Let z=x+iy be a complex number. If 1+¯¯¯zz is a real number, then
A
x=−12 or y=−12
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B
x=−12 or y=1
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C
x=0 or y=−12
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D
x=−12 or y=0
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Solution
The correct option is Dx=−12 or y=0 Given : z=x+iy, then ¯¯¯z=x−iy
Now, 1+¯¯¯zz=1+x−iyx+iy=1+x−iyx+iy×x−iyx−iy=x(1+x)−y2+i(−2xy−y)x2+y2
As 1+¯¯¯zz is a real number, so −2xy−yx2+y2=0⇒y(2x+1)=0⇒y=0 or x=−12