Let S=zC: |z-6iz-2i|=1, |z-8+2iz+2i|=35.

Question

Let $S=\left\{z\in\mathbb{C}: \left|\dfrac{z-6i}{z-2i}\right|=1,\ \left|\dfrac{z-8+2i}{z+2i}\right|=\dfrac35\right\}$. Then $\displaystyle\sum_{z\in S}|z|^2$ is equal to:

Accepted Answer

The first equation $|z-6i|=|z-2i|$ means $z$ is equidistant from $(0,2)$ and $(0,6)$, so $z$ lies on the perpendicular bisector: $z=x+4i,\ x\in\mathbb R$.

Substitute into the second equation, $|z-8+2i|=\dfrac35|z+2i|$:
$$|x+4i-8+2i| = \frac35|x+4i+2i| \implies |(x-8)+6i| = \frac35|x+6i|.$$

$$5\sqrt{(x-8)^2+36}=3\sqrt{x^2+36}$$
$$25\big(x^2-16x+64+36\big)=9(x^2+36)$$
$25x^2-400x+2500 = 9x^2+324$
$$16x^2-400x+2176=0 \implies x^2-25x+136=0 \implies (x-17)(x-8)=0.$$

So $x=17$ or $x=8$, giving $z_1=17+4i,\ z_2=8+4i$.

$$\sum_{z\in S}|z|^2=|17+4i|^2+|8+4i|^2 = (289+16)+(64+16)=305+80=385.$$

Solutions · 1