Question

Pentru r>0, fie multimea M={z e C ; |z|=1 si |z-3i|=r}. Fie A={r>0; M are un singur element}. Sa se determine suma S a elementelor multimii A.

Answer 1

$M = \Big\{z\in \mathbb{C};\,\,\,|z| =1\,\,\,\text{si}\,\,\, |z-3i| = r\Big\} \\ \\ A = \Big\{r>0;\,\,\,\text{card}(M) = 1\Big\} \\ \\ \\ |z|=1\Rightarrow \sqrt{z\overline{z}} = 1\Rightarrow z\overline{z} = 1 \\ \\ |z-3i| =r\Rightarrow \sqrt{(z-3i)(\overline{z-3i})} = r\Rightarrow (z-3i)(\overline{z-3i}) = r^2 \Rightarrow \\ \\ \Rightarrow (z-3i)[\overline{z}+(\overline{-3i})] = r^2 \Rightarrow (z-3i)(\overline{z}+3i) = r^2 \Rightarrow$

$\Rightarrow z\overline{z}+3iz-3i\overline{z}+9 = r^2 \Rightarrow 1+3i(z-\overline{z})+9 = r^2 \\ \\ z = a+bi\Rightarrow z-\overline{z} = 2bi \\ \\ \Rightarrow r^2 = 10+3i(2bi) \Rightarrow r^2=10-6b \Rightarrow b = \dfrac{10-r^2}{6} \\ \\\\ |z| = 1 \Rightarrow a^2+b^2 = 1 \Rightarrow a^2+\Big(\dfrac{10-r^2}{6}\Big)^2 = 1 \Rightarrow \\ \\ \Rightarrow a^2+\Big(\dfrac{10-r^2}{6}\Big)^2-1 = 0 \\ \\\\ \text{card}(M) = 1 \Rightarrow a^2 = 0 \Rightarrow a = 0 \Rightarrow\Big(\dfrac{10-r^2}{6}\Big)^2-1 = 0 \Rightarrow$

$\Big(\dfrac{10-r^2}{6}\Big)^2= 1 \Rightarrow \pm\Big(\dfrac{10-r^2}{6}\Big) = 1 \Rightarrow \\ \\ \Rightarrow \dfrac{10-r^2}{6} = 1 \Rightarrow 10-r^2 = 6 \Rightarrow r^2 = 4 \Rightarrow \boxed{r = 2} \\ \\ \Rightarrow -\dfrac{10-r^2}{6} = 1 \Rightarrow 10-r^2 = -6 \Rightarrow r^2 = 16 \Rightarrow \boxed{r = 4} \\ \\ \Rightarrow S = 2+4 \Rightarrow \boxed{S = 6}$