191 câu Bài tập số phức mức độ cơ bản, nâng cao cực hay có lời giải chi tiết(P5)

Đáp án C

$$\begin{gathered} \mathrm{z}=\frac{(2+\mathrm{i})(1-2\mathrm{i})}{\mathrm{i}}=-3-4\mathrm{i} \\ \Rightarrow \overline{\mathrm{z}}=-3+4\mathrm{i} \\ \to \mathrm{Phần} \mathrm{ảo} \mathrm{của} \overline{\mathrm{z}} \mathrm{là} 4 \end{gathered}$$

Đáp án A

$$\begin{gathered} \mathbf{TH}\mathbf{1}: \mathrm{x}-2\mathrm{y}=4\iff \mathrm{x}=4+2\mathrm{y} \\ \Rightarrow {(4+2\mathrm{y})}^2+{\mathrm{y}}^2=5 \\ \iff 5{\mathrm{y}}^2+16\mathrm{y}+11=0 \\ \iff \mathrm{y}=-1 \mathrm{hoặc} \mathrm{y}=\frac{-11}{5} \\ \Rightarrow \mathrm{x}=2 \mathrm{hoặc} \mathrm{x}=\frac{-2}{5} \\ \mathbf{TH}\mathbf{2}: \mathrm{x}-2\mathrm{y}=-4\iff \mathrm{x}=2\mathrm{y}-4 \\ \Rightarrow {(2\mathrm{y}-4)}^2+{\mathrm{y}}^2=5 \\ \iff 5{\mathrm{y}}^2-16\mathrm{y}+11=0 \\ \iff \mathrm{y}=\frac{11}{5} \mathrm{hoặc} \mathrm{y}=1 \\ \Rightarrow \mathrm{x}=\frac{2}{5} \mathrm{hoặc} \mathrm{y}=-2 \end{gathered}$$

KL: Có 4 số phức thỏa mãn bài toán

Đáp án C

$$\begin{gathered} \mathrm{Giải} \mathrm{PT}: {\mathrm{z}}^2-2\mathrm{z}+3=0\iff \left\{\begin{array}{l}{\mathrm{z}}_1=1+\mathrm{i}\sqrt{2}\\ {\mathrm{z}}_2=1-\mathrm{i}\sqrt{2}\end{array}\right. \\ \mathrm{Ta} \mathrm{thấy} {\mathrm{z}}_1 \mathrm{và} {\mathrm{z}}_2 \mathrm{là} 2 \mathrm{số} \mathrm{phức} \mathrm{liên} \mathrm{hợp} \mathrm{của} \mathrm{nhau} \mathrm{và} \left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|=\sqrt{3} \\ \Rightarrow \mathrm{S}=\frac{\overline{{\mathrm{z}}_1}}{\left|{\mathrm{z}}_2\right|}+\frac{\overline{{\mathrm{z}}_2}}{\left|{\mathrm{z}}_1\right|}=\frac{{\mathrm{z}}_2+{\mathrm{z}}_1}{\sqrt{3}}=\frac{2}{\sqrt{3}} \end{gathered}$$

Đáp án D

$$\begin{gathered} \mathrm{PT} \mathrm{có} \mathrm{nghiệm}: \\ \left\{\begin{array}{l}\mathrm{z}=\frac{1}{\mathrm{i}}\\ \mathrm{z}=\frac{1}{{\mathrm{i}}^2}\\ ...\\ \mathrm{z}=\frac{1}{{\mathrm{i}}^{20}}\end{array}\right. \\ \mathrm{Ta} \mathrm{thấy}: {\mathrm{i}}^{2\mathrm{n}+2}={(-1)}^{\mathrm{n}+1} \left(\mathrm{lũy} \mathrm{thừa} \mathrm{bậc} \mathrm{chẵn} \mathrm{của} \mathrm{i} \mathrm{luôn} \mathrm{bằng} \pm 1\right) \\ \mathrm{và} {\mathrm{i}}^{2\mathrm{n}+1}=\mathrm{i}.{(-1)}^{\mathrm{n}} \left(\mathrm{lũy} \mathrm{thừa} \mathrm{bậc} \mathrm{lẻ} \mathrm{của} \mathrm{i} \mathrm{luôn} \mathrm{bằng} \pm \mathrm{i} \right) \\ \Rightarrow \mathrm{PT}\hspace{0.17em}\mathrm{luôn} \mathrm{chỉ} \mathrm{có} 4 \mathrm{nghiệm} \mathrm{là} \mathrm{z}=\pm 1 \mathrm{và} \mathrm{z}=\pm \frac{1}{\mathrm{i}} \end{gathered}$$

