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

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

Đáp án B

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} (\mathrm{x};\mathrm{y}\in \mathrm{ℝ}) \\ \mathrm{Ta} \mathrm{có}:\left|\mathrm{z}+1+\mathrm{i}\right|=3 \\ \iff \left|\mathrm{x}+1+\left(\mathrm{y}+1\right)\mathrm{i}\right|=3 \\ \iff {\left(\mathrm{x}+1\right)}^2+{\left(\mathrm{y}+1\right)}^2=9 \\ \Rightarrow \mathrm{Tập} \mathrm{hợp} \mathrm{điểm} \mathrm{M} \mathrm{biểu} \mathrm{diễn} \mathrm{z} \mathrm{nằm} \mathrm{trên} \mathrm{đường} \mathrm{tròn} \mathrm{tâm} \mathrm{I}(-1;-1) \mathrm{có} \mathrm{R}=3 \end{gathered}$$

Gọi A(3;-2) biểu diễn số phức 3-2i

$\Rightarrow \overrightarrow{\mathrm{AM}}=\left(\mathrm{x}-3;\mathrm{y}+2\right) \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{w}=\mathrm{z}+2\mathrm{i}-3$

$\Rightarrow \left|w\right|=AM$

$$\begin{gathered} Từ hình vẽ ta thấy AM đạt GTLN khi điểm M trùng với điểm B \\ \Rightarrow {\left|w\right|}_{max}=AB=AI+IB=AI+R \\ Mà \overrightarrow{AI}=(-4;1)\Rightarrow AI=\sqrt{17} \\ \Rightarrow {\left|w\right|}_{max}=\sqrt{17}+3 \end{gathered}$$

Đáp án B

Đáp án B

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} (\mathrm{x};\mathrm{y}\in \mathrm{ℝ}) \Rightarrow \overline{\mathrm{z}}=\mathrm{x}-\mathrm{yi} \\ \mathrm{Theo} \mathrm{đề} \mathrm{bài} \mathrm{z}=\overline{\mathrm{z}}\iff \left\{\begin{array}{l}\mathrm{x}=\mathrm{x}\\ \mathrm{y}=-\mathrm{y}\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}\in \mathrm{ℝ}\\ \mathrm{y}=0\end{array}\right. \\ \mathrm{KL}: \mathrm{Tập} \mathrm{hợp} \mathrm{điểm} \mathrm{M} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z} \mathrm{thỏa} \mathrm{mãn} \mathrm{z}=\overline{\mathrm{z}} \mathrm{là} \mathrm{trục} \mathrm{Ox} \end{gathered}$$

Đáp án D

$$\begin{gathered} \mathrm{PT}\iff \mathrm{x}=-\mathrm{i}; \mathrm{x}=-{\mathrm{i}}^2 ; ... ; \mathrm{x}=-{\mathrm{i}}^{100} \\ \mathrm{Nhận} \mathrm{xét}: \\ {\mathrm{i}}^{2\mathrm{n}+2}={\mathrm{i}}^2.{\mathrm{i}}^{2\mathrm{n}}={(-1)}^{\mathrm{n}+1} \mathrm{với} \mathrm{n}\in \mathrm{ℝ} \\ {\mathrm{i}}^{2\mathrm{n}+1}=\mathrm{i}.{\mathrm{i}}^{2\mathrm{n}}=(-1).{\mathrm{i}}^{\mathrm{n}+1} \mathrm{với} \mathrm{n}\in \mathrm{ℝ} \\ \Rightarrow \mathrm{PT} \mathrm{luôn} \mathrm{chỉ} \mathrm{có} 4 \mathrm{nghiệm} \mathrm{là} \mathrm{z}=\pm 1 \mathrm{và} \mathrm{z}=\pm \mathrm{i} \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{PT}\hspace{0.17em}\mathrm{có} \mathrm{nghiệm} \mathrm{là} 2 \mathrm{số} \mathrm{phức} \mathrm{liên} \mathrm{hợp} \mathrm{khi} \mathrm{PT}\hspace{0.17em}\mathrm{đó} \mathrm{không} \mathrm{có} \mathrm{nghiệm} \mathrm{thực} \mathrm{hoặc} \mathrm{có} \mathrm{nghiệm} \mathrm{thực} \mathrm{kép} \\ \Rightarrow ∆\text{'}\le 0 \\ \iff {\mathrm{m}}^2-1\le 0 \\ \iff -1\le \mathrm{m}\le 1 \end{gathered}$$

