JEE MAIN - Mathematics (2021 - 27th August Evening Shift - No. 15)
Let z1 and z2 be two complex numbers such that $$\arg ({z_1} - {z_2}) = {\pi \over 4}$$ and z1, z2 satisfy the equation | z $$-$$ 3 | = Re(z). Then the imaginary part of z1 + z2 is equal to ___________.
Answer
6
Explanation
Let z1 = x1 + iy ; z2 = x2 + iy2
z1 $$-$$ z2 = (x1 $$-$$ x2) + i(y1 $$-$$ y2)
$$\therefore$$ $$\arg ({z_1} - {z_2}) = {\pi \over 4}$$ $$\Rightarrow$$ $${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}$$
$${y_1} - {y_2} = {x_1} - {x_2}$$ ....... (1)
$$|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2$$ .... (2)
$$|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2$$ .... (3)
sub (2) & (3)
$${({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2$$
$$({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})$$
$$ = ({x_1} - {x_2})({x_1} + {x_2})$$
$${x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6$$
z1 $$-$$ z2 = (x1 $$-$$ x2) + i(y1 $$-$$ y2)
$$\therefore$$ $$\arg ({z_1} - {z_2}) = {\pi \over 4}$$ $$\Rightarrow$$ $${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}$$
$${y_1} - {y_2} = {x_1} - {x_2}$$ ....... (1)
$$|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2$$ .... (2)
$$|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2$$ .... (3)
sub (2) & (3)
$${({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2$$
$$({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})$$
$$ = ({x_1} - {x_2})({x_1} + {x_2})$$
$${x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6$$
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