Đáp án A

$$\begin{gathered} \mathrm{z}=-1+\mathrm{i} \mathrm{có} \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{là} \mathrm{M}(-1;1) \\ \mathrm{M}\text{'} \mathrm{đối} \mathrm{xứng} \mathrm{với} \mathrm{M} \mathrm{qua} ∆ \Rightarrow \left\{\begin{array}{l}\mathrm{H} \mathrm{là} \mathrm{trung} \mathrm{điểm} \mathrm{của} \mathrm{MM}\text{'} (\mathrm{H}\in ∆)\\ \mathrm{MH}\perp ∆\end{array}\right. \\ \mathrm{Gọi} \mathrm{H} (\mathrm{t}; \mathrm{t}+1)\in ∆. \mathrm{Ta} \mathrm{có} \overrightarrow{{\mathrm{n}}_{∆}}=(1;-1) \Rightarrow \overrightarrow{{\mathrm{u}}_{∆}}=(1;1); \overrightarrow{\mathrm{MH}}=(\mathrm{t}+1;\mathrm{t}) \\ \mathrm{Do} \mathrm{MH}\perp ∆\Rightarrow \overrightarrow{{\mathrm{u}}_{∆}}\perp \overrightarrow{\mathrm{MH}} \\ \Rightarrow \mathrm{t}+1+\mathrm{t}=0\Rightarrow \mathrm{t}=\frac{-1}{2} \Rightarrow \mathrm{H}\left(\frac{-1}{2};\frac{1}{2}\right) \\ \Rightarrow \mathrm{M}\text{'}\left(0;0\right) \\ \Rightarrow \mathrm{w}=0 \end{gathered}$$

Đáp án B

$$\begin{gathered} {\mathrm{z}}^3+8=0\iff \left\{\begin{array}{l}{\mathrm{z}}_1=-2\\ {\mathrm{z}}_2=1-\mathrm{i}\sqrt{3}\\ {\mathrm{z}}_3=1+\mathrm{i}\sqrt{3}\end{array}\right. \\ \mathrm{Gọi} \mathrm{M}\left(-2;0\right); \mathrm{N}\left(1;-\sqrt{3}\right); \mathrm{P}\left(1;\sqrt{3}\right) \mathrm{biểu} \mathrm{diễn} 3 \mathrm{số} \mathrm{phức} {\mathrm{z}}_1; {\mathrm{z}}_2 \mathrm{và} {\mathrm{z}}_3 \\ \overrightarrow{\mathrm{MN}}=\left(3;-\sqrt{3}\right)\Rightarrow \mathrm{MN}=2\sqrt{3} \\ \overrightarrow{\mathrm{MP}}=\left(3;\sqrt{3}\right)\Rightarrow \mathrm{MP}=2\sqrt{3} \\ \overrightarrow{\mathrm{NP}}=\left(0;2\sqrt{3}\right)\Rightarrow \mathrm{NP}=2\sqrt{3} \\ \Rightarrow △\mathrm{MNP} \mathrm{đều} \\ \Rightarrow {\mathrm{S}}_{△\mathrm{MNP}}=\frac{{\left(2\sqrt{3}\right)}^2.\sqrt{3}}{4}=3\sqrt{3} \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} (\mathrm{x};\mathrm{y}\in \mathrm{ℝ}) \\ \mathrm{Ta} \mathrm{có}: \left|\mathrm{z}+\overline{\mathrm{z}}-\mathrm{i}\right|=1 \\ \iff \left|\mathrm{x}+\mathrm{yi}+\mathrm{x}-\mathrm{yi}-\mathrm{i}\right|=1 \\ \iff \left|2\mathrm{x}-\mathrm{i}\right|=1 \\ \iff 4{\mathrm{x}}^2+1=1 \\ \iff \mathrm{x}=0 \\ \mathrm{KL}: \left\{\mathrm{M}\right\} \mathrm{là} \mathrm{trục} \mathrm{tung} \end{gathered}$$

