191 câu Bài tập số phức mức độ cơ bản, nâng cao cực hay có lời giải chi tiết(P4)

Đáp án B

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} (\mathrm{x};\mathrm{y}\in \mathrm{ℝ}) \mathrm{có} \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{là} \mathrm{M}(\mathrm{x};\mathrm{y}) \\ \mathrm{Ta} \mathrm{có}: \left|\mathrm{z}-3\right|=\left|\mathrm{z}+4\mathrm{i}\right| \\ \iff \left|\mathrm{x}-3+\mathrm{yi}\right|=\left|\mathrm{x}+(\mathrm{y}+4)\mathrm{i}\right| \\ \iff {\left(\mathrm{x}-3\right)}^2+{\mathrm{y}}^2={\mathrm{x}}^2+{\left(\mathrm{y}+4\right)}^2 \\ \iff -6\mathrm{x}+9=8\mathrm{y}+16 \\ \iff 6\mathrm{x}+8\mathrm{y}+7=0 \\ \Rightarrow \mathrm{Tập} \mathrm{hợp} \mathrm{các} \mathrm{điểm} \mathrm{biểu} \mathrm{dễn} \mathrm{số} \mathrm{phức} \mathrm{z} \mathrm{thỏa} \mathrm{mãn} \mathrm{điều} \mathrm{kiện} \mathrm{trên} \mathrm{nằm} \mathrm{trên} \mathrm{đường} \mathrm{thẳng} \left(\mathrm{d}\right): 6\mathrm{x}+8\mathrm{y}+7=0 \\ \mathrm{Gọi} \mathrm{A}( 3;-4) \mathrm{là} \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} 3-4\mathrm{i} \\ \Rightarrow \overrightarrow{\mathrm{AM}}=\left(\mathrm{x}-3;\mathrm{y}+4\right) \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{w}=\mathrm{z}+4\mathrm{i}-3 \\ \Rightarrow \left|\mathrm{w}\right|=\mathrm{AM} \end{gathered}$$

$$\begin{gathered} \mathrm{Từ} \mathrm{hình} \mathrm{vẽ} \mathrm{ta} \mathrm{thấy} {\mathrm{AM}}_{\min }\iff \mathrm{AM}\perp \left(\mathrm{d}\right) \mathrm{hay} \mathrm{AM}\equiv \mathrm{AH} \\ \mathrm{Mặt} \mathrm{khác} \mathrm{AH}=\mathrm{d}(\mathrm{A};(\mathrm{d}\left)\right)=\frac{\left|6.3+8.(-4)+7\right|}{\sqrt{6^2+8^2}}=\frac{7}{10} \\ \Rightarrow {\left|\mathrm{z}\right|}_{\min }=\mathrm{AH}=\frac{7}{10} \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: {\left(\frac{1}{2}+\mathrm{i}.\frac{\sqrt{3}}{2}\right)}^2=\frac{-1}{2}+\mathrm{i}.\frac{\sqrt{3}}{2} \\ \Rightarrow \mathrm{z}={\left(\frac{-1}{2}+\mathrm{i}.\frac{\sqrt{3}}{2}\right)}^2=\frac{-1}{2}-\mathrm{i}.\frac{\sqrt{3}}{2} \\ \Rightarrow \mathrm{Phần} \mathrm{ảo} \mathrm{của} \mathrm{z} \mathrm{là} \mathrm{b}=-\frac{\sqrt{3}}{2} \end{gathered}$$

Đáp án D

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \Rightarrow \overline{\mathrm{z}}=\mathrm{a}-\mathrm{bi} \\ \Rightarrow \left\{\begin{array}{l}{\mathrm{z}}^2={\mathrm{a}}^2-{\mathrm{b}}^2+2\mathrm{abi}\\ {\left|\mathrm{z}\right|}^2={\mathrm{a}}^2+{\mathrm{b}}^2=\left|\overline{\mathrm{z}}\right|\\ {\overline{\mathrm{z}}}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{abi}\end{array}\right.\Rightarrow \mathrm{đáp} \mathrm{án} \mathrm{A},\mathrm{B},\mathrm{C} \mathrm{sai} \\ \mathrm{Mặt} \mathrm{khác}: \left|\overline{{\mathrm{z}}^2}\right|=\left|{\mathrm{a}}^2-{\mathrm{b}}^2-2\mathrm{abi}\right|=\sqrt{{\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}^2+4{\mathrm{a}}^2{\mathrm{b}}^2} \\ =\sqrt{{\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}^2}={\mathrm{a}}^2+{\mathrm{b}}^2=\left|{\mathrm{z}}^2\right| \\ \Rightarrow \mathrm{Đáp} \mathrm{án} \mathrm{D} \mathrm{đúng} \end{gathered}$$

