191 câu Bài tập số phức mức độ cơ bản, nâng cao cực hay có lời giải chi tiết(P2)

Đáp án A

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \Rightarrow \overline{\mathrm{z}}=\mathrm{a}-\mathrm{bi} \\ \left|\overline{\mathrm{z}}-1+\mathrm{i}\right|=\left|\mathrm{a}-1+(1-\mathrm{b})\mathrm{i}\right|=\sqrt{{(\mathrm{a}-1)}^2+{(1-\mathrm{b})}^2} \\ \left|\mathrm{z}+3+\mathrm{i}\right|=\left|\mathrm{a}+3+(\mathrm{b}+1)\mathrm{i}\right|=\sqrt{{(\mathrm{a}+3)}^2+{(\mathrm{b}+1)}^2} \\ \mathrm{Từ} \mathrm{giả} \mathrm{thiết} \mathrm{đề} \mathrm{bài} \Rightarrow {(\mathrm{a}-1)}^2+{(1-\mathrm{b})}^2={(\mathrm{a}+3)}^2+{(\mathrm{b}+1)}^2 \\ \iff -2\mathrm{a}+1+1-2\mathrm{b}=6\mathrm{a}+9+2\mathrm{b}+1 \\ \iff 8\mathrm{a}+4\mathrm{b}+8=0 \\ \iff 2\mathrm{a}+\mathrm{b}+2=0 \\ \iff \mathrm{b}=-2-2\mathrm{a} \\ \Rightarrow \left|\mathrm{z}\right|=\sqrt{{\mathrm{a}}^2+{(-2-2\mathrm{a})}^2}=\sqrt{{\mathrm{a}}^2+4+8\mathrm{a}+4{\mathrm{a}}^2}=\sqrt{5{\mathrm{a}}^2+8\mathrm{a}+4} \\ \mathrm{Đặt} \mathrm{f}\left(\mathrm{x}\right)=5{\mathrm{a}}^2+8\mathrm{a}+4 \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=10\mathrm{a}+8=0\iff \mathrm{a}=\frac{-4}{5} \\ \Rightarrow \mathrm{f}{\left(\mathrm{x}\right)}_{\min }=\mathrm{f}\left(\frac{-4}{5}\right)=\frac{4}{5} \\ \Rightarrow {\left|\mathrm{z}\right|}_{\min }=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}} \end{gathered}$$

Đáp án B

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ})\Rightarrow \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z} \mathrm{là} \mathrm{A}(\mathrm{a};\mathrm{b}) \\ \Rightarrow \mathrm{w}=\frac{\mathrm{z}}{1-\mathrm{i}}=\frac{\mathrm{z}(1+\mathrm{i})}{1-{\mathrm{i}}^2}=\frac{(\mathrm{a}+\mathrm{bi})(1+\mathrm{i})}{2}=\frac{\mathrm{a}-\mathrm{b}+(\mathrm{a}+\mathrm{b})\mathrm{i}}{2} \\ \Rightarrow \mathrm{Điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{w} \mathrm{là} \mathrm{B}\left(\frac{\mathrm{a}-\mathrm{b}}{2};\frac{\mathrm{a}+\mathrm{b}}{2}\right) \\ \mathrm{Ta} \mathrm{có}: \\ \overrightarrow{\mathrm{AB}}=\left(\frac{-\mathrm{a}-\mathrm{b}}{2};\frac{\mathrm{a}-\mathrm{b}}{2}\right)\Rightarrow \mathrm{AB}=\sqrt{\frac{{\left(\mathrm{a}+\mathrm{b}\right)}^2}{4}+\frac{{\left(\mathrm{a}-\mathrm{b}\right)}^2}{4}} \\ \overrightarrow{\mathrm{OA}}=(\mathrm{a};\mathrm{b})\Rightarrow \mathrm{OA}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2} \\ \overrightarrow{\mathrm{OB}}=\left(\frac{\mathrm{a}-\mathrm{b}}{2};\frac{\mathrm{a}+\mathrm{b}}{2}\right)\Rightarrow \mathrm{OB}=\sqrt{\frac{{\left(\mathrm{a}-\mathrm{b}\right)}^2}{4}+\frac{{\left(\mathrm{a}+\mathrm{b}\right)}^2}{4}} \\ \Rightarrow △\mathrm{ABO} \mathrm{cân} \mathrm{tại} \mathrm{B} \left(1\right) \\ \mathrm{Mặt} \mathrm{khác}: \\ \overrightarrow{\mathrm{AB}}.\overrightarrow{\mathrm{BO}}=\frac{-\mathrm{a}-\mathrm{b}}{2}.\frac{-(\mathrm{a}-\mathrm{b})}{2}+\frac{\mathrm{a}-\mathrm{b}}{2}.\frac{-(\mathrm{a}+\mathrm{b})}{2} \\ =\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{4}-\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{4}=0 \\ \Rightarrow △\mathrm{ABO} \mathrm{vuông} \mathrm{tại} \mathrm{B} \left(2\right) \\ \mathrm{Từ} \left(1\right) \mathrm{và} \left(2\right)\Rightarrow △\mathrm{ABO} \mathrm{vuông} \mathrm{cân} \mathrm{tại} \mathrm{B} \end{gathered}$$