Đáp án B

$$\begin{gathered} \mathrm{Giả} \mathrm{sử} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \Rightarrow \overline{\mathrm{z}}=\mathrm{a}-\mathrm{bi} \\ \mathrm{z}=\overline{\mathrm{z}}\iff \left\{\begin{array}{l}\mathrm{a}=\mathrm{a}\\ \mathrm{b}=-\mathrm{b}\end{array}\iff \left\{\begin{array}{l}\mathrm{a}\in \mathrm{ℝ}\\ \mathrm{b}=0\end{array}\right.\right. \\ \mathrm{Trong} 4 \mathrm{đáp} \mathrm{án} \mathrm{ta} \mathrm{thấy} \mathrm{đáp} \mathrm{án} \mathrm{B} \mathrm{thỏa} \mathrm{mãn} \mathrm{điều} \mathrm{kiện} \end{gathered}$$

Đáp án A

$$\begin{gathered} \mathrm{Giả} \mathrm{sử} \mathrm{z}=\mathrm{a}+\mathrm{bi} \left(\mathrm{a};\mathrm{b}\in \mathrm{ℝ}\right) \\ \mathrm{z}\in \mathrm{ℝ}\Rightarrow \mathrm{b}=0 \\ \Rightarrow \mathrm{iz}=\mathrm{i}\left(\mathrm{a}+\mathrm{bi}\right)=\mathrm{ai}-\mathrm{b}=\mathrm{ai} \notin \mathrm{ℝ} \mathrm{khi} \mathrm{a}\ne 0 \\ \mathrm{Ta} \mathrm{có}:{\mathrm{z}}^2=1 \mathrm{có} 2 \mathrm{nghiệm} \mathrm{phức} \mathrm{nhưng} \mathrm{PT} {\mathrm{z}}^4=1 \mathrm{có} 4 \mathrm{nghiệm} \mathrm{phức} \mathrm{phân} \mathrm{biệt} \\ \Rightarrow 2 \mathrm{PT} \mathrm{không} \mathrm{tương} \mathrm{đương} \mathrm{nhau} \\ {\left(\mathrm{z}-1\right)}^3=-1\iff \left\{\begin{array}{l}\mathrm{z}=0\\ \mathrm{z}-1=\frac{1}{2}+\frac{\mathrm{i}\sqrt{3}}{2}\\ \mathrm{z}-1=\frac{1}{2}-\frac{\mathrm{i}\sqrt{3}}{2}\end{array}\right. \\ \Rightarrow \mathrm{PT} \mathrm{có} 3 \mathrm{nghiệm} \\ \mathrm{z}+\overline{\mathrm{z}}=\mathrm{a}+\mathrm{bi}+\mathrm{a}-\mathrm{bi}=2\mathrm{a}=0\iff \mathrm{z}=0 \mathrm{hoặc} \mathrm{z}=\mathrm{bi} \mathrm{với} \mathrm{b}\in \mathrm{ℝ} \end{gathered}$$

Đáp án C

Thay từng nghiệm của đáp án để so sánh:

$$\begin{gathered} \mathrm{Với} \mathrm{z}=1\Rightarrow \mathrm{iz}+1=1+\mathrm{i}\ne 0 \Rightarrow \mathrm{A} \mathrm{loại} \\ \mathrm{Với} \mathrm{z}=-1\Rightarrow \mathrm{iz}+1=1-\mathrm{i}\ne 0 \Rightarrow \mathrm{B} \mathrm{loại} \\ \mathrm{Với} \mathrm{z}=\mathrm{i}\Rightarrow \left\{\begin{array}{l}\mathrm{iz}+1=1+{\mathrm{i}}^2=0 \\ {\mathrm{z}}^4-1={\mathrm{i}}^4-1=1-1=0\end{array}\right.\Rightarrow \mathrm{C} \mathrm{đúng} \end{gathered}$$

Đáp án D

Số phức a+bi có điểm biểu diễn trên mặt phẳng phức là M(a;b)

$\Rightarrow$Điểm M(-2;3) biểu diễn số phức z = -2+3i

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} \left(\mathrm{a};\mathrm{b}\in \mathrm{ℝ}\right) \\ {\mathrm{z}}^2={\left|\mathrm{z}\right|}^2\iff {\left(\mathrm{a}+\mathrm{bi}\right)}^2={\left|\mathrm{a}+\mathrm{bi}\right|}^2 \\ \iff {\mathrm{a}}^2-{\mathrm{b}}^2+2\mathrm{abi}={\mathrm{a}}^2+{\mathrm{b}}^2 \\ \iff \left\{\begin{array}{l}{\mathrm{a}}^2-{\mathrm{b}}^2={\mathrm{a}}^2+{\mathrm{b}}^2\\ 2\mathrm{ab}=0\end{array}\right. \\ \iff \left\{\begin{array}{l}{\mathrm{b}}^2=0\\ 2\mathrm{ab}=0\end{array}\iff \left\{\begin{array}{l}\mathrm{a}\in \mathrm{ℝ}\\ \mathrm{b}=0\end{array}\right.\right. \end{gathered}$$

