alpha \pm \beta|^2 = (\alpha \pm \beta)(\overline{\alpha \pm \beta}) = (\alpha \pm \beta)(\bar\alpha \pm \bar\beta) $$
It suffices to show that $|z|^2 = z \cdot \overline z$, where $z$ is a complex number. And this is easily shown by writing $z= x+iy$ and expanding.
\begin{align*} z\cdot \overline z &= (x+iy)(x-iy)\\ &= x^2 -ixy +ixy -i^2 y^2\\ &= x^2 - (-1)y^2\\ &= x^2 + y^2 \end{align*} And recall that $$ |z| = \sqrt{x^2+y^2}. $$
Therefore $z\cdot \overline z = x^2+y^2 = |z|^2.$
Why even consider using the conjugates?
Because that form of the expression can be helpful in proving other equalities. A good example is the one in the image you linked.
