The equation $z^2-i{|z-1|}^2=0,$ where $i=\sqrt{-1}$ has 

On putting       z = x + iy, we have

${(x+iy)}^2-i{|x+iy-1|}^2=0$

$x^2-y^2+2ixy-i({(x-1)}^2+y^2)=0$

On comparing real and imaginary parts, we get

$x^2-y^2=0and\text{\hspace{0.17em}\hspace{0.17em}}2xy={(x-1)}^2+y^2$

Case 1  When y= x, then $2xy={(x-1)}^2+y^2$ reduces to

$2x^2={(x-1)}^2+x^2$

0=-2x+1

$x=\frac{1}{2}=y$

$z=x+iy=\frac{1+i}{2}$

Case II   when y =-x. then $2xy={(x-1)}^2+y^2$ reduces to

$-2x^2={(x-1)}^2+x^2$

$\Rightarrow {(x-1)}^2+3x^2=0\text{\hspace{0.17em}\hspace{0.17em}}.$ which is not possible

From Eqs. (i) and (ii) , we get $z=\frac{1+i}{2}$

i.e ,  no real  and no purely imaginary   roots and $\left|z\right|=\frac{1}{\sqrt{2}}<1$