Gọi 

- Hướng dẫn giải

Đáp án cần chọn là: B

Ta có:

$$\begin{gathered} z^2+4z+5=0\iff {\left(z+2\right)}^2=-1\iff {\left(z+2\right)}^2=i^2 \\ \iff \left\{\begin{array}{l}{z}_1=-2+i\\ z_2=-2-i\end{array}\Rightarrow \left\{\begin{array}{l}{z}_1+1=i-1\\ z_2+1=-i-1\end{array}\right.\right. \end{gathered}$$

Khi đó ta được:

$$\begin{gathered} \left\{\begin{array}{l}{\left(z_1+1\right)}^2={\left(i-1\right)}^2=-2i\\ {\left(z_2+1\right)}^2={\left(-i-1\right)}^2=2i\end{array}\Rightarrow \left\{\begin{array}{l}{\left(z_1+1\right)}^4=-4\\ {\left(z_2+1\right)}^4=-4\end{array}\right.\right. \\ \Rightarrow {\left(z_1+1\right)}^{100}+{\left(z_2+1\right)}^{100}={\left(-4\right)}^{25}+{\left(-4\right)}^{25}=2.{\left(-2^2\right)}^{25}=-2^{51} \end{gathered}$$