Đáp án A

$$\begin{gathered} \mathrm{Xét} \mathrm{đáp} \mathrm{án} \mathrm{A}: \\ \mathrm{z}=1+\cos \frac{\mathrm{\pi }}{7}+\mathrm{i}.\sin \frac{\mathrm{\pi }}{7} \\ \Rightarrow \left|\mathrm{z}\right|={\left(1+\cos \frac{\mathrm{\pi }}{7}\right)}^2+{\left(\sin \frac{\mathrm{\pi }}{7}\right)}^2=1+2\cos \frac{\mathrm{\pi }}{7}+{\left(\cos \frac{\mathrm{\pi }}{7}\right)}^2+{\left(\sin \frac{\mathrm{\pi }}{7}\right)}^2 \\ =2+2\cos \frac{\mathrm{\pi }}{7}\ne 1 \\ \Rightarrow \mathrm{Đáp} \mathrm{án} \mathrm{A} \mathrm{đúng} \\ \mathrm{Xét} \mathrm{đáp} \mathrm{án} \mathrm{B}: \\ \mathrm{z}=\cos \frac{2000\mathrm{\pi }}{3}+\mathrm{i}.\sin \frac{2000\mathrm{\pi }}{3} \\ \Rightarrow \left|\mathrm{z}\right|=\sqrt{{\left(\cos \frac{2000\mathrm{\pi }}{3}\right)}^2+{\left(\sin \frac{2000\mathrm{\pi }}{3}\right)}^2}=1 \Rightarrow \mathrm{Đáp} \mathrm{án} \mathrm{B} \mathrm{loại} \\ \mathrm{Xét} \mathrm{đáp} \mathrm{án} \mathrm{C}: \\ \mathrm{z}=\frac{1-2\mathrm{i}\sqrt{2}}{3} \\ \Rightarrow \left|\mathrm{z}\right|=\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(\frac{-2\sqrt{2}}{3}\right)}^2}=\sqrt{\frac{1}{9}+\frac{8}{9}}=1 \Rightarrow \mathrm{Đáp} \mathrm{án} \mathrm{C} \mathrm{loại} \\ \mathrm{Xét} \mathrm{đáp} \mathrm{án} \mathrm{D}: \\ \mathrm{z}=\frac{4-\mathrm{i}}{4+\mathrm{i}}=\frac{15}{17}+\frac{8}{17}\mathrm{i} \\ \Rightarrow \left|\mathrm{z}\right|=\sqrt{{\left(\frac{15}{17}\right)}^2+{\left(\frac{8}{17}\right)}^2}=\sqrt{\frac{289}{289}}=1 \Rightarrow \mathrm{Đáp} \mathrm{án} \mathrm{D} \mathrm{loại} \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} (\mathrm{x};\mathrm{y}\in \mathrm{ℝ}) \\ \mathrm{Ta} \mathrm{có}: \mathrm{w}=\frac{\mathrm{z}}{1-\mathrm{i}}=\frac{\mathrm{z}(1+\mathrm{i})}{1-{\mathrm{i}}^2}=\frac{\left(\mathrm{x}+\mathrm{yi}\right)\left(1+\mathrm{i}\right)}{2}=\frac{\mathrm{x}-\mathrm{y}+(\mathrm{x}+\mathrm{y})\mathrm{i}}{2} \\ \Rightarrow \mathrm{M}(\mathrm{x};\mathrm{y}); \mathrm{N}\left(\frac{\mathrm{x}-\mathrm{y}}{2};\frac{\mathrm{x}+\mathrm{y}}{2}\right) \\ \Rightarrow \overrightarrow{\mathrm{MN}}=\left(\frac{-\mathrm{x}-\mathrm{y}}{2};\frac{\mathrm{x}-\mathrm{y}}{2}\right) \\ \Rightarrow \mathrm{MN}=\sqrt{\frac{{\left(\mathrm{x}+\mathrm{y}\right)}^2}{4}+\frac{{\left(\mathrm{x}-\mathrm{y}\right)}^2}{4}}=\frac{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}{\sqrt{2}}=\frac{\left|\mathrm{z}\right|}{\sqrt{2}} \\ \mathrm{Mặt} \mathrm{khác}: \mathrm{OM}=\left|\mathrm{z}\right| \mathrm{và} \mathrm{ON}=\sqrt{\frac{{\left(\mathrm{x}-\mathrm{y}\right)}^2}{4}+\frac{{\left(\mathrm{x}+\mathrm{y}\right)}^2}{4}}=\frac{\left|\mathrm{z}\right|}{\sqrt{2}} \\ \mathrm{Theo} \mathrm{đề} \mathrm{bài} \mathrm{chu} \mathrm{vi} ∆\mathrm{OMN}=2 \\ \iff \mathrm{OM}+\mathrm{ON}+\mathrm{MN}=2 \\ \iff \left|\mathrm{z}\right|+\frac{\left|\mathrm{z}\right|}{\sqrt{2}}+\frac{\left|\mathrm{z}\right|}{\sqrt{2}}=2 \\ \iff \left|\mathrm{z}\right|\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=2 \\ \iff \left|\mathrm{z}\right|=-2+2\sqrt{2} \end{gathered}$$

