If Z1,Z2 are complex numbers such that ReZ1=Z1-1,ReZ2=Z2-1 and argz1-z2=π6, then Imz1+z2 is equal to

Let |z-1|=Re(z), Let z1=x1+iy1,z2=x2+iy2⇒(x-1)2+y2=x2⇒y2=2x-1∴x1,y1,x2,y2 lies on y2=2x-1⇒y12-y22=2x1-x2 Since argz1-z2=π6,y1-y2x1-x2=13Hence, y1−y2y1+y2x1−x2=2y1+y2=23Imz1+z2=23

If Z1,Z2 are complex numbers such that ReZ1=Z1-1,ReZ2=Z2-1 and argz1-z2=π6, then Imz1+z2 is equal to

Let |z-1|=Re(z), Let z1=x1+iy1,z2=x2+iy2⇒(x-1)2+y2=x2⇒y2=2x-1∴x1,y1,x2,y2 lies on y2=2x-1⇒y12-y22=2x1-x2 Since argz1-z2=π6,y1-y2x1-x2=13Hence, y1−y2y1+y2x1−x2=2y1+y2=23Imz1+z2=23

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$If Z_{1}, Z_{2} are complex numbers such that Re (Z_{1}) = |Z_{1} - 1|, Re (Z_{2}) = |Z_{2} - 1| and \arg (z_{1} - z_{2}) = \frac{π}{6}, then$ $Im (z_{1} + z_{2}) is equal to$

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$\frac{1}{\sqrt{3}}$

$\frac{\sqrt{3}}{2}$

$\frac{2}{\sqrt{3}}$

$2 \sqrt{3}$

answer is D.

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Detailed Solution

$\begin{array}{l} Let | z - 1 | = Re (z), Let z_{1} = x_{1} + i y_{1}, z_{2} = x_{2} + i y_{2} \\ \Rightarrow (x - 1)^{2} + y^{2} = x^{2} \\ \Rightarrow y^{2} = 2 x - 1 \\ ∴ (x_{1}, y_{1}), (x_{2}, y_{2}) lies on y^{2} = 2 x - 1 \\ \Rightarrow y_{1}^{2} - y_{2}^{2} = 2 (x_{1} - x_{2}) \\ Since \arg (z_{1} - z_{2}) = \frac{π}{6}, \frac{y_{1} - y_{2}}{x_{1} - x_{2}} = \frac{1}{\sqrt{3}} \end{array}$

Hence,

$\begin{matrix} \frac{(y_{1} - y_{2}) (y_{1} + y_{2})}{x_{1} - x_{2}} = 2 \\ y_{1} + y_{2} = 2 \sqrt{3} \\ Im (z_{1} + z_{2}) = 2 \sqrt{3} \end{matrix}$