Đáp án B

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

Đáp án D

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

Đáp án B

Có $z^3+1=\left(z+1\right)\left(z^2-z+1\right)=0$

$\iff \mathrm{z}=1 \mathrm{và} \mathrm{z}=\frac{1\pm \mathrm{i}\sqrt{3}}{2}$

Đáp án D

Đáp án C

Lưu ý 

$$\begin{gathered} Đặt z=a+bi \Rightarrow \overline{z}=a-bi \left(a;b\in \mathrm{ℝ}\right) \\ \Rightarrow \left\{\begin{array}{l}z+\overline{z}=2a\in \mathrm{ℝ}\\ z-\overline{z}=2bi\notin \mathrm{ℝ}\iff b\ne 0\\ z.\overline{z}=a^2+b^2\in \mathrm{ℝ}\end{array}\right. \end{gathered}$$

Đáp án D

$\mathrm{z}=\left|\mathrm{z}\right|\iff \mathrm{a}+\mathrm{bi}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\Rightarrow \mathrm{b}=0$ $\mathrm{và} \mathrm{a}=\sqrt{{\mathrm{a}}^2}=\left|\mathrm{a}\right|\iff \mathrm{a}\ge 0$

Đáp án A

$$\begin{gathered} {\left|\mathrm{z}\right|}_{\min }=\mathrm{d}\left(\mathrm{O};\left(\mathrm{d}\right)\right)=\frac{\left|3\right|}{\sqrt{2^2+{(-1)}^2}}=\frac{3}{\sqrt{5}} \\ \mathrm{với} \mathrm{d}: 2\mathrm{x}-\mathrm{y}+3=0 \end{gathered}$$

Đáp án B

$$\begin{gathered} {\left|\mathrm{z}\right|}_{\min }=\mathrm{d}\left(\mathrm{O};\left(∆\right)\right)=\frac{\left|4\right|}{\sqrt{3^2+{\left(-1\right)}^2}}=\frac{4}{\sqrt{10}}=\sqrt{\frac{8}{5}} \\ \mathrm{với} ∆: 3\mathrm{x}-\mathrm{y}+4=0 \end{gathered}$$

Đáp án C

$$\begin{gathered} {\mathrm{z}}_1={\mathrm{z}}_2\iff \left\{\begin{array}{l}{\mathrm{a}}_1={\mathrm{a}}_2\\ {\mathrm{b}}_1={\mathrm{b}}_2\end{array}\right.\mathrm{với} {\mathrm{a}}_1; {\mathrm{a}}_2 \mathrm{là} \mathrm{phần} \mathrm{thực} \mathrm{của} {\mathrm{z}}_1\mathrm{và} {\mathrm{z}}_2 \Rightarrow \mathrm{A} \mathrm{sai} \\ \left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|\iff {\mathrm{a}}_1^2+{\mathrm{b}}_1^2={\mathrm{a}}_2^2+{\mathrm{b}}_2^2 \Rightarrow \mathrm{B}\hspace{0.17em}\mathrm{sai} \\ {\mathrm{z}}_1+{\mathrm{z}}_2={\mathrm{a}}_1+{\mathrm{a}}_2+({\mathrm{b}}_1+{\mathrm{b}}_2)\mathrm{i}\in \mathrm{ℝ}\iff {\mathrm{b}}_1+{\mathrm{b}}_2=0 \Rightarrow \mathrm{C} \mathrm{đúng} \\ {\mathrm{z}}_1.{\mathrm{z}}_2=\left({\mathrm{a}}_1+{\mathrm{ib}}_1\right)\left({\mathrm{a}}_2+{\mathrm{ib}}_2\right)={\mathrm{a}}_1{\mathrm{a}}_2-{\mathrm{b}}_1{\mathrm{b}}_2+\left({\mathrm{a}}_1{\mathrm{b}}_2+{\mathrm{a}}_2{\mathrm{b}}_1\right)\mathrm{i} \\ {\mathrm{z}}_1.{\mathrm{z}}_2\in \mathrm{ℝ}\iff {\mathrm{a}}_1{\mathrm{b}}_2+{\mathrm{a}}_2{\mathrm{b}}_1=0 \Rightarrow \mathrm{D} \mathrm{sai} \end{gathered}$$