Đáp án D

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} \left(\mathrm{x};\mathrm{y}\in \mathrm{ℝ}\right) \\ \mathrm{Từ} \mathrm{giả} \mathrm{thiết}: \left|\mathrm{z}-1+2\mathrm{i}\right|=\left|\mathrm{z}+3-\mathrm{i}\right| \\ \iff {\left(\mathrm{x}-1\right)}^2+{(\mathrm{y}+2)}^2={\left(\mathrm{x}+3\right)}^2+{(\mathrm{y}-1)}^2\iff -2\mathrm{x}+1+4\mathrm{y}+4=6\mathrm{x}+9-2\mathrm{y}+1 \\ \iff 8\mathrm{x}-6\mathrm{y}+5=0 \\ \Rightarrow \mathrm{Tập} \mathrm{hợp} \mathrm{điểm} \mathrm{M} \mathrm{thỏa} \mathrm{mãn} \mathrm{giả} \mathrm{thiết} \mathrm{nằm} \mathrm{trên} \mathrm{đường} \mathrm{thẳng} \left(\mathrm{d}\right): 8\mathrm{x}-6\mathrm{y}+5=0 \\ \mathrm{Mặt} \mathrm{khác}: \left|\mathrm{z}\right|=\mathrm{OM} \\ {\mathrm{OM}}_{\min }\iff \mathrm{OM}\perp \mathrm{d} \mathrm{hay} {\mathrm{OM}}_{\min }=\mathrm{d}\left(\mathrm{O};\left(\mathrm{d}\right)\right)=\frac{\left|5\right|}{\sqrt{8^2+{(-6)}^2}}=\frac{1}{2} \end{gathered}$$

Đáp án D

Đáp án C

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: \frac{1-\mathrm{i}}{1+\mathrm{i}}=-\mathrm{i} \\ \Rightarrow {\left(\frac{1-\mathrm{i}}{1+\mathrm{i}}\right)}^{15}={\left(-\mathrm{i}\right)}^{15}={\left(-1\right)}^{15}.{\left({\mathrm{i}}^2\right)}^7.\mathrm{i}=\mathrm{i}=\mathrm{a}+\mathrm{bi} \\ \Rightarrow \mathrm{a}=0 \mathrm{và} \mathrm{b}=1 \end{gathered}$$

