đặc \(z=a+bi\) (\(a;b\in R\) ; \(i^2=-1\))
ta có : \(8z^2-4z+1=0\Leftrightarrow8\left(a+bi\right)^2-4\left(a+bi\right)+1=0\)
\(\Leftrightarrow8\left(a^2+2abi-b^2\right)-4\left(a+bi\right)+1=0\)
\(\Leftrightarrow8a^2+16abi-8b^2-4a+4bi+1=0\)
\(\Leftrightarrow\left(8a^2-8b^2-4a+1\right)+\left(16ab+4b\right)i=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}8a^2-8b^2-4a+1=0\\16ab+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}b=\dfrac{-1}{4}\\a=\dfrac{1}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}b=\dfrac{1}{4}\\a=\dfrac{1}{4}\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow z=\dfrac{1}{4}+\dfrac{1}{4}i;z=\dfrac{1}{4}-\dfrac{1}{4}i\)
vậy ................................................................................................................................