Let z1 and z2 be two complex numbers such that arg⁡z1−z2=π4 and z1,z2 satisfy the equation |z−3|=Re⁡(z) Then the imaginary part of z1+z2 is equal to

z1=x1+iy1;z2=x2+iy2z1−z2=x1−x2+iy1−y2∴arg⁡z1−z2=π4⇒tan−1⁡y1−y2x1−x2=π4y1−y2=x1−x2 …..(1)z1−3=Re⁡z1⇒x1−32+y12=x12⋯(2)z2−3=Re⁡z2⇒x2−32+y22=x22 sub (2) and (3) x1−32−x2−32+y12−y22=x12−x22x1−x2x1+x2−6+y1−y2y1+y2=x1−x2x1+x2x1+x2−6+y1+y2=x1+x2⇒y1+y2=6

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$Let z_{1} and z_{2} be two complex numbers such that \arg (z_{1} - z_{2}) = \frac{π}{4} and z_{1}, z_{2} satisfy the equation | z - 3 | = Re (z) Then the imaginary part of z_{1} + z_{2} is equal to$

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answer is 6.

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Detailed Solution

$\begin{array}{l} z_{1} = x_{1} + i y_{1}; z_{2} = x_{2} + i y_{2} \\ z_{1} - z_{2} = (x_{1} - x_{2}) + i (y_{1} - y_{2}) \\ ∴ \arg (z_{1} - z_{2}) = \frac{π}{4} \Rightarrow \tan^{- 1} (\frac{y_{1} - y_{2}}{x_{1} - x_{2}}) = \frac{π}{4} \\ y_{1} - y_{2} = x_{1} - x_{2} \dots . . (1) \\ |z_{1} - 3| = Re (z_{1}) \Rightarrow {(x_{1} - 3)}^{2} + y_{1}^{2} = x_{1}^{2} \dots (2) \\ |z_{2} - 3| = Re (z_{2}) \Rightarrow {(x_{2} - 3)}^{2} + y_{2}^{2} = x_{2}^{2} \end{array}$
$\begin{array}{l} s u b (2) a n d (3) {(x_{1} - 3)}^{2} - {(x_{2} - 3)}^{2} + y_{1}^{2} - y_{2}^{2} = x_{1}^{2} - x_{2}^{2} \\ (x_{1} - x_{2}) (x_{1} + x_{2} - 6) + (y_{1} - y_{2}) (y_{1} + y_{2}) = (x_{1} - x_{2}) (x_{1} + x_{2}) \\ x_{1} + x_{2} - 6 + y_{1} + y_{2} = x_{1} + x_{2} \Rightarrow y_{1} + y_{2} = 6 \end{array}$