191 câu Bài tập số phức mức độ cơ bản, nâng cao cực hay có lời giải chi tiết(P6)

Đáp án C

Phương trình 

$\iff \left\{\begin{array}{l}{\mathrm{z}}_{1,2}=-1\pm \mathrm{i}\sqrt{2}\\ {\mathrm{z}}_{3,4}=1\pm \mathrm{i}\sqrt{2}\end{array}\right.$

$$\begin{gathered} \mathrm{Ta} \mathrm{thấy} \left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|=\left|{\mathrm{z}}_3\right|=\left|{\mathrm{z}}_4\right|=\sqrt{3} \\ \mathrm{Mặt} \mathrm{khác} \left|\overline{{\mathrm{z}}_1}\right|=\left|{\mathrm{z}}_2\right|, \left|\overline{{\mathrm{z}}_2}\right|=\left|{\mathrm{z}}_1\right|, \left|\overline{{\mathrm{z}}_3}\right|=\left|{\mathrm{z}}_4\right|, \left|\overline{{\mathrm{z}}_4}\right|=\left|{\mathrm{z}}_3\right| \\ ( \mathrm{do} \mathrm{cặp} {\mathrm{z}}_1, {\mathrm{z}}_2 \mathrm{và} {\mathrm{z}}_3, {\mathrm{z}}_{4 } \mathrm{là} \mathrm{các} \mathrm{số} \mathrm{phức} \mathrm{liên} \mathrm{hợp} \mathrm{của} \mathrm{nhau} ) \\ \Rightarrow \mathrm{T}=4\sqrt{3} \end{gathered}$$

Đáp án D

$$\begin{gathered} {\mathrm{z}}^3+1=0\iff \left(\mathrm{z}+1\right)\left({\mathrm{z}}^2-\mathrm{z}+1\right)=0 \\ \iff \left\{\begin{array}{l}\mathrm{z}=-1\\ \mathrm{z}=\frac{1}{2}\pm \mathrm{i}.\frac{\sqrt{3}}{2}\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}\mathrm{Điểm} \mathrm{M}\left(-1;0\right) \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z}=-1\\ \mathrm{Điểm} \mathrm{N}\left(\frac{1}{2};\frac{\sqrt{3}}{2}\right) \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z}=\frac{1}{2}+\mathrm{i}.\frac{\sqrt{3}}{2}\\ \mathrm{Điểm} \mathrm{P}\left(\frac{1}{2};\frac{-\sqrt{3}}{2}\right) \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z}=\frac{1}{2}-\mathrm{i}.\frac{\sqrt{3}}{2}\end{array}\right. \\ \mathrm{Ta} \mathrm{có}: \\ \overrightarrow{\mathrm{MN}}=\left(\frac{3}{2};\frac{\sqrt{3}}{2}\right)\Rightarrow \mathrm{MN}=\sqrt{3} \\ \overrightarrow{\mathrm{MP}}=\left(\frac{3}{2};\frac{-\sqrt{3}}{2}\right)\Rightarrow \mathrm{MP}=\sqrt{3} \\ \overrightarrow{\mathrm{NP}}=\left(0;-\sqrt{3}\right)\Rightarrow \mathrm{NP}=\sqrt{3} \\ \Rightarrow △\mathrm{MNP} \mathrm{đều} \end{gathered}$$

Đáp án D

$$\begin{gathered} \mathrm{Điểm} \mathrm{M}\left(2;3\right) \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z}=2+3\mathrm{i} \Rightarrow \overline{\mathrm{z}}=2-3\mathrm{i} \\ \Rightarrow \left\{\begin{array}{l}\mathrm{Đáp} \mathrm{án} \mathrm{A} \mathrm{và} \mathrm{B} \mathrm{sai}\\ \mathrm{Điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \overline{\mathrm{z}} \mathrm{là} \mathrm{N}\left(2;-3\right)\Rightarrow \mathrm{Đáp} \mathrm{án} \mathrm{C} \mathrm{sai} \\ \left|\mathrm{z}\right|=\left|\overline{\mathrm{z}}\right|=\sqrt{13}\Rightarrow \mathrm{Đáp} \mathrm{án} \mathrm{D} \mathrm{đúng}\end{array}\right. \end{gathered}$$

Đáp án A

$$\begin{gathered} \mathrm{z}=-4=-4+0\mathrm{i} \\ \Rightarrow \overline{\mathrm{z}}=-4-0\mathrm{i}=-4 \end{gathered}$$

