If z1 , z2 are complex numbers such that Re(z1) = |z1 -- 1|, Re(z2) = |z2 -- 1| , and arg(…

Question

If z1
, z2
 are complex numbers such that

Re(z1) = |z1 -- 1|, Re(z2) = |z2 -- 1| , and

arg(z1 - z2) = $${\pi  \over 6}$$,  then Im(z1
 + z2
) is equal to :

Accepted Answer

Let $${z_1} = {x_1} + i{y_1},\,{z_2} = {x_2} + i{y_2}$$

Given Re(z1) = |z1 -- 1|

$$ \therefore $$ x1 = |(x1 - 1) + iy1|

$$ \Rightarrow $$ x1 = $$\sqrt {{{\left( {{x_1} - 1} \right)}^2} + y_1^2} $$

$$ \Rightarrow $$ ${x_1}^2 = {({x_1} - 1)^2} + {y_1}^2$

$$ \Rightarrow {y_1}^2 - 2{x_1} + 1 = 0$$

Also given Re(z2) = |z2 -- 1|

$$ \therefore $$ x2 = |(x2 - 1) + iy2|

$$ \Rightarrow $$ ${x_2}^2 = {({x^2} - 1)^2} + {y_2}^2$

$$ \Rightarrow $$ $y_2^2 - 2{x_2} - 1 = 0$

Performing equation (1) - (2),

$$({y_1}^2 - {y_2}^2) + 2({x_2} - {x_1}) = 0$$

$$ \Rightarrow $$ $$({y_1} + {y_2})({y_1} - {y_2}) = 2({x_1} - {x_2})$$

$$ \Rightarrow $$ $${y_1} + {y_2} = 2\left( {{{{x_1} - {x_2}} \over {{y_1} - {y_2}}}} \right)$$

Now given, $$\arg ({z_1} - {z_2}) = {\pi  \over 6}$$

$$ \Rightarrow $$ $${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi  \over 6}$$

$$ \Rightarrow {{{y_1} - {y_2}} \over {{x_1} - {x_2}}} = {1 \over {\sqrt 3 }}$$

$$ \therefore $$ $${y_1} + {y_2} = 2\sqrt 3 $$

Solutions · 1