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Question

Find how many nonzero complex numbers z satisfy z=iz2.

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Solution

z=iz2
Let z=x+iy
z2=(x+iy)2
=x2y2+2ixy
iz2=i(x2y2)2xy
x+2xy=0
x(1+2y)=0
x=0,y=12
or, x2y2=y
When x=0
y(y+1)=0
y=0,y=1
When y=12
x214=12
x2=14(not possible)
So, answer is 2

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