Đáp án D

$$\begin{gathered} \frac{\left(1-\mathrm{i}\right)\mathrm{z}}{1-\mathrm{iz}}=\mathrm{i} \\ \iff \left(1-\mathrm{i}\right)\mathrm{z}=\mathrm{i}\left(1-\mathrm{iz}\right) \\ \iff \left(1-\mathrm{i}\right)\mathrm{z}=\mathrm{i}-{\mathrm{i}}^2\mathrm{z}=\mathrm{z}+\mathrm{i} \\ \iff \mathrm{z}\left(1-\mathrm{i}-1\right)=\mathrm{i} \\ \iff -\mathrm{zi}=\mathrm{i} \\ \iff \mathrm{z}=-1 \end{gathered}$$

Đáp án B

$$\begin{gathered} \mathrm{Đặt} \mathrm{P}=\mathrm{i}+{\mathrm{i}}^2+...+{\mathrm{i}}^{20} \Rightarrow \mathrm{S}=\frac{1+\mathrm{P}}{21} \\ \mathrm{Ta} \mathrm{thấy} \mathrm{P} \mathrm{là} \mathrm{tổng} \mathrm{của} 1 \mathrm{cấp} \mathrm{số} \mathrm{nhân} \mathrm{với} {\mathrm{U}}_1=\mathrm{i}; {\mathrm{U}}_{20}={\mathrm{i}}^{20} \mathrm{và} \mathrm{công} \mathrm{bội} \mathrm{q}=\mathrm{i} \\ \Rightarrow \mathrm{P}={\mathrm{U}}_1.\frac{1-{\mathrm{q}}^{\mathrm{n}}}{1-\mathrm{q}}=\mathrm{i}.\frac{1-{\mathrm{i}}^{20}}{1-\mathrm{i}} \\ \mathrm{Mặt} \mathrm{khác}: {\mathrm{i}}^{20}={\left({\mathrm{i}}^2\right)}^{10}=1 \\ \Rightarrow \mathrm{P}=\mathrm{i}.\frac{1-1}{1-\mathrm{i}}=0 \\ \Rightarrow \mathrm{S}=\frac{1}{21} \end{gathered}$$

Đáp án A

$$\begin{gathered} \mathrm{Gọi} \mathrm{z}=\mathrm{x}+\mathrm{yi} \left(\mathrm{x};\mathrm{y}\in \mathrm{ℝ}\right)\Rightarrow \overline{\mathrm{z}}=\mathrm{x}-\mathrm{yi} \\ \mathrm{Ta} \mathrm{có}: \left|\overline{\mathrm{z}}-2\mathrm{i}\right|=4 \\ \iff \left|\mathrm{x}-\left(\mathrm{y}+2\right)\mathrm{i}\right|=4 \\ \iff {\mathrm{x}}^2+{\left(\mathrm{y}+2\right)}^2=16 \\ \mathrm{KL}: \left\{\mathrm{M}\right\} \mathrm{là} \mathrm{đường} \mathrm{tròn} \left(\mathrm{C}\right): {\mathrm{x}}^2+{\left(\mathrm{y}+2\right)}^2=16 \end{gathered}$$

Đáp án A

$$\begin{gathered} \mathrm{z}=\frac{2-5\mathrm{i}}{\mathrm{i}}=\frac{\mathrm{i}\left(2-5\mathrm{i}\right)}{{\mathrm{i}}^2}=\frac{5+2\mathrm{i}}{-1}=-5-2\mathrm{i} \\ \Rightarrow \mathrm{phần} \mathrm{ảo} \mathrm{của} \mathrm{z} \mathrm{là} \mathrm{l}=-2 \end{gathered}$$