Đáp án A

$$\begin{gathered} {\mathrm{z}}^2-\mathrm{zw}+{\mathrm{w}}^2=0 \\ \iff {\left(\frac{\mathrm{z}}{\mathrm{w}}\right)}^2-\frac{\mathrm{z}}{\mathrm{w}}+1=0 \\ \iff \frac{\mathrm{z}}{\mathrm{w}}=\frac{1}{2}\pm \mathrm{i}\frac{\sqrt{3}}{2} \\ \Rightarrow \left|\frac{\mathrm{z}}{\mathrm{w}}\right|=\left|\frac{1}{2}\pm \mathrm{i}\frac{\sqrt{3}}{2}\right| \\ \iff \frac{\left|\mathrm{z}\right|}{\left|\mathrm{w}\right|}=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=1 \\ \iff \frac{2}{\left|\mathrm{w}\right|}=1\Rightarrow \left|\mathrm{w}\right|=2 \end{gathered}$$

Đáp án B

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: {\mathrm{z}}^2+1=\mathrm{iz} \iff \mathrm{z}+\frac{1}{\mathrm{z}}=\mathrm{i} (*) \\ \mathrm{Mà} \mathrm{S}={\mathrm{z}}^4+\frac{1}{{\mathrm{z}}^4}={\left({\mathrm{z}}^2+\frac{1}{{\mathrm{z}}^2}\right)}^2-2={\left[{\left(\mathrm{z}+\frac{1}{\mathrm{z}}\right)}^2-2\right]}^2-2 \\ \Rightarrow \mathrm{S}={\left({\mathrm{i}}^2-2\right)}^2-2={\left(-3\right)}^2-2=9-2=7 \end{gathered}$$

Đáp án A

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: \frac{1+\mathrm{i}}{1-\mathrm{i}}=\frac{{\left(1+\mathrm{i}\right)}^2}{1-{\mathrm{i}}^2}=\frac{2\mathrm{i}}{2}=\mathrm{i} \\ \mathrm{Phương} \mathrm{trình} \mathrm{trở} \mathrm{thành}: {\mathrm{i}}^{\mathrm{n}}=\mathrm{i} \\ \mathrm{Mặt} \mathrm{khác}: {\mathrm{i}}^{2\mathrm{n}+1}=\mathrm{i}.{\left(-1\right)}^{\mathrm{n}}\Rightarrow \mathrm{Loại} \mathrm{B} \mathrm{và} \mathrm{D} \\ \mathrm{Xét} 2 \mathrm{đáp} \mathrm{án} \mathrm{A} \mathrm{và} \mathrm{C} \\ {\mathrm{i}}^{97}=\mathrm{i}.{\left({\mathrm{i}}^2\right)}^{48}=\mathrm{i}.{\left(-1\right)}^{48}=\mathrm{i} ( \mathrm{A} \mathrm{đúng}) \\ {\mathrm{i}}^{99}=\mathrm{i}.{\left({\mathrm{i}}^2\right)}^{49}=\mathrm{i}.{\left(-1\right)}^{49}=-\mathrm{i} (\mathrm{C} \mathrm{sai}) \end{gathered}$$

