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Question

If z is a complex number, then the number of solution(s) for the equation z2=¯¯¯z is

A
1
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B
2
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C
4
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D
5
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Solution

The correct option is C 4
Let z=x+iy
As z2=¯¯¯z
(x+iy)2=xiyx2y2+2ixy=xiy
Equality of complex numbers holds when
x2y2=x, 2xy=y2xy=yy(2x+1)=0y=0,x=12

When y=0, then
x2y2=xx(x1)=0x=0,1

When x=12, then
x2y2=x14y2=12y2=34y=±32

Therefore,
(x,y)=(0,0),(1,0),(12,32),(12,32)

Hence, there are 4 complex numbers satisfying the given equation.

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