Find the sum of the elements of $A=\bigg\{ z\in\mathbb{C}\bigg|z\cdot \overline{z}=2, \bigg| \dfrac{2z+3}{z-3i} \bigg|=1 \bigg\}$.

Question

Given the set:

$$A=\bigg\{ z\in\mathbb{C}\bigg|z\cdot \overline{z}=2, \bigg| \dfrac{2z+3}{z-3i} \bigg|=1 \bigg\}$$

I have to find the following sum:

$$S=\sum_{z\in A}z$$

I tried to find $z$ by writing it as:

$$z=a+bi$$

with $a, b \in \mathbb{R}$. $z \cdot \overline{z} = 2$ means that

$$\hspace{9cm} a^2+b^2=2 \hspace{8.3cm}(1)$$

So we have the first condition. Then what I did was: $$\bigg| \dfrac{2z+3}{z-3i} \bigg| = 1$$

$$\bigg| \dfrac{(2z+3)(z+3i)}{z^2-9i^2} \bigg| = 1$$

$$\bigg| \dfrac{2z^2+6zi+3z+9i}{z^2+9} \bigg| = 1$$

$$\bigg| \dfrac{2z^2+3z+(6z+9)i}{z^2+9} \bigg| = 1$$

But I got stuck here. I don't see how I could find the elements of $A$.

Answer 1

$z \cdot \overline{z}=2$ means that $|z|^2=2$. From the second equation:

$$|2z+3|^2=|z-3i|^2\Rightarrow (2z+3)(2\overline{z}+3)=(z-3i)(\overline{z}+3i)$$

and upon expanding

$$4|z|^2+6(z+\overline{z})+9=|z|^2+3i(z-\overline{z})+9$$

or equivalently

$$2+2(z+\overline{z})=i(z-\overline{z})$$

Multiply by $z$ and use $z\cdot \overline{z}=2$, to arrive at the second degree equation:

$$(2-i)z^2+2z+4+2i=0$$

Can you end it from here?

Answer 2

Here's a more conceptual approach that might lead to insights wider than just this problem.

Define $m(z) = (2z+3)/(z-3i)$. This is a Möbius transformation, and the second condition picks out the complex numbers $z$ such that $m(z)$ lies on the unit circle. There's a nice way to find this set explicitly, because the inverse of any Möbius transformation $(az+b)/(cz+d)$ is the related Möbius transformation $(dz-b)/(-cz+a)$.

In this case, $m^{-1}(z) = (-3i-3)/(-z+2)$, and so the set of $z$ satisfying the second condition is simply the image of the unit circle under $m^{-1}$. Furthermore, the image of a circle-or-line under a Möbius transformation is always a circle-or-line; so the individual calculations $$ m^{-1}(i) = 0, \quad m^{-1}(1) = -3-3i, \quad m^{-1}(-1) = -1+i $$ are enough to tell us that the image of the unit circle under $m^{-1}$ is exactly the circle through those three complex numbers—which turns out to be the circle of radius $\sqrt5$ centered at $-2-i$.

So the set of $z$ satisfying both original conditions is comprised simply of the intersection between this circle and the circle of radius $\sqrt2$ centered at $0$. This becomes a standard analytic geometry problem (whose exact solution I won't spoil for you).