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QJEEM21Ma2e2e: JEE Main 2021 Mathematics - Class 11 JEEM21Ma2e2e Solution: Let z 1 and z 2 be two complex numbers such that and z 1 , z 2…

Let z1 and z2 be two complex numbers such that arg⁡(z1−z2)=π4\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 ___________.

Step-by-Step Verified Solution

Let z1 = x1 + iy ; z2 = x2 + iy2

z1 −- z2 = (x1 −- x2) + i(y1 −- y2)

∴\therefore arg⁡(z1−z2)=π4\arg ({z_1} - {z_2}) = {\pi \over 4} ⇒\Rightarrow tan⁡−1(y1−y2x1−x2)=π4{\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}

y1−y2=x1−x2{y_1} - {y_2} = {x_1} - {x_2} ....... (1)

∣z1−3∣ =Re(z1)⇒(x1−3)2+y12=x12|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2 .... (2)

∣z2−3∣ =Re(z2)⇒(x2−3)2+y22=x22|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2 .... (3)

sub (2) & (3)

(x1−3)2−(x2−3)2+y12−y22=x12−x22{({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2

(x1−x2)(x1+x2−6)+(y1−y2)(y1+y2)({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})

=(x1−x2)(x1+x2)= ({x_1} - {x_2})({x_1} + {x_2})

x1+x2−6+y1+y2=x1+x2⇒y1+y2=6{x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6

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