Đáp án D

Có $z^3 = 8$ 

Đáp án B

Có 

Đáp án A

Phương trình có bốn nghiệm

Đáp án D

Đáp án D

$$\begin{gathered} \mathrm{Gọi} {\mathrm{z}}_1={\mathrm{a}}_1+{\mathrm{ib}}_1 \mathrm{và} {\mathrm{z}}_2={\mathrm{a}}_2+{\mathrm{ib}}_2 \left({\mathrm{a}}_1, {\mathrm{b}}_1, {\mathrm{a}}_2, {\mathrm{b}}_2\in \mathrm{ℝ}\right) \\ \left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|\iff {\mathrm{a}}_1^2+{\mathrm{b}}_1^2={\mathrm{a}}_2^2+{\mathrm{b}}_2^2 \\ {\mathrm{z}}_1^2={\mathrm{z}}_2^2 \iff \left\{\begin{array}{l}{\mathrm{a}}_1^2-{\mathrm{b}}_1^2={\mathrm{a}}_2^2-{\mathrm{b}}_2^2\\ {\mathrm{a}}_1{\mathrm{b}}_1={\mathrm{a}}_2{\mathrm{b}}_2\end{array}\right. \\ {\mathrm{z}}_1={\mathrm{z}}_2\iff \left\{\begin{array}{l}{\mathrm{a}}_1={\mathrm{a}}_2\\ {\mathrm{b}}_1={\mathrm{b}}_2\end{array}\right. \\ \overline{{\mathrm{z}}_1}=\overline{{\mathrm{z}}_2}\iff \left\{\begin{array}{l}{\mathrm{a}}_1={\mathrm{a}}_2\\ -{\mathrm{b}}_1=-{\mathrm{b}}_2\iff {\mathrm{b}}_1={\mathrm{b}}_2\end{array}\right. \end{gathered}$$

Đáp án A

$\mathrm{z}={\left(\frac{1}{2}-\frac{\mathrm{i}\sqrt{3}}{2}\right)}^4={\left[{\left(\frac{1}{2}-\frac{\mathrm{i}\sqrt{3}}{2}\right)}^2\right]}^2={\left(\frac{-1}{2}-\frac{\mathrm{i}\sqrt{3}}{2}\right)}^2=\frac{-1}{2}+\frac{\mathrm{i}\sqrt{3}}{2}$

Đáp án B

$$\begin{gathered} {\left(z-i\right)}^4=1 \\ \iff {\left(z-i\right)}^4-1=0 \\ \iff \left[{\left(z-i\right)}^2-1\right]\left[{\left(z-i\right)}^2+1\right]=0 \\ \iff \left\{\begin{array}{l}{\left(z-i\right)}^2=1\\ {\left(z-i\right)}^2=-1\end{array}\iff \left\{\begin{array}{l}z-i=\pm 1\\ z-i=\pm i\end{array}\right.\right.\iff \left\{\begin{array}{l}z=\pm 1+i\\ z=2i và z=0\end{array}\right. \end{gathered}$$

Đáp án D

$$\begin{gathered} {\mathrm{z}}^4+3{\mathrm{z}}^2+4=0 \\ \iff \left\{\begin{array}{l}{\mathrm{z}}^2=\frac{-3}{2}+\mathrm{i}\frac{\sqrt{7}}{2}\\ {\mathrm{z}}^2=\frac{-3}{2}+\mathrm{i}\frac{\sqrt{7}}{2}\end{array}\right. \\ \mathrm{T}={\mathrm{z}}_1^3+{\mathrm{z}}_2^3+{\mathrm{z}}_3^3+{\mathrm{z}}_4^3=\left({\mathrm{z}}_1+{\mathrm{z}}_2\right)\left[{\left({\mathrm{z}}_1+{\mathrm{z}}_2\right)}^2-3{\mathrm{z}}_1{\mathrm{z}}_2\right]+\left({\mathrm{z}}_3+{\mathrm{z}}_4\right)\left[{\left({\mathrm{z}}_3+{\mathrm{z}}_4\right)}^2-3{\mathrm{z}}_3{\mathrm{z}}_4\right] \\ \Rightarrow \mathrm{T}=0 \mathrm{do} {\mathrm{z}}_1+{\mathrm{z}}_2={\mathrm{z}}_3+{\mathrm{z}}_4=0 \end{gathered}$$

Đáp án A

$$\begin{gathered} Đặt z=x+yi \left(x;y\in \mathrm{ℝ}\right) \Rightarrow \overline{z}=x-yi \\ Ta có: 4\left|\overline{z}+1-3i\right|=3 \\ \iff \left|x+1-\left(y+3\right)i\right|=\frac{3}{4} \\ \iff {\left(x+1\right)}^2+{\left(y+3\right)}^2=\frac{3}{4} \end{gathered}$$

