Tập nghiệm của phương trình \(\left(\frac{z+i}{z-i}\right)^4=1\) là
\(\left\{-1;0;1\right\}\). \(\left\{0;1\right\}\). \(\left\{-1;0;2\right\}\). \(\left\{-1;1;2\right\}\). Hướng dẫn giải:\(\left(\frac{z+i}{z-i}\right)^4=1\Leftrightarrow\left[\left(\frac{z+i}{z-i}\right)^2-1\right]\left[\left(\frac{z+i}{z-i}\right)^2+1\right]=0\)
TH1: \(\left(\frac{z+i}{z-i}\right)^2=1\Leftrightarrow\frac{z+i}{z-i}=\pm1\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}z+i=z-i\\z+i=-z+i\end{array}\right.\)
\(\Leftrightarrow z=0\).
TH2: \(\left(\frac{z+i}{z-i}\right)^2=-1\Leftrightarrow\left(\frac{z+i}{z-i}\right)^2-i^2=0\Leftrightarrow\left(\frac{z+i}{z-i}-i\right)\left(\frac{z+i}{z-i}+i\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{z+i}{z-i}=i\\\frac{z+i}{z-i}=-i\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}z+i=i\left(z-i\right)\\z+i=-i\left(z-i\right)\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}z\left(1-i\right)=-i^2-i\\z\left(1+i\right)=i^2-i\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}z\left(1-i\right)=1-i\\z\left(1+i\right)=-1-i\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}z=1\\z=-1\end{array}\right.\).
Đáp số: \(\left\{-1;0;1\right\}\) .