Đáp án C

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \mathrm{PT}: {\mathrm{z}}^2+\overline{\mathrm{z}}=0\iff {\left(\mathrm{a}+\mathrm{bi}\right)}^2+\mathrm{a}-\mathrm{bi}=0 \\ \Rightarrow {\mathrm{a}}^2-{\mathrm{b}}^2+\mathrm{a}+(2\mathrm{ab}-\mathrm{b})\mathrm{i}=0 \\ \iff \left\{\begin{array}{l}{\mathrm{a}}^2-{\mathrm{b}}^2+\mathrm{a}=0 \left(1\right)\\ 2\mathrm{ab}-\mathrm{b}=0 \left(2\right)\end{array}\right. \\ \mathrm{Từ} \left(2\right)\Rightarrow \mathrm{b}=0 \mathrm{hoặc} \mathrm{a}=\frac{1}{2} \\ \mathrm{TH}1: \mathrm{b}=0 \mathrm{thì} \left(1\right) \mathrm{trở} \mathrm{thành} {\mathrm{a}}^2+\mathrm{a}=0\iff \mathrm{a}=0 \mathrm{hoặc} \mathrm{a}=-1 \\ \Rightarrow \mathrm{Có} 2 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \mathrm{là} \mathrm{z}=0 \mathrm{và} \mathrm{z}=-1 \\ \mathrm{TH}2: \mathrm{a}=\frac{1}{2} \mathrm{thì} \left(1\right) \mathrm{trở} \mathrm{thành} \frac{3}{4}-{\mathrm{b}}^2=0\iff \mathrm{b}=\pm \frac{\sqrt{3}}{2} \\ \Rightarrow \mathrm{Có} 2 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \mathrm{là} \mathrm{z}=\frac{1}{2}+\mathrm{i}\frac{\sqrt{3}}{2}\mathrm{và} \mathrm{z}=\frac{1}{2}-\mathrm{i}\frac{\sqrt{3}}{2} \\ \mathrm{KL}: \mathrm{có} 4 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \mathrm{bài} \mathrm{toán} \end{gathered}$$

Đáp án A

$$\begin{gathered} \mathrm{z}=4-\mathrm{i}-\frac{3}{\mathrm{i}}=4-\mathrm{i}-\frac{3\mathrm{i}}{{\mathrm{i}}^2}=4-\mathrm{i}+3\mathrm{i}=4+2\mathrm{i} \\ \Rightarrow \mathrm{Phần} \mathrm{ảo} \mathrm{của} \mathrm{z} \mathrm{là} \mathrm{b}=2 \end{gathered}$$

Đáp án D

$$\begin{gathered} {\mathrm{z}}^4-3{\mathrm{z}}^2-10=0\iff {\mathrm{z}}^2=5 \mathrm{hoặc} {\mathrm{z}}^2=-2 \\ \mathrm{Với} {\mathrm{z}}^2=5\iff {\mathrm{z}}_1=\sqrt{5}; {\mathrm{z}}_2=-\sqrt{5} \\ \mathrm{với} {\mathrm{z}}^2=-2\iff {\mathrm{z}}_3=\mathrm{i}\sqrt{2}; {\mathrm{z}}_4=-\mathrm{i}\sqrt{2} \\ \mathrm{Vậy} \mathrm{S}={\mathrm{z}}_1^4+{\mathrm{z}}_2^4+{\mathrm{z}}_3^4+{\mathrm{z}}_4^4={\left(\sqrt{5}\right)}^4+{\left(-\sqrt{5}\right)}^4+{\left(\mathrm{i}\sqrt{2}\right)}^4+{\left(-\mathrm{i}\sqrt{2}\right)}^4 \\ \mathrm{S}=2.25+2.4=50+8=58 \end{gathered}$$

