Answer
$z=(1+\sqrt 2)i$ or $z=(1-\sqrt 2)i$
Work Step by Step
$z^2-2iz+1=0~~$ ($a=1,b=-2i,c=1$):
$z=\frac{-b±\sqrt {b^2-4ac}}{2a}=\frac{-(-2i)±\sqrt {(-2i)^2-4(1)(1)}}{2(1)}=\frac{2i±\sqrt {-8}}{2}=\frac{2i±2\sqrt 2i}{2}$
$z=(1+\sqrt 2)i$ or $z=(1-\sqrt 2)i$
