$z \in \mathbb{C}^-$ where $\mathbb{C}^- := \{z \in \mathbb{C} : Re(z) < 0\}$ identity

Question

Why is $|2-z|^2 = 4 + |z|^2 - 4Re(z)$,

where $z \in \mathbb{C}^-$ and $\mathbb{C}^- := \{z \in \mathbb{C} : Re(z) < 0\}$.

Why is this true?

Answer 1

hint

$$|2-z|^2=(2-z)(2-\bar{z})$$

$$=4+z\bar{z}-2(z+\bar{z})$$ $$=4+|z|^2-2(x+iy+x-iy)$$

Answer 2

This identity is true for any $z\in\mathbb C$

$|2-z|^2=(2-z)\overline{(2-z)}=(2-z)(2-\overline z)=4 + z\overline z - 2(z+\overline z)=4+|z|^2-4Re(z)$