How can $|z-(i \bar w)| = |z-(-i \bar w)|$ where z and w are complex numbers?

Question

How can $|z-(i \bar w)| = |z-(-i \bar w)|$ where z and w are complex numbers ? This is a step of the solution of a problem I'm doing and I don't understand why this is right. Isn't this like saying that $|z-a| = |z-(-a)| = |z+a| $ where z, a are complex numbers? The question specifies that |z|<= 1, |w|<=1 and gives the condition $|z+iw| = |z-i \bar w| = 2$, from which the sentence I'm asking about is derived.

Answer 1

What is true is that

$$|z-(i \bar w)| =|\overline{z-(i \bar w)}|= |\bar z-(-i w)|$$

but in general

$$|z-(-i \bar w)|\neq |\bar z-(-i w)|$$

Answer 2

You're correct, if no further conditions are imposed on $z$ and $w$. For example, if $z = i$ and $w = 1$, we would have

$$ 0 = |i-(i\cdot1)| = |i-(-i\cdot1)| = |2i| = 2 $$

which is false.

Answer 3

Recall the things $\overline{z_1z_2} = \overline{z}_1\overline{z}_2$, and $|z| = \sqrt{z\overline{z}}$, you will go $$|z - is| = \sqrt{(z-is)\overline{(z-is)}} = \sqrt{\overline{(\overline{z}+i\overline{s})}(\overline{z}+i\overline{s})} = |\overline{z}+i\overline{s}|.$$ You see, $$|z - i\overline{w}| = |z+i\overline{w}|$$ is hold for all complex numbers $z,\overline{w}$ that imaginary part of each of they is $0$.