Complex Numbers · Locus of complex numbers

Let

S = \left\{z \in \mathbb{C} : \left\lvert\dfrac{z - 6i}{z - 2i}\right\rvert = 1 \text{ and } \left\lvert\dfrac{z - 8 + 2i}{z + 2i}\right\rvert = \dfrac35\right\}

. Then

\displaystyle\sum_{z \in S}\lvert z\rvert^2

is equal to

We corrected a printing error in the original paper

Source printed the summation index as “ $z \in s$ ” (lower-case $s$ ; the set is $S$ ).

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