Đáp án B

$$\begin{gathered} \mathrm{z}=\frac{\mathrm{z}-4\mathrm{i}}{{\mathrm{i}}^3}\iff -\mathrm{zi}=\mathrm{z}-4\mathrm{i} \\ \iff -\mathrm{zi}-\mathrm{z}=-4\mathrm{i} \\ \iff \mathrm{z}(-\mathrm{i}-1)=-4\mathrm{i} \\ \iff \mathrm{z}=\frac{-4\mathrm{i}}{-\mathrm{i}-1}=\frac{4\mathrm{i}(\mathrm{i}-1)}{{\mathrm{i}}^2-1}=\frac{-4-4\mathrm{i}}{-2}=2+2\mathrm{i} \\ \Rightarrow \mathrm{Phần} \mathrm{thực} \mathrm{của} \mathrm{số} \mathrm{phức} \mathrm{z} \mathrm{là} 2 \end{gathered}$$

Đáp án D

$$\begin{gathered} \mathrm{Gọi} \mathrm{M}(\mathrm{a};\mathrm{b}) \mathrm{là} \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z}=\mathrm{a}+\mathrm{bi} \\ \Rightarrow \mathrm{w}=-\overline{\mathrm{z}}=-(\mathrm{a}-\mathrm{bi})=-\mathrm{a}+\mathrm{bi} \\ \Rightarrow \mathrm{Điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{w} \mathrm{là} \mathrm{N}(-\mathrm{a};\mathrm{b}) \\ \mathrm{Ta} \mathrm{thấy} \mathrm{tung} \mathrm{độ} 2 \mathrm{điểm} \mathrm{M} \mathrm{và} \mathrm{N} \mathrm{bằng} \mathrm{nhau} \mathrm{vào} \mathrm{hoành} \mathrm{độ} 2 \mathrm{điểm} \mathrm{này} \mathrm{đối} \mathrm{nhau} \\ \Rightarrow \mathrm{M} \mathrm{và} \mathrm{N} \mathrm{đối} \mathrm{xứng} \mathrm{nhau} \mathrm{qua} \mathrm{trục} \mathrm{Oy} \end{gathered}$$

Đáp án C

Thay từng các đáp án, nếu thấy 2 vế của phương trình bằng nhau thì đáp án đó là nghiệm của phương trình

