191 câu Bài tập số phức mức độ cơ bản, nâng cao cực hay có lời giải chi tiết(P3)

Đáp án D

Đáp án A

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \Rightarrow \overline{\mathrm{z}}=\mathrm{a}-\mathrm{bi} \\ \mathrm{Từ} \mathrm{giả} \mathrm{thiết}\Rightarrow \left|\mathrm{a}+5-(\mathrm{b}+12)\mathrm{i}\right|=9 \\ \iff {(\mathrm{a}+5)}^2+{(\mathrm{b}+12)}^2=81 \\ \Rightarrow \mathrm{Tập} \mathrm{hợp} \mathrm{các} \mathrm{điểm} \mathrm{biếu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z} \mathrm{năm} \mathrm{trên} \mathrm{đường} \mathrm{tròn} \mathrm{tâm} \mathrm{I}(-5;-12) \mathrm{có} \mathrm{R}=9 \end{gathered}$$

$$\begin{gathered} \mathrm{Từ} \mathrm{hình} \mathrm{vẽ} \mathrm{ta} \mathrm{thấy} {\left|\mathrm{z}\right|}_{\min } =\mathrm{OA}=\mathrm{OI}-\mathrm{IA}=\mathrm{OI}-\mathrm{R} \\ \Rightarrow {\left|\mathrm{z}\right|}_{\min }=\sqrt{{(-5)}^2+{(-12)}^2}-9=13-9=4 \end{gathered}$$

Đáp án B

$$\begin{gathered} \mathrm{w}=\frac{\mathrm{z}-1+2\mathrm{i}}{\mathrm{i}}=\frac{\mathrm{i}(\mathrm{a}+\mathrm{bi}-1+2\mathrm{i})}{{\mathrm{i}}^2}=\frac{\mathrm{i}(\mathrm{a}-1)-(\mathrm{b}+2)}{-1} \\ \Rightarrow \mathrm{w}=\mathrm{b}+2-\mathrm{i}(\mathrm{a}-1) \\ \Rightarrow \mathrm{Phần} \mathrm{ảo} \mathrm{của} \mathrm{w} \mathrm{là} 1-\mathrm{a} \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} \mathrm{và} \mathrm{w}=\mathrm{c}+\mathrm{di} (\mathrm{a};\mathrm{b};\mathrm{c};\mathrm{d}\in \mathrm{ℝ}) \\ \mathrm{Xét} \mathrm{đáp} \mathrm{án} \mathrm{A}: \\ \mathrm{z}=\overline{\mathrm{w}}\iff \left\{\begin{array}{l}\mathrm{a}=\mathrm{c}\\ \mathrm{b}=-\mathrm{d}\end{array}\iff \left\{\begin{array}{l}\mathrm{a}=\mathrm{c}\\ -\mathrm{b}=\mathrm{d}\end{array}\iff \overline{\mathrm{z}}=\mathrm{w} ( \mathrm{đúng} )\right.\right. \\ \mathrm{Xét} \mathrm{đáp} \mathrm{án} \mathrm{B} \mathrm{và} \mathrm{D} \\ \mathrm{z}=\overline{\mathrm{w}}\iff \left\{\begin{array}{l}\mathrm{a}=\mathrm{c}\\ \mathrm{b}=-\mathrm{d}\end{array}\right. \\ \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{c}}^2+{\mathrm{d}}^2\iff \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=\sqrt{{\mathrm{c}}^2+{\mathrm{d}}^2} \mathrm{hay} \left|\mathrm{z}\right|=\left|\overline{\mathrm{w}}\right| ( \mathrm{B} \mathrm{đúng}) \\ \mathrm{Mặt} \mathrm{khác}: {\mathrm{z}}^2=\overline{{\mathrm{w}}^2}\iff {(\mathrm{a}+\mathrm{bi})}^2=\overline{{(\mathrm{c}+\mathrm{di})}^2} \\ \iff {\mathrm{a}}^2-{\mathrm{b}}^2+2\mathrm{abi}={\mathrm{c}}^2-{\mathrm{d}}^2-2\mathrm{cdi} \\ \Rightarrow {\mathrm{c}}^2-{\mathrm{d}}^2-2\mathrm{cdi}={\mathrm{c}}^2-{\mathrm{d}}^2-2\mathrm{cdi} ( \mathrm{D} \mathrm{đúng}) \\ \Rightarrow \mathrm{C} \mathrm{sai} \end{gathered}$$