Đáp án D

$$\begin{gathered} \mathrm{Đặt} {\mathrm{z}}_1=\mathrm{a}+{\mathrm{ib}}_1 \mathrm{và} {\mathrm{z}}_2=\mathrm{c}+{\mathrm{ib}}_2 \\ \Rightarrow \left\{\begin{array}{l}{\mathrm{z}}_1^{\text{'}}=\left(1+\mathrm{i}\right){\mathrm{z}}_1=\left(1+\mathrm{i}\right)\left(\mathrm{a}+{\mathrm{ib}}_1\right)=\mathrm{a}+{\mathrm{ib}}_1+\mathrm{ai}-{\mathrm{b}}_1=\mathrm{a}-{\mathrm{b}}_1+\mathrm{i}(\mathrm{a}+{\mathrm{b}}_1)\\ {\mathrm{z}}_2^{\text{'}}=\left(1+\mathrm{i}\right){\mathrm{z}}_2=\left(1+\mathrm{i}\right)\left(\mathrm{c}+{\mathrm{ib}}_2\right)=\mathrm{c}+{\mathrm{ib}}_2+\mathrm{ci}-{\mathrm{b}}_2=\mathrm{c}-{\mathrm{b}}_2+\mathrm{i}(\mathrm{c}+{\mathrm{b}}_2)\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}{\mathrm{b}}_1^{\text{'}}=\mathrm{a}+{\mathrm{b}}_1\\ {\mathrm{b}}_2^{\text{'}}=\mathrm{c}+{\mathrm{b}}_2\end{array}\right. \\ \mathrm{Nếu} {\mathrm{b}}_1={\mathrm{b}}_2\Rightarrow {\mathrm{b}}_1^{\text{'}}-{\mathrm{b}}_2^{\text{'}}=\mathrm{a}-\mathrm{c}\iff {\mathrm{b}}_1^{\text{'}}=\left(\mathrm{a}-\mathrm{c}\right)+{\mathrm{b}}_2^{\text{'}} \Rightarrow \mathrm{A} \mathrm{sai} \\ \mathrm{Nếu} {\mathrm{b}}_1^{\text{'}}={\mathrm{b}}_2^{\text{'}} \Rightarrow \mathrm{a}+{\mathrm{b}}_1=\mathrm{c}+{\mathrm{b}}_2 ( \mathrm{a}=\mathrm{c} \mathrm{thì} {\mathrm{b}}_1={\mathrm{b}}_2 )\Rightarrow \mathrm{B} \mathrm{sai} \\ \mathrm{Nếu} {\mathrm{b}}_1=-{\mathrm{b}}_2 \Rightarrow {\mathrm{b}}_1^{\text{'}}+{\mathrm{b}}_2^{\text{'}}=\mathrm{a}+\mathrm{c}+{\mathrm{b}}_1+{\mathrm{b}}_2=\mathrm{a}+\mathrm{c}\Rightarrow \mathrm{C} \mathrm{sai} \\ \overline{{\mathrm{z}}_1}=\overline{{\mathrm{z}}_2}\Rightarrow \left\{\begin{array}{l}\mathrm{a}=\mathrm{c}\\ {\mathrm{b}}_1={\mathrm{b}}_2\end{array}\right. \\ \Rightarrow {\mathrm{b}}_1^{\text{'}}-{\mathrm{b}}_2^{\text{'}}=\mathrm{a}-\mathrm{c}+{\mathrm{b}}_1-{\mathrm{b}}_2=0\iff {\mathrm{b}}_1^{\text{'}}={\mathrm{b}}_2^{\text{'}}\Rightarrow \mathrm{D} \mathrm{đúng} \end{gathered}$$

Đáp án B

$$\begin{gathered} \mathrm{Gọi} {\mathrm{M}}_{\mathrm{o}}\left(\mathrm{x};\mathrm{y}\right) \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z}=\mathrm{x}+\mathrm{yi} \\ \Rightarrow \left\{\begin{array}{l}-\mathrm{z}=-\mathrm{x}-\mathrm{yi} \Rightarrow \mathrm{M}\left(-\mathrm{x};-\mathrm{y}\right)\\ \overline{\mathrm{z}}=\mathrm{x}-\mathrm{yi} \Rightarrow \mathrm{M}\text{'}\left(\mathrm{x};-\mathrm{y}\right)\end{array}\right. \\ \mathrm{Dễ} \mathrm{thấy} 2 \mathrm{điểm} \mathrm{M} \mathrm{và} \mathrm{M}\text{'} \mathrm{đối} \mathrm{xúng} \mathrm{với} \mathrm{nhau} \mathrm{qua} \mathrm{trục} \mathrm{Oy} \end{gathered}$$

