Cho số phức z = a + bi(a,b thuộc R). Biết {M} biểu diễn số phức

* Hướng dẫn giải

Đáp án C

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: \mathrm{F}=4\mathrm{a}+3\mathrm{b} \Rightarrow \mathrm{a}=\frac{\mathrm{F}-3\mathrm{b}}{4} \\ \mathrm{Lại} \mathrm{có}: {\left(\mathrm{a}-4\right)}^2+{\left(\mathrm{b}-3\right)}^2=9 \\ \iff {\left(\frac{\mathrm{F}-3\mathrm{b}}{4}-4\right)}^2+{\left(\mathrm{b}-3\right)}^2=9 \\ \iff {\left(\mathrm{F}-3\mathrm{b}-16\right)}^2+16{\left(\mathrm{b}-3\right)}^2-144=0 \\ \iff {\left(\mathrm{F}-16\right)}^2-6\mathrm{b}\left(\mathrm{F}-16\right)+9{\mathrm{b}}^2+16{\mathrm{b}}^2-96\mathrm{b}+144-144=0 \\ \iff 25{\mathrm{b}}^2-6\mathrm{bF}+{\mathrm{F}}^2-32\mathrm{F}+256=0 \\ ∆\text{'}=9{\mathrm{F}}^2-25\left({\mathrm{F}}^2-32\mathrm{F}+256\right)=-16{\mathrm{F}}^2+800\mathrm{F}-6400=-16\left({\mathrm{F}}^2-50\mathrm{F}+400\right) \\ ∆\text{'}\ge 0\iff -16\left({\mathrm{F}}^2-50\mathrm{F}+400\right)\ge 0\iff 10\le \mathrm{F}\le 40 \end{gathered}$$