The locus of a point on the argand plane represented by the complex number z, when z satisfies the condition | z-1+iz+1-i |= Re ( z-1+iz+1-i ) is

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Accepted Answer

Correct answer: A. (a) Given condition is ( z-1+iz+1-i )= ( Re ⁡ ( z-1+iz+1-i ) ) ,z -1+i Let z=x+ iy, then z-1+iz+1-i= (x-1 )+i (y+1 ) (x+1 )+i (y-1 ) (x+1 )-i (y-1 ) (x+1 )-i (y-1 )= ((x^2-1)+(y^2-1))+i [ (x+1 ) (y+1 ) (x+1 )^2+ (y-1 )^2 Since |z |= x^2+y^2 So, ( z-1+iz+1-i )= ((x^2-)+(y^2-1))^2+ (2 (x+y ) )^2( (x+1 )^2+ (y-1 )^2)^2 and ( Re ⁡ ( z-1+iz+1-i ) )= ((x^2-1)+(y^2-1)) (x+1 )^2+ (y-1 )^2 Now, according to given condition ((x^2-1)+(y^2-1))^2+ [2 (x+y ) ]^2 (x+1 )^2+ (y-1 )^2= ((x^2-1)+(y^2-1)) (x+1 )^2+ (y-1 )^2 ⇒((x^2-1)+(y^2-1))^2+4 (x+y )^2=((x^2-1)+(y^2-1))^2 ⇒x+y=0 is a straight line, but as z -1+i so locus of z is a straight line does not contain the point (-1+i ). Hence, option (a) is correct.

$The locus of a point on the argand plane represented by the complex number z, when z satisfies the condition \frac{z - 1 + i}{z + 1 - i} = Re (\frac{z - 1 + i}{z + 1 - i}) is$

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