NDA Math Complex Numbers

Given z2=i

−

z

Let us suppose that z=x+iy
⇒(x+iy)2=i(x−iy)
⇒x2−y2+2xyi=ix+y
Comparing real and imaginary part of both sides
x2−y2=y and 2xy=x
Taking 2xy=x
(2y−1)x=0
⇒x=0 and y=‌‌

1

2

if x=0
∴y+y2=0   (∵x2−y2=y)
y(y+1)=0
y=0 and −1
x=0   y=0 and −1
if y=

1

2

x2−‌‌

1

4

=‌‌

1

2

⇒x2=‌‌

3

4

⇒x=±‌‌

√3

2

Since given numbers are non zero complex numbers.
So, z1=0+(−1)i=−i
Hence z1+z2+z3=0