$$\begin{gathered} \mathrm{Đặt} {\mathrm{z}}_1=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \mathrm{Vì} {\mathrm{z}}_1; {\mathrm{z}}_2 \mathrm{là} 2 \mathrm{số} \mathrm{phức} \mathrm{có} \mathrm{phần} \mathrm{thực} \mathrm{bằng} \mathrm{nhau} \\ \Rightarrow {\mathrm{z}}_2=\mathrm{a}+\mathrm{ci} \\ \mathrm{Mặt} \mathrm{khác}: \left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|\iff \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=\sqrt{{\mathrm{a}}^2+{\mathrm{c}}^2} \\ \iff {\mathrm{b}}^2={\mathrm{c}}^2\iff \mathrm{b}=\mathrm{c} \mathrm{hoặc} \mathrm{b}=-\mathrm{c} \\ \mathrm{TH}1: \mathrm{b}=\mathrm{c} \mathrm{thì} {\mathrm{z}}_1={\mathrm{z}}_2 \\ \mathrm{TH}2: \mathrm{b}=-\mathrm{c} \iff -\mathrm{b}=\mathrm{c} \Rightarrow \overline{{\mathrm{z}}_1}={\mathrm{z}}_2 \mathrm{hoặc} \overline{{\mathrm{z}}_2}={\mathrm{z}}_1 \end{gathered}$$Đáp án C

Đáp án B

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} (\mathrm{x};\mathrm{y}\in \mathrm{ℝ}) \\ \mathrm{Ta} \mathrm{có}: {\mathrm{z}}^2+{\left|\mathrm{z}\right|}^2=0\iff {\left(\mathrm{x}+\mathrm{yi}\right)}^2+{\mathrm{x}}^2+{\mathrm{y}}^2=0 \\ \iff 2{\mathrm{x}}^2+2\mathrm{xyi}=0 \\ \iff \left\{\begin{array}{l}2{\mathrm{x}}^2=0\\ 2\mathrm{xy}=0\end{array}\iff \left\{\begin{array}{l}\mathrm{x}=0\\ \mathrm{y}\in \mathrm{ℝ}\end{array}\right.\right. \\ \mathrm{KL}: \left\{\mathrm{M}\right\} \mathrm{là} \mathrm{trục} \mathrm{Oy} \end{gathered}$$

Đáp án B

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \mathrm{Hệ} \left\{\begin{array}{l}\left|\mathrm{z}-1+2\mathrm{i}\right|=\left|\mathrm{z}+3-\mathrm{i}\right| \\ \left|\mathrm{z}-1+3\mathrm{i}\right|=4 \end{array}\right. \\ \iff \left\{\begin{array}{l}\left|\mathrm{a}-1+(\mathrm{b}+2)\mathrm{i}\right|=\left|\mathrm{a}+3+(\mathrm{b}-1)\mathrm{i}\right|\\ \left|\mathrm{a}-1+(\mathrm{b}+3)\mathrm{i}\right|=4\end{array}\right. \\ \iff \left\{\begin{array}{l}{\left(\mathrm{a}-1\right)}^2+{\left(\mathrm{b}+2\right)}^2={\left(\mathrm{a}+3\right)}^2+{\left(\mathrm{b}-1\right)}^2\\ {\left(\mathrm{a}-1\right)}^2+{\left(\mathrm{b}+3\right)}^2=16\end{array}\right. \\ \iff \left\{\begin{array}{l}8\mathrm{a}-6\mathrm{b}+5=0 (*)\\ {\left(\mathrm{a}-1\right)}^2+{\left(\mathrm{b}+3\right)}^2=16 (**)\end{array}\right. \\ \mathrm{Từ} (*)\iff \mathrm{b}=\frac{5+8\mathrm{a}}{6} \mathrm{thay} \mathrm{vào} (**): \\ {\left(\mathrm{a}-1\right)}^2+{\left(\frac{5+8\mathrm{a}}{6}+3\right)}^2=16 \\ \iff {\left(\mathrm{a}-1\right)}^2+{\left(\frac{8\mathrm{a}+23}{6}\right)}^2=16 \\ \iff 36{\mathrm{a}}^2-72\mathrm{a}+36+64{\mathrm{a}}^2+368\mathrm{a}+529-576=0 \\ \iff 100{\mathrm{a}}^2+296\mathrm{a}-11=0 \left(1\right) \\ \mathrm{PT}\left(1\right) \mathrm{có} 2 \mathrm{nghiệm} \mathrm{a} \mathrm{phân} \mathrm{biệt} \\ \Rightarrow \mathrm{Có} 2 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \end{gathered}$$