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \mathrm{PT}\iff \mathrm{a}+\mathrm{bi}+\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}+\mathrm{a}-\mathrm{bi}=0 \\ \iff 2\mathrm{a}+\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=0 \\ \iff \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=-2\mathrm{a} \\ \iff \left\{\begin{array}{l}\mathrm{a}<0\\ {\mathrm{a}}^2+{\mathrm{b}}^2=4{\mathrm{a}}^2\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{a}<0\\ 3{\mathrm{a}}^2={\mathrm{b}}^2\end{array}\right. \\ \iff -\mathrm{a}=\frac{\left|\mathrm{b}\right|}{\sqrt{3}} \\ \mathrm{Chọn} \mathrm{b}=\pm \sqrt{3}\Rightarrow \mathrm{a}=-1 \\ \Rightarrow \mathrm{z}=-1+\mathrm{i}\sqrt{3} \mathrm{hoặc} \mathrm{z}=-1-\mathrm{i}\sqrt{3} \end{gathered}$$Đáp án C

Đáp án C

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

Đáp án A

$$\begin{gathered} z^6=1 \\ \iff {\mathrm{z}}^6-1=0 \\ \iff \left({\mathrm{z}}^3+1\right)\left({\mathrm{z}}^3-1\right)=0 \\ \iff \left(\mathrm{z}+1\right)\left(\mathrm{z}-1\right)\left({\mathrm{z}}^2-\mathrm{z}+1\right)\left({\mathrm{z}}^2+\mathrm{z}+1\right)=0 \\ \iff \left\{\begin{array}{l}\mathrm{z}=1\\ \mathrm{z}=-1\\ \mathrm{z}=\frac{1}{2}+\mathrm{i}\frac{\sqrt{3}}{2}\\ \mathrm{z}=\frac{1}{2}-\mathrm{i}\frac{\sqrt{3}}{2}\\ \mathrm{z}=\frac{-1}{2}+\mathrm{i}\frac{\sqrt{3}}{2}\\ \mathrm{z}=\frac{-1}{2}-\mathrm{i}\frac{\sqrt{3}}{2}\end{array}\right. \\ \Rightarrow \mathrm{Các} \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{nghiệm} \mathrm{của} \mathrm{PT} \mathrm{là} \mathrm{A}\left(-1;0\right); \mathrm{B}\left(1;0\right); \mathrm{F}\left(\frac{1}{2};\frac{\sqrt{3}}{2}\right); \mathrm{E}\left(\frac{1}{2};-\frac{\sqrt{3}}{2}\right); \mathrm{C}\left(\frac{-1}{2};\frac{\sqrt{3}}{2}\right) \mathrm{và} \mathrm{D}\left(\frac{-1}{2};\frac{\sqrt{3}}{2}\right) \end{gathered}$$

Từ hình vẽ ta thấy 6 điểm trên lập thành 1 hình lục giác đều

$\Rightarrow S=2S_{ACFB}=2.\frac{1}{2}.\frac{\sqrt{3}}{2}(1+2)=\frac{3\sqrt{3}}{2}$

Đáp án D

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \Rightarrow \overline{\mathrm{z}}=\mathrm{a}-\mathrm{bi} \\ \mathrm{Ta} \mathrm{có}:\left\{\begin{array}{l}\mathrm{z}+\left(1-2\mathrm{i}\right)\overline{\mathrm{z}}=5\mathrm{i} \left(1\right)\\ 2\mathrm{z}-(1-2\mathrm{i})\overline{\mathrm{z}}=1+4\mathrm{i} \left(2\right)\end{array}\right. \\ \mathrm{Lấy} \left(1\right)+\left(2\right)\Rightarrow 3\mathrm{z}=1+9\mathrm{i}\iff \mathrm{z}=\frac{1}{3}+3\mathrm{i}. \mathrm{Thay} \mathrm{vào} \left(1\right) \\ \left(\frac{1}{3}+3\mathrm{i}\right)+\left(1-2\mathrm{i}\right)\overline{\mathrm{z}}=5\mathrm{i} \\ \iff \left(1-2\mathrm{i}\right)\overline{\mathrm{z}}=\frac{-1}{3}+2\mathrm{i} \\ \iff \overline{\mathrm{z}}=\frac{{\displaystyle \frac{-1}{3}}+2\mathrm{i}}{1-2\mathrm{i}}=\frac{-13}{15}+\frac{4}{15}\mathrm{i} \\ \mathrm{Ta} \mathrm{thấy} \overline{\mathrm{z}} \mathrm{không} \mathrm{phải} \mathrm{số} \mathrm{phức} \mathrm{liên} \mathrm{hợp} \mathrm{của} \mathrm{z} \\ \Rightarrow \mathrm{không} \mathrm{có} \mathrm{số} \mathrm{phức} \mathrm{z} \mathrm{thỏa} \mathrm{mãn} \mathrm{bài} \mathrm{toán} \end{gathered}$$