If Re(z-1/2z+i) = 1, where z = x + iy, then the point (x, y)

Question 4:- If Re(z-1/2z+i) = 1, where z = x + iy, then the point (x, y) lies on a :

(A) Straight line whose slope is -2/3

(B) Straight line whose slope 3/2

(C) Circle whose diameter is √5/2

(D) Circle whose centre is at (-1/2, -3/2)

Solution:- Since, z = x + iy

$\dfrac{z-1}{2z + i} = \dfrac{x + iy-1}{2x + 2iy+i}$

⇒ $\dfrac{(x-1) + iy}{2x + i(2y+1)}\times \dfrac{2x - i(2y+1)}{2x - i(2y+1)}$

⇒ $\dfrac{2x(x-1)-(x-1)(2y + 1)i+2xyi +y(2y+1)}{4x^2 + (2y+1)^2}$

⇒ $Re(\dfrac{z-1}{2z + i}) = \dfrac{2x(x-1)+y(2y+1)}{4x^2 + (2y+1)^2} =1$

⇒ $\dfrac{2x(x-1)+y(2y+1)}{4x^2 + (2y+1)^2} =1$

⇒ 2x² – 2x + 2y² + y = 4x² + 4y² + 1 + 4y

⇒ 2x² + 2y² + 2x + 3y + 1 = 0

⇒ x² + y² + x + 3/2 y + 1/2 = 0

This is a equation of circle

g = -1/2, f = -3/2, c = 1/2

Centre of circle = [-1/2, -3/4]

Radius = √g² + f² – c

= √(-1/2)² + (-3/2)² – 1/2

⇒ $\sqrt{\frac{1}{4} + \frac{9}{16} -\frac{1}{2}}$

⇒ $\sqrt{\frac{4 + 9- 8}{16}}$

⇒ $\dfrac{\sqrt{5}}{4}$

Answer is (C) Circle whose diameter is √5/2

Question 1:- If z is complex number such that |z| ≥ 2 . then the minimum value of |z + 1/2|.

(a) is equal to 5/2

(b) lies int interval (1, 2)

(c) is strictly greater than 5/2

(d) is strictly greater than 3/2 but less than 5/2

Solution:- See full solution

Question 2:- Let complex numbers α and $1/\overline{\alpha}$ lies on circles $(x - x_0)^2 + (y-y_0)^2 = r^2$ and $(x - x_0)^2 + (y-y_0)^2 = 4r^2$ , respectively.

If $z_0 = x_0 + iy_0$ satisfies the equation $2|z_0|^2 = r^2 + 2$ , then |α| is equal to

(2013 Adv.)

Solution:- See full solution

Question 3:- If w = α + iβ where β ≠ 0 and z ≠ 1 satisfies that condition that $\left(\dfrac{w - \overline{w}z}{1 - z}\right)$ is purely real, then the set of values of z is