Đáp án A

$$\begin{gathered} \mathrm{Giải} \mathrm{PT}: 3{\mathrm{z}}^2-4\mathrm{z}+2=0\iff \left\{\begin{array}{l}{\mathrm{z}}_1=\frac{2}{3}+\frac{\sqrt{2}}{3}\mathrm{i}\\ {\mathrm{z}}_2=\frac{2}{3}-\frac{\sqrt{2}}{3}\mathrm{i}\end{array}\right. \\ \Rightarrow \mathrm{S}={\mathrm{z}}_1{\left(\overline{{\mathrm{z}}_2}\right)}^2+{\mathrm{z}}_2{\left(\overline{{\mathrm{z}}_1}\right)}^2 \\ =\left(\frac{2}{3}+\frac{\sqrt{2}}{3}\mathrm{i}\right){\left(\frac{2}{3}+\frac{\sqrt{2}}{3}\mathrm{i}\right)}^2+\left(\frac{2}{3}-\frac{\sqrt{2}}{3}\mathrm{i}\right){\left(\frac{2}{3}-\frac{\sqrt{2}}{3}\mathrm{i}\right)}^2 \\ ={\left(\frac{2}{3}+\frac{\sqrt{2}}{3}\mathrm{i}\right)}^3+{\left(\frac{2}{3}-\frac{\sqrt{2}}{3}\mathrm{i}\right)}^3 \\ =\frac{-8}{27} \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ})\iff \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{z} \mathrm{là} \mathrm{M}(\mathrm{a};\mathrm{b}) \\ \Rightarrow \mathrm{w}=\frac{\mathrm{z}}{\mathrm{i}}=\frac{\mathrm{zi}}{{\mathrm{i}}^2}=\frac{(\mathrm{a}+\mathrm{bi})\mathrm{i}}{-1}=\frac{-\mathrm{b}+\mathrm{ai}}{-1}=\mathrm{b}-\mathrm{ai} \\ \Rightarrow \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{z} \mathrm{là} \mathrm{N}(\mathrm{b};-\mathrm{a}) \\ \mathrm{Giả} \mathrm{sử} \mathrm{d} \mathrm{và} ∆ \mathrm{đi} \mathrm{qua} \mathrm{gốc} \mathrm{tọa} \mathrm{độ} \mathrm{thì} \\ \mathrm{PTĐT} \mathrm{d} \mathrm{có} \mathrm{VTCP} \mathrm{là} \overrightarrow{\mathrm{OM}}=(\mathrm{a};\mathrm{b}) \\ \mathrm{PTĐT}∆ \mathrm{có} \mathrm{VTCP} \mathrm{là} \overrightarrow{\mathrm{ON}}=(\mathrm{b};-\mathrm{a}) \\ \Rightarrow \overrightarrow{\mathrm{OM}}.\overrightarrow{\mathrm{ON}}=\mathrm{ab}-\mathrm{ab}=0 \\ \Rightarrow \overrightarrow{\mathrm{OM}}\perp \overrightarrow{\mathrm{ON}} \\ \Rightarrow \mathrm{d}\perp ∆ \end{gathered}$$

Đáp án B

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} (\mathrm{x};\mathrm{y}\in \mathrm{ℝ}) \\ \mathrm{Theo} \mathrm{ĐB}: \left|\mathrm{z}-1+3\mathrm{i}\right|=\left|\mathrm{x}-1+(\mathrm{y}-3)\mathrm{i}\right|=\sqrt{{\left(\mathrm{x}-1\right)}^2+{(\mathrm{y}-3)}^2}=2 \\ \Rightarrow {\left(\mathrm{x}-1\right)}^2+{(\mathrm{y}-3)}^2=4 \\ \mathrm{Tập} \mathrm{hợp} \mathrm{các} \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{z} \mathrm{nằm} \mathrm{trên} \mathrm{đường} \mathrm{tròn} \mathrm{tâm} \mathrm{I}(1;3) \mathrm{và} \mathrm{R}=2 \end{gathered}$$

Từ hình vẽ suy ra: ${\left|z\right|}_{max}=OI+R=\sqrt{10}+2$