Question

cho pt \(\left(z^2+3z-1\right)^2+\left(2z+3\right)^2=0\) tính \(\left|2z+1+i\right|\)

Answer 1

\(\left(z^2+1+3z-2\right)^2+\left(2z-3\right)^2=0\\ \Leftrightarrow\left(z^2+1\right)^2+2\left(z^2+1\right)\left(3z-2\right)+\left(3z-2\right)^2+\left(2z-3\right)^2=0\\ \Leftrightarrow\left(z^2+1\right)^2+2\left(z^2+1\right)\left(3z-2\right)+\left[\left(3z-2\right)^2+\left(2z-3\right)^2\right]=0\\\Leftrightarrow\left(z^2+1\right)^2+2\left(z^2+1\right)\left(3z-2\right)+13\left(z^2+1\right)=0\Leftrightarrow\left(z^2+1\right)\left(z^2+6z+10\right)=0\)

Giải ra được:

\(\left[\begin{matrix}z=\pm i\\z=-3+i\\z=-3-i\end{matrix}\right.\)