Số phức z=a+bi biết z=1+i+i^2

* Hướng dẫn giải

Đáp án D

Ta có $1+x+x^2+\mathrm{...}+x^{2017}=\frac{x^{2017}-1}{x-1}$

$\begin{array}{l}\Rightarrow 1+2x+\mathrm{...}+2017x^{2016}=\frac{2017x^{2016}\left(x-1\right)-x^{2017}+1}{{\left(x-1\right)}^2}\\ \Rightarrow z=1+i+i^2+2i^3+3i^4+\mathrm{...}+2017i^{2018}\\ =1+i+i^2\left(1+2i+3i^2+\mathrm{...}+2017i^{2016}\right)\\ =1+i-\frac{2017i^{2016}\left(i-1\right)-i^{2017}+1}{{\left(i-1\right)}^2}\\ =1+i+1008+1008i=1009+1009i\end{array}$