Answer :

Squaring both side

-i = (x + iy)^{2}

= (x^{2}-y^{2})+2ixy

x^{2}-y^{2} = 0

2xy = -1

As we know all that,

(x^{2}+y^{2})^{2} = (x^{2}-y^{2})^{2}+4x^{2}y^{2}

(x^{2}+y^{2})^{2} = 0+1

(x^{2}+y^{2})^{2} = 1

x^{2}+y^{2} = 1

X^{2}-y^{2} = 0 ……………………. (1)

x^{2}+y^{2} = 1 ……………….(2)

From (1)

x^{2} = y^{2}…………….(3)

2x^{2} = 1 (because x^{2} = y^{2})

2xy = -1

it means xy<0

Either x<0 , y > 0

Or x>0, y<0

X and y have different sign

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