Đáp án C

$$\begin{gathered} {\left(z-2\right)}^4=16 \\ \Rightarrow \left[{\left(z-2\right)}^2+4\right]\left[{\left(z-2\right)}^2-4\right]=0 \\ \iff \left\{\begin{array}{l}{\left(z-2\right)}^2+4=0\\ {\left(z-2\right)}^2-4=0\end{array}\right. \\ \iff \left\{\begin{array}{l}{\left(z-2\right)}^2=-4=4i^2\\ {\left(z-2\right)}^2=4\end{array}\right. \\ \iff \left\{\begin{array}{l}z-2=\pm 2i\\ x-2=\pm 2\end{array}\right.\iff \left\{\begin{array}{l}z=2\pm 2i\\ z=4 và z=0\end{array}\right. \end{gathered}$$

Đáp án A

Đáp án D

$$\begin{gathered} \mathrm{Đặt} \mathrm{P}=\left(1-\mathrm{i}\right)+{\left(1-\mathrm{i}\right)}^2+...+{\left(1-\mathrm{i}\right)}^{10} \\ \Rightarrow \mathrm{S}+1+\mathrm{P}=\mathrm{a}+\mathrm{bi} \\ \mathrm{Ta} \mathrm{thấy} \mathrm{P} \mathrm{là} \mathrm{tổng} \mathrm{của} 1 \mathrm{cấp} \mathrm{số} \mathrm{nhân} \mathrm{với} {\mathrm{U}}_1=1-\mathrm{i}, \mathrm{q}=1-\mathrm{i}, \mathrm{n}=10 \\ \Rightarrow \mathrm{P}={\mathrm{U}}_1.\frac{1-{\mathrm{q}}^{\mathrm{n}}}{1-\mathrm{q}}=\left(1-\mathrm{i}\right).\frac{1-{\left(1-\mathrm{i}\right)}^{10}}{1-\left(1-\mathrm{i}\right)} \\ \mathrm{Mặt} \mathrm{khác}: {\left(1-\mathrm{i}\right)}^{10}={\left[{\left(1-\mathrm{i}\right)}^2\right]}^5={\left(-2\mathrm{i}\right)}^5=-32{\mathrm{i}}^5=-32\mathrm{i} \\ \Rightarrow \mathrm{P}=\frac{\left(1-\mathrm{i}\right)\left(1+32\mathrm{i}\right)}{\mathrm{i}}=31-33\mathrm{i} \\ \Rightarrow \mathrm{S}=1+31-33\mathrm{i}=32-33\mathrm{i}=\mathrm{a}+\mathrm{bi} \\ \Rightarrow \left\{\begin{array}{l}\mathrm{a}=32\\ \mathrm{b}=-33\end{array}\right. \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: \mathrm{F}=4\mathrm{a}+3\mathrm{b} \Rightarrow \mathrm{a}=\frac{\mathrm{F}-3\mathrm{b}}{4} \\ \mathrm{Lại} \mathrm{có}: {\left(\mathrm{a}-4\right)}^2+{\left(\mathrm{b}-3\right)}^2=9 \\ \iff {\left(\frac{\mathrm{F}-3\mathrm{b}}{4}-4\right)}^2+{\left(\mathrm{b}-3\right)}^2=9 \\ \iff {\left(\mathrm{F}-3\mathrm{b}-16\right)}^2+16{\left(\mathrm{b}-3\right)}^2-144=0 \\ \iff {\left(\mathrm{F}-16\right)}^2-6\mathrm{b}\left(\mathrm{F}-16\right)+9{\mathrm{b}}^2+16{\mathrm{b}}^2-96\mathrm{b}+144-144=0 \\ \iff 25{\mathrm{b}}^2-6\mathrm{bF}+{\mathrm{F}}^2-32\mathrm{F}+256=0 \\ ∆\text{'}=9{\mathrm{F}}^2-25\left({\mathrm{F}}^2-32\mathrm{F}+256\right)=-16{\mathrm{F}}^2+800\mathrm{F}-6400=-16\left({\mathrm{F}}^2-50\mathrm{F}+400\right) \\ ∆\text{'}\ge 0\iff -16\left({\mathrm{F}}^2-50\mathrm{F}+400\right)\ge 0\iff 10\le \mathrm{F}\le 40 \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{z}=\left(2+3\mathrm{i}\right)\left(\mathrm{i}+1\right)=-1+5\mathrm{i} \\ \Rightarrow \mathrm{Điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z} \mathrm{là} \mathrm{M}\left(-1;5\right) \end{gathered}$$

