If z_1,z_2 are complex numbers such that Re(z_1)=|z_1-1| and Re(z_2)=|z_2-1| and arg(z_1-z_2)= π 6, then Im(z_1+z_2) is equal to :

Let ${z}_{1}={x}_{1}+i{y}_{1}$z1​=x1​+iy1​ and ${z}_{2}={x}_{2}+i{y}_{2}$z2​=x2​+iy2​

Given $|{z}_{1}-1|=Re({z}_{1})$∣z1​−1∣=Re(z1​)

$$\Rightarrow {({x}_{1}-1)}^{2}+{y}_{1}^{2}={x}_{1}^{2}$$⇒(x1​−1)2+y12​=x12​

$$\Rightarrow {y}_{1}^{2}-2{x}_{1}+1=0..............(1)$$⇒y12​−2x1​+1=0..............(1)

$$|{z}_{2}-1|=Re({z}_{2})$$∣z2​−1∣=Re(z2​)

$$\Rightarrow {({x}_{2}-1)}^{2}+{y}_{2}^{2}={x}_{2}^{2}$$⇒(x2​−1)2+y22​=x22​

$$\Rightarrow {y}_{2}^{2}-2{x}_{2}+1=0...........(2)$$⇒y22​−2x2​+1=0...........(2)

Subtracting equation $(2)$(2) from equation $(1)$(1), we get

$${y}_{1}^{2}-{y}_{2}^{2}-2({x}_{1}-{x}_{2})=0$$y12​−y22​−2(x1​−x2​)=0

$$\Rightarrow ({y}_{1}-{y}_{2})({y}_{1}+{y}_{2})=2({x}_{1}-{x}_{2})$$⇒(y1​−y2​)(y1​+y2​)=2(x1​−x2​)

$$\Rightarrow {y}_{1}+{y}_{2}=2(\frac{{x}_{1}-{x}_{2}}{{y}_{1}-{y}_{2}})............(3)$$⇒y1​+y2​=2(y1​−y2​x1​−x2​​)............(3)

Also given, $arg({z}_{1}-{z}_{2})=\frac{\pi }{6}$arg(z1​−z2​)=6π​

$$\Rightarrow {\mathrm{tan}}^{-1}(\frac{{y}_{1}-{y}_{2}}{{x}_{1}-{x}_{2}})=\frac{\pi }{6}$$⇒tan−1(x1​−x2​y1​−y2​​)=6π​

$$\Rightarrow \frac{{y}_{1}-{y}_{2}}{{x}_{1}-{x}_{2}}=\frac{1}{\sqrt{3}}.........(4)$$⇒x1​−x2​y1​−y2​​=3​1​.........(4)

Putting equation $(4)$(4) in equation $(3)$(3), we get

$${y}_{1}+{y}_{2}=2\sqrt{3}\Rightarrow Im({z}_{1}+{z}_{2})=2\sqrt{3}$$y1​+y2​=23​⇒Im(z1​+z2​)=23​