Which $z\in\mathbb{C}$ satisfy the equation $|z-i|=\sqrt{2}|\bar{z}+1|$?

Question

Let $S$ be the set of all complex numbers $z$ satisfying the rule $$|z-i|=\sqrt{2}|\bar{z}+1|$$

Show that $S$ contains points on a circle.

My attempt,

By substituting $z = x + yi$, and squaring both sides. But I can't get the circle form.

Answer 1

Let $z=x+iy$. Note that $\overline{z}=x-iy$.

$$\begin{align} |z-i|&=\sqrt{2}|\overline{z}+1|\\ |z-i|^2&=2|\overline{z}+1|^2\\ |x+i(y-1)|^2&=2|(x+1)-iy|^2\\ x^2+(y-1)^2&=2(x+1)^2+y^2)\\ x^2+4x+2+y^2+2y-1&=0\\ (x+2)^2+(y+1)^2&=2^2 \end{align}$$

Answer 2

Altenative if you are familiar with Circles of Apollonius,

$$|z-i| = \sqrt{2}|\bar{z}+1|=\sqrt{2}|z+1|=\sqrt{2}|z-(-1)|$$

Here $i$ and $-1$ are the foci.

Answer 3

Another way is using the complex equation of a circle and complex square completion:

• $|z-a| = r$ is the circle with radius $r$ around $a = a_x + ia_y$ $$|z-a|^2 = |z|^2 + |a|^2 - 2Re(\bar z a)$$
• Furthermore, note that $|\bar z+1| = |z+1|$
• $|z-i|=\sqrt{2}|\bar{z}+1|$ becomes $$|z|^2 + 1 -2Re(\bar z i)=2(|z|^2 + 1 +2Re(\bar z \cdot 1))\Leftrightarrow$$ $$|z|^2 + 1 + 2Re(\bar z (2+i)) \stackrel{square\; compl.}{=} |z-(-2-i)|^2+1-5 = |z-(-2-i)|^2-4 = 0 \Leftrightarrow$$ $$\boxed{|z-(-2-i)|^2 = 4}$$ This is a circle around $a= -2-i$ with radius $r= 2$.