Đáp án D

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} (\mathrm{x};\mathrm{y}\in \mathrm{ℝ}) \Rightarrow \overline{\mathrm{z}}=\mathrm{x}-\mathrm{yi} \\ \mathrm{Ta} \mathrm{có}: \mathrm{z}= \left(2-\mathrm{z}\right)(\mathrm{i}+\overline{\mathrm{z}})=\left(2-\mathrm{x}-\mathrm{yi}\right)(\mathrm{x}-\mathrm{yi}+\mathrm{i}) \\ =\mathrm{x}(2-\mathrm{x})+(2-\mathrm{x})(1-\mathrm{y})\mathrm{i}-\mathrm{xyi}-\mathrm{y}(1-\mathrm{y}){\mathrm{i}}^2 \\ =\mathrm{x}(2-\mathrm{x})+\mathrm{y}(1-\mathrm{y})+\left(2-2\mathrm{y}-\mathrm{x}\right)\mathrm{i} \\ \mathrm{z}\in \mathrm{ℝ}\iff 2-2\mathrm{y}-\mathrm{x}=0\iff \mathrm{x}+2\mathrm{y}-2=0 \\ \mathrm{KL}: \mathrm{Tập} \mathrm{hợp} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z} \mathrm{có} \mathrm{dạng} \mathrm{đường} \mathrm{thẳng} \end{gathered}$$

Đáp án C

$$\begin{gathered} {\left(-1+\mathrm{iz}\right)}^3=1\iff \left\{\begin{array}{l}{\mathrm{z}}_1=-2\mathrm{i}\\ -1+{\mathrm{iz}}_2=\frac{-1}{2}+\mathrm{i}.\frac{\sqrt{3}}{2}\\ -1+{\mathrm{iz}}_3=\frac{-1}{2}-\mathrm{i}.\frac{\sqrt{3}}{2}\end{array}\right. \\ \iff \left\{\begin{array}{l}{\mathrm{z}}_1=-2\mathrm{i}\\ {\mathrm{z}}_2=\frac{\sqrt{3}}{2}-\frac{1}{2}\mathrm{i}\\ {\mathrm{z}}_3=\frac{-\sqrt{3}}{2}-\frac{1}{2}\mathrm{i}\end{array}\right. \\ \Rightarrow \mathrm{S}=2\mathrm{i}+\frac{\sqrt{3}}{2}+\frac{1}{2}\mathrm{i}+\frac{-\sqrt{3}}{2}+\frac{1}{2}\mathrm{i}=2\mathrm{i}+\mathrm{i}=3\mathrm{i} \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \mathrm{Ta} \mathrm{dễ} \mathrm{dàng} \mathrm{chứng} \mathrm{minh} \mathrm{được} \mathrm{tập} \mathrm{hợp} \mathrm{điểm} \mathrm{M} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z} \\ \mathrm{nằm} \mathrm{trên} \mathrm{đường} \mathrm{tròn} \left(\mathrm{C}\right): {(\mathrm{x}-1)}^2+{(\mathrm{y}+2)}^2=9 \\ \mathrm{Gọi} \mathrm{I}(1;-2) \mathrm{là} \mathrm{tâm} \mathrm{của} \left(\mathrm{C}\right) \Rightarrow \mathrm{OI}=\sqrt{5} \\ {\left|\mathrm{z}\right|}_{\max }=\mathrm{OI}+\mathrm{R}=\sqrt{5}+3 \end{gathered}$$

Đáp án C

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

Đáp án A

Đáp án B

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} \mathrm{và} \mathrm{w}=\mathrm{c}+\mathrm{di} (\mathrm{a};\mathrm{b};\mathrm{c};\mathrm{d}\in \mathrm{ℝ}) \\ \mathrm{z}=\overline{\mathrm{w}}\iff \left\{\begin{array}{l}\mathrm{a}=\mathrm{c}\\ \mathrm{b}=-\mathrm{d}\end{array}\iff {\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{c}}^2+{\mathrm{d}}^2\right. \\ \iff \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=\sqrt{{\mathrm{c}}^2+{\mathrm{d}}^2} \mathrm{hay} \left|\overline{\mathrm{z}}\right|=\left|\mathrm{w}\right| \\ \mathrm{z}=-\overline{\mathrm{w}} \iff \left\{\begin{array}{l}\mathrm{a}=-\mathrm{c}\\ \mathrm{b}=\mathrm{d}\end{array}\iff \right.\left\{\begin{array}{l}\mathrm{a}=-\mathrm{c}\\ -\mathrm{b}=-\mathrm{d}\end{array}\right.\mathrm{hay} \overline{\mathrm{z}}=-\mathrm{w} \\ \mathrm{Mặt} \mathrm{khác}: \mathrm{z}=\mathrm{w}\to {\mathrm{z}}^3={\mathrm{w}}^3 \\ \mathrm{KL}: \mathrm{Có} \mathrm{ý} 1; 2 \mathrm{và} 5 \mathrm{đúng} \end{gathered}$$

