Cho

A

- Hướng dẫn giải

Ta có 

$$\begin{gathered} \frac{1+i}{1-i}=\frac{\left(1+i\right)\left(1+i\right)}{\left(1+i\right)\left(1-i\right)} \\ =\frac{i^2+2i+1}{1-i^2}=1 \\ \Rightarrow z=i^{2021}={\left(i^2\right)}^{1010}i=1 \end{gathered}$$

Do đó 

$$\begin{gathered} M=z^k+z^{k+1}+z^{k+2}+z^{k+3} \\ =i^k+i^{k+1}+i^{k+2}+i^{k+3} \\ =i^k\left(1+i+i^2+i^3\right) \\ =i^k\left(1+i-1-i\right)=0 \end{gathered}$$

Đáp án A