Đáp án D

Có 

Đáp án B

$$\begin{gathered} {\mathrm{z}}^2+1=\mathrm{i}.\mathrm{z}. \mathrm{Chia} \mathrm{cả} 2 \mathrm{vế} \mathrm{cho} \mathrm{z}\ne 0 \mathrm{ta} \mathrm{được}: \mathrm{z}+\frac{1}{\mathrm{z}}=\mathrm{i} \\ \mathrm{Ta} \mathrm{có}:\mathrm{T}= {\mathrm{z}}^4+\frac{1}{{\mathrm{z}}^4}={\left({\mathrm{z}}^2+\frac{1}{{\mathrm{z}}^2}\right)}^2-2={\left[{\left(\mathrm{z}+\frac{1}{\mathrm{z}}\right)}^2-2\right]}^2-2 \\ \Rightarrow \mathrm{T}={\left({\mathrm{i}}^2-2\right)}^2-2={\left(-3\right)}^2-2=7 \end{gathered}$$

Đáp án A

$$\begin{gathered} {\mathrm{z}}^2-2\mathrm{z}+2=0 \iff \left\{\begin{array}{l}{\mathrm{z}}_1=1+\mathrm{i}\\ {\mathrm{z}}_2=1-\mathrm{i}\end{array}\right. \\ \mathrm{S}={\mathrm{z}}_1^{10}+{\mathrm{z}}_2^{10}={\left[{\left(1+\mathrm{i}\right)}^2\right]}^5+{\left[{\left(1-\mathrm{i}\right)}^2\right]}^5={\left(2\mathrm{i}\right)}^5+{\left(-2\mathrm{i}\right)}^5 \\ =2^5.{\mathrm{i}}^5+{\left(-2\right)}^5.{\mathrm{i}}^5=0 \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} \left(\mathrm{a},\mathrm{b}\in \mathrm{ℝ}\right) \\ {\mathrm{z}}^2+\overline{\mathrm{z}}=0 \\ \iff {\left(\mathrm{a}+\mathrm{bi}\right)}^2+\mathrm{a}-\mathrm{bi}=0 \\ \iff {\mathrm{a}}^2-{\mathrm{b}}^2+\mathrm{a}+\left(2\mathrm{ab}-\mathrm{b}\right)\mathrm{i}=0 \\ \iff \left\{\begin{array}{l}{\mathrm{a}}^2-{\mathrm{b}}^2+\mathrm{a}=0\\ \mathrm{b}(2\mathrm{a}-1)=0\end{array}\right. \\ \mathrm{TH}1: \mathrm{b}=0 \Rightarrow {\mathrm{a}}^2+\mathrm{a}=0\iff \left\{\begin{array}{l}\mathrm{a}=0\\ \mathrm{a}=-1\end{array}\Rightarrow \left\{\begin{array}{l}\mathrm{z}=0\\ \mathrm{z}=-1\end{array}\right.\right. \\ \mathrm{TH}2: \mathrm{a}=\frac{1}{2}\Rightarrow \frac{3}{4}-{\mathrm{b}}^2=0\iff \mathrm{b}=\pm \frac{\sqrt{3}}{2} \\ \Rightarrow \left\{\begin{array}{l}\mathrm{z}=\frac{1}{2}+\mathrm{i}.\frac{\sqrt{3}}{2}\\ \mathrm{z}=\frac{1}{2}-\mathrm{i}.\frac{\sqrt{3}}{2}\end{array}\right. \end{gathered}$$

Đáp án B

Đáp án C

$$\begin{gathered} \frac{\left(4-2\mathrm{i}\right)\mathrm{z}}{5\left(1-\mathrm{i}\right)}=\frac{-1+3\mathrm{i}}{2-\mathrm{i}}=-1+\mathrm{i} \\ \iff \left(4-2\mathrm{i}\right)\mathrm{z}=5\left(1-\mathrm{i}\right)\left(-1+\mathrm{i}\right)=10\mathrm{i} \\ \iff \mathrm{z}=\frac{10\mathrm{i}}{4-2\mathrm{i}}=-1+2\mathrm{i} \end{gathered}$$