Đáp án C

Đáp án D

Đáp án D

$$\begin{gathered} \mathrm{Gọi} \mathrm{M}(\mathrm{x};\mathrm{y}) \mathrm{là} \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z} \mathrm{thỏa} \mathrm{mãn} \mathrm{đề} \mathrm{bài} \\ \mathrm{A}(0;-1) \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} -\mathrm{i} \mathrm{và} \mathrm{I}(0;0) \mathrm{là} \mathrm{tâm} \mathrm{của} \left(\mathrm{T}\right) \\ \Rightarrow \overrightarrow{\mathrm{AM}}=(\mathrm{x};\mathrm{y}+1) \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z}+\mathrm{i} \\ \Rightarrow \left|\mathrm{z}+\mathrm{i}\right|=\mathrm{AM} \\ {\mathrm{AM}}_{\max }=\mathrm{AI}+\mathrm{R}=1+2=3 \end{gathered}$$

Đáp án C

$\mathrm{z}={(1+\mathrm{i}\sqrt{3})}^3=-8\Rightarrow \mathrm{Phần} \mathrm{ảo} \mathrm{của} \mathrm{z} \mathrm{là} 0$

Đáp án A

$z.\overline{z}=(3-5i)(3+5i)=9-25i^2=9+25=34$

Đáp án B

$$\begin{gathered} \mathrm{Ta} \mathrm{thấy} \mathrm{nghiệm} \mathrm{của} \mathrm{PT} \mathrm{là} \mathrm{z}=-\mathrm{i}, \mathrm{z}=-{\mathrm{i}}^2, ... , \mathrm{z}=-{\mathrm{i}}^{10} \\ \mathrm{NX}: \\ {\mathrm{i}}^{2\mathrm{n}+2}={(-1)}^{\mathrm{n}+1} \mathrm{và} {\mathrm{i}}^{2\mathrm{n}+1}=\mathrm{i}.{(-1)}^{\mathrm{n}} \\ \mathrm{KL}: \mathrm{PT} \mathrm{luôn} \mathrm{chỉ} \mathrm{có} 4 \mathrm{nghiệm} \mathrm{z}=\pm 1 \mathrm{và} \mathrm{z}=\pm \mathrm{i} \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{Gọi} \mathrm{A}(1;-2) \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z} \\ \mathrm{B}(\mathrm{x};\mathrm{y}) \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{w}=\mathrm{x}+\mathrm{yi} \\ \mathrm{A} \mathrm{và} \mathrm{B} \mathrm{đối} \mathrm{xứng} \mathrm{nhau} \mathrm{qua} \mathrm{trục} \mathrm{Oy} \Rightarrow \left\{\begin{array}{l}\mathrm{x}=-1\\ \mathrm{y}=-2\end{array}\right. \\ \Rightarrow \mathrm{w}=-1-2\mathrm{i} \end{gathered}$$

Đáp án D

$$\begin{gathered} z+1-3i=2\iff z=1+3i \\ \Rightarrow \left\{M\right\}=\left\{\right(1;3\left\}\right) \end{gathered}$$

Đáp án A

$$\begin{gathered} Gọi M(x;y)\in ∆ \\ Ta có: \left|z\right|=OM \\ O{M}_{min}\iff OM\perp d \\ \Rightarrow O{M}_{min}=d(O;∆)=\frac{\left|-2\right|}{\sqrt{3^2+4^2}}=\frac{2}{5} \end{gathered}$$