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

Đáp án D

số phức z = -1+2i có điểm biểu diễn là M(-1;2)

Số phức w = -2+i có điểm biểu diễn là N(-2;1)

Ta thấy M và N đối xứng nhau qua đường thẳng y=-x

Đáp án D

Đáp án A

$$\begin{gathered} \mathrm{Giải} \mathrm{PT}: {\mathrm{z}}^2-3\mathrm{z}+4=0 \iff \left\{\begin{array}{l}{\mathrm{z}}_1=\frac{3}{2}+\frac{\sqrt{7}}{2}\mathrm{i}\\ {\mathrm{z}}_2=\frac{3}{2}-\frac{\sqrt{7}}{2}\mathrm{i}\end{array}\right. \\ \mathrm{Ta} \mathrm{có}:\overline{ {\mathrm{z}}_1^4}=\overline{{\left(\frac{3}{2}+\frac{\sqrt{7}}{2}\mathrm{i}\right)}^4}=\overline{\frac{-31}{2}+\frac{3\sqrt{7}}{2}\mathrm{i}}=\frac{-31}{2}-\frac{3\sqrt{7}}{2}\mathrm{i} \\ \overline{ {\mathrm{z}}_2^4}=\overline{{\left(\frac{3}{2}-\frac{\sqrt{7}}{2}\mathrm{i}\right)}^4}=\overline{\frac{-31}{2}-\frac{3\sqrt{7}}{2}\mathrm{i}}=\frac{-31}{2}+\frac{3\sqrt{7}}{2}\mathrm{i} \\ \mathrm{Bấm} \mathrm{máy} \mathrm{tính} \mathrm{ta} \mathrm{tìm} \mathrm{được} \mathrm{F}=-9 \end{gathered}$$

Đáp án A

$$\begin{gathered} Ta có: \frac{1+i}{1-i}=i \\ \Rightarrow \left|\frac{\left(1+i\right)z}{1-i}+2\right|=1\iff \left|iz+2\right|=1 \\ \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} (\mathrm{x};\mathrm{y}\in \mathrm{ℝ}) \\ \Rightarrow \left|\mathrm{i}(\mathrm{x}+\mathrm{yi})+2\right|=1 \\ \iff \left|-\mathrm{y}+2+\mathrm{xi}\right|=1 \\ \iff {\mathrm{x}}^2+{\left(\mathrm{y}-2\right)}^2=1 \\ \Rightarrow \mathrm{Tập} \mathrm{hợp} \mathrm{điểm} \mathrm{M} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z} \mathrm{nằm} \mathrm{trên} \mathrm{đường} \mathrm{tròn} \mathrm{tâm} \mathrm{I}(0;2) \mathrm{có} \mathrm{R}=1 \\ \Rightarrow {\left|\mathrm{z}\right|}_{\max }=\mathrm{OI}+\mathrm{R}=2+1=3 \end{gathered}$$

Đáp án B

$$\begin{gathered} \mathrm{z}=\frac{1}{2-\mathrm{i}\sqrt{3}}=\frac{2+\mathrm{i}\sqrt{3}}{4-3{\mathrm{i}}^2}=\frac{2+\mathrm{i}\sqrt{3}}{7} \\ \Rightarrow \mathrm{Phần} \mathrm{ảo} \mathrm{của} \mathrm{z} \mathrm{là} \mathrm{b}=\frac{\sqrt{3}}{7} \end{gathered}$$

Đáp án C