Answer
$z=\frac{1+\sqrt 5}{2}i$
$z=\frac{1-\sqrt 5}{2}i$
Work Step by Step
$z^2-iz+1=0~~$ ($a=1,b=-i,c=1$):
$z=\frac{-b±\sqrt {b^2-4ac}}{2a}=\frac{-(-i)±\sqrt {(-i)^2-4(1)(1)}}{2(1)}=\frac{i±\sqrt {-5}}{2}=\frac{i±\sqrt 5i}{2}$
$z=\frac{1+\sqrt 5}{2}i$ or $z=\frac{1-\sqrt 5}{2}i$
