$(\frac{z+i}{z-i})^4=1\Leftrightarrow (1+\frac{2i}{z-i})^4=1$
Số $1$ có $4$ căn bậc $4$ là $1 , -1 , i , -i$
Nếu $1+\frac{2i}{z-i}=1\Rightarrow \frac{2i}{z-i}=0$ Không tồn tại $z$ thỏa mãn
Nếu $1+\frac{2i}{z-i}=-1\Rightarrow \frac{2i}{z-i}=-2\Rightarrow z-i=-i\Rightarrow z=0$
Nếu $1+\frac{2i}{z-i}=i\Rightarrow \frac{2i}{z-i}=i-1\Rightarrow z-i=\frac{2i}{i-1}=\frac{2i.(i+1)}{(i-1)(i+1)}$
$=\frac{2i^2+2i}{i^2-1}=\frac{-2+2i}{-2}=1-i$
$\Rightarrow z=1$
Nếu $1+\frac{2i}{z-i}=-i\Rightarrow \frac{2i}{z-i}=-i-1\Rightarrow z-i=-\frac{2i}{i+1}=\frac{2i(i-1)}{(i+1)(i-1)}$
$=-\frac{2i^2-2i}{i^2-1}=-\frac{-2-2i}{-2}=-1-i$
$\Rightarrow z=-1$