"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 :"

Question

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

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

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

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

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 :

Solution

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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 :

Solution

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More Complex Numbers and Quadratic Equations questions

Drill 25 more like these. Every day. Free.

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 :