Đáp án D

Mỗi số phức đều có điểm biểu diễn khác nhau 

nếu z₁ = z₂ thì điểm biểu diễn 2 số phức z₁ , z₂ trùng nhau

Đáp án B

Nhận xét:

Phương trình bậc 2 giải ra trên trên tập số phức thu được 2 nghiệm phức là 2 số phức liên hợp của nhau

Đáp án C

Số phức z = 1-i có điểm biểu diễn là M(1;-1)

Đáp án C

Số phức z = 1-i có điểm biểu diễn là M(1;-1)

Đáp án A

Phương trình 

hoặc 4 nghiệm là 

$\Rightarrow \mathrm{T}=2(-125+27)=-196$

Đáp án D

$$\begin{gathered} \mathrm{Tập} \mathrm{hợp} \mathrm{điểm} \mathrm{M} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z} \mathrm{năm} \mathrm{trên} \mathrm{đường} \mathrm{thẳng} \left(\mathrm{d}\right): 3\mathrm{x}+2\mathrm{y}-5=0 \\ \mathrm{Để} \mathrm{z} \mathrm{có} \mathrm{phần} \mathrm{thực} \mathrm{bằng} \mathrm{phần} \mathrm{ảo} \mathrm{thi} \mathrm{x}=\mathrm{y} \\ \Rightarrow 3\mathrm{x}+2\mathrm{x}-5=0\iff \mathrm{x}=1\Rightarrow \mathrm{y}=1 \\ \Rightarrow \mathrm{z}=1+\mathrm{i} \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{Gọi} \mathrm{M}(\mathrm{x};\mathrm{y}) \mathrm{là} \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z}=\mathrm{x}+\mathrm{yi} \\ \mathrm{Ta} \mathrm{có}: \left|\mathrm{z}+1\right|+\left|\mathrm{z}-1\right|=6 \\ \iff \left|\mathrm{x}+1+\mathrm{yi}\right|+\left|\mathrm{x}-1+\mathrm{yi}\right|=6 \\ \iff \sqrt{{\left(\mathrm{x}+1\right)}^2+{\mathrm{y}}^2}+\sqrt{{\left(\mathrm{x}-1\right)}^2+{\mathrm{y}}^2}=6 \\ \mathrm{Gọi} {\mathrm{F}}_1\left(-1;0\right) \mathrm{và} {\mathrm{F}}_2\left(1;0\right) \Rightarrow {\mathrm{MF}}_1+{\mathrm{MF}}_2=6 \\ \Rightarrow \left\{\mathrm{M}\right\} \mathrm{năm} \mathrm{trên} \mathrm{elip} \left(\mathrm{E}\right) \mathrm{có} \mathrm{tiêu} \mathrm{cự} {\mathrm{F}}_1{\mathrm{F}}_2=2=2\mathrm{c}\Rightarrow \mathrm{c}=1, \mathrm{độ} \mathrm{dài} \mathrm{trục} \mathrm{lớn} 2\mathrm{a}=6\Rightarrow \mathrm{a}=3 \\ \Rightarrow \mathrm{Độ} \mathrm{dài} \mathrm{trục} \mathrm{bé}: 2\mathrm{b}=2\sqrt{{\mathrm{a}}^2-{\mathrm{c}}^2}=2\sqrt{3^2-1^2}=4\sqrt{2}\Rightarrow \mathrm{b}=2\sqrt{2} \\ \left(\mathrm{E}\right) \mathrm{có} \mathrm{phương} \mathrm{trình} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \Rightarrow \left\{\mathrm{M}\right\} \mathrm{là} \mathrm{elip}: \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{8}=1 \end{gathered}$$

Đáp án B

$$\begin{gathered} \mathrm{z}={\left(1-5\mathrm{i}\right)}^2=-24-10\mathrm{i} \\ \Rightarrow \mathrm{Phần} \mathrm{thực} \mathrm{của} \mathrm{z} \mathrm{là} -24 \end{gathered}$$

Đáp án A

$$\begin{gathered} \mathrm{z}=1-2\mathrm{i} \mathrm{có} \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{là} \mathrm{M}(1;-2) \\ \mathrm{Vì} \mathrm{M}\text{'} \mathrm{đối} \mathrm{xứng} \mathrm{với} \mathrm{M} \mathrm{qua} \mathrm{trục} \mathrm{Ox} \Rightarrow \mathrm{M}\text{'}(1;2) \\ \Rightarrow \mathrm{w}=1+2\mathrm{i} \end{gathered}$$