Biết $z_1,z_2$  là hai nghiệm phức của phương trình $z^2-3z+4=0$. Tính $F = \frac{\overline{z_1^4}}{z_2}+\frac{\overline{z_2^4}}{z_1}$

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \mathrm{Ta} \mathrm{có} \left|\mathrm{z}\right|+{\mathrm{z}}^2=0 \\ \iff \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}+{(\mathrm{a}+\mathrm{bi})}^2=0\iff \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}+{\mathrm{a}}^2-{\mathrm{b}}^2+2\mathrm{abi}=0 \\ \iff \left\{\begin{array}{l}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}+{\mathrm{a}}^2-{\mathrm{b}}^2=0 (*)\\ 2\mathrm{ab}=0\end{array}\right. \\ \mathrm{TH}1: \mathrm{a}=0 \mathrm{thì} (*)\iff \sqrt{{\mathrm{b}}^2}-{\mathrm{b}}^2=0 \\ \iff \left|\mathrm{b}\right|={\mathrm{b}}^2 \\ \iff {\mathrm{b}}^2=\mathrm{b} \mathrm{hoặc} {\mathrm{b}}^2=-\mathrm{b} \\ \iff \left\{\begin{array}{l}\mathrm{b}=0\\ \mathrm{b}=\pm 1\end{array}\right. \\ \Rightarrow \mathrm{Có} 3 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \\ \mathrm{TH}2: \mathrm{b}=0 \mathrm{thì} (*)\iff \sqrt{{\mathrm{a}}^2}+{\mathrm{a}}^2=0\iff \mathrm{a}=0 \\ \mathrm{KL}: \mathrm{có} 3 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \mathrm{yêu} \mathrm{cầu} \mathrm{bài} \mathrm{toán} \mathrm{là} \mathrm{z}=0; \mathrm{z}=\pm \mathrm{i} \end{gathered}$$

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: {\left(1+\mathrm{i}\right)}^2=2\mathrm{i} \\ \mathrm{Xét} \mathrm{đáp} \mathrm{án} \mathrm{A}: \\ {\left(1+\mathrm{i}\right)}^{2000}={\left(2\mathrm{i}\right)}^{1000}=2^{1000}.{\mathrm{i}}^{1000}=2^{1000}.{\left({\mathrm{i}}^2\right)}^{500}=2^{1000} \\ \Rightarrow \mathrm{Phần} \mathrm{thực} \mathrm{là} 2^{1000} >0 \left(\mathrm{Loại}\right) \\ \mathrm{Xét} \mathrm{đáp} \mathrm{án} \mathrm{B}: \\ {\left(1+\mathrm{i}\right)}^{2018}={\left(2\mathrm{i}\right)}^{1009}=2^{1009}.\mathrm{i}.{\mathrm{i}}^{1008}=2^{1009}.\mathrm{i}.{\left({\mathrm{i}}^2\right)}^{504}=2^{1009}\mathrm{i} \\ \Rightarrow \mathrm{Phần} \mathrm{thực} \mathrm{là} 0 \left(\mathrm{loại}\right) \\ \mathrm{Xét} \mathrm{đáp} \mathrm{án} \mathrm{C}: \\ {\left(1+\mathrm{i}\right)}^{2019}={\left[{\left(1+\mathrm{i}\right)}^2\right]}^{1009}.\left(1+\mathrm{i}\right)=2^{1009}.\mathrm{i}.\left(1+\mathrm{i}\right) \\ =-2^{1009}+2^{1009}.\mathrm{i} \\ \Rightarrow \mathrm{Phần} \mathrm{thực} \mathrm{là} -2^{1009}<0 \left(\mathrm{chọn}\right) \\ \mathrm{Xét} \mathrm{đáp} \mathrm{án} \mathrm{D}: \\ {\left(1+\mathrm{i}\right)}^{2040}={\left(2\mathrm{i}\right)}^{1020}=2^{1020}.{\mathrm{i}}^{1020}=2^{1020}.{\left({\mathrm{i}}^2\right)}^{510}=2^{1020} \\ \Rightarrow \mathrm{Phần} \mathrm{thực} \mathrm{là} 2^{1020}>0 \left(\mathrm{loại}\right) \end{gathered}$$Đáp án C