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Question 83 of 122

Q.If P represents the variable complex number zz and if ∣2z−1∣=2∣z∣|2z-1| = 2|z| then the locus of P is :

(a) the straight line x=14x = \dfrac{1}{4}
(b) the straight line y=14y = \dfrac{1}{4}
(c) the straight line z=12z = \dfrac{1}{2}
(d) the circle x2+y2−4x−1=0x^2 + y^2 - 4x - 1 = 0
Puducherry TnboardTamil Nadu HSC (DGE) Board 2017MCQ· 1mImportance★★★★★
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Substitute z=x+iyz=x+iy, square both sides of ∣2z−1∣=2∣z∣|2z-1|=2|z|, and simplify — the y2y^2 terms cancel, leaving a vertical line x=1/4x=1/4.

  1. Let z=x+iyz=x+iy, so 2z−1=(2x−1)+2iy2z-1 = (2x-1)+2iy.
  2. ∣2z−1∣=2∣z∣|2z-1|=2|z| means (2x−1)2+(2y)2=2x2+y2\sqrt{(2x-1)^2+(2y)^2} = 2\sqrt{x^2+y^2}.
  3. Square both sides: (2x−1)2+4y2=4(x2+y2)=4x2+4y2(2x-1)^2+4y^2 = 4(x^2+y^2) = 4x^2+4y^2.
  4. Expand the left side: 4x2−4x+1+4y2=4x2+4y24x^2-4x+1+4y^2 = 4x^2+4y^2. …

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