Số phức nào dưới đây có phần thực là một số âm? (1+i)^2000

* Hướng dẫn giải

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