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JEE MainMathematicsModerateComplex Numbers

If z_1, z_2, z_3, are three collinear points in the argand
plane such that
∣(z3−z1)(z3−z2)∣=                A2\left| \left( z_{3} - z_{1} \right)\left( z_{3} - z_{2} \right) \right| = {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A}^{2}and
z2−z3=λ⋅A2z¯1−z¯3z_{2} - z_{3} = \frac{\lambda \cdot A^{2}}{{\overset{¯}{z}}_{1} - {\overset{¯}{z}}_{3}}then
value of |λ| is

Show answer and solution
z1z2z3 are collinear so z_{1}z_{2}z_{3}\ are\ collinear\ so\
z3−z1∣z3−z1∣=z2−z3∣z2−z3∣⋅ei0 or z2−z3∣z2−z3∣⋅eiπ\frac{z_{3} - z_{1}}{\left| z_{3} - z_{1} \right|} = \frac{z_{2} - z_{3}}{\left| z_{2} - z_{3} \right|} \cdot e^{i0}\ or\ \frac{z_{2} - z_{3}}{\left| z_{2} - z_{3} \right|} \cdot e^{i\pi}
⇒z3−z1z¯3−z¯1=z2−z3z¯2−z¯3\Rightarrow \frac{z_{3} - z_{1}}{{\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{1}} = \frac{z_{2} - z_{3}}{{\overset{¯}{z}}_{2} - {\overset{¯}{z}}_{3}}
z3−z1=(z2−z3)(z¯3−z¯1)z¯2−z¯3z_{3} - z_{1} = \frac{\left( z_{2} - z_{3} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{1} \right)}{{\overset{¯}{z}}_{2} - {\overset{¯}{z}}_{3}}
 Now  ∣(z3−z1)(z3−z2)∣=A2\ Now\ \ \left| \left( z_{3} - z_{1} \right)\left( z_{3} - z_{2} \right) \right| = A^{2}
(z3−z1)(z3−z2)(z¯3−z¯1)(z¯3−z¯2)=A4\left( z_{3} - z_{1} \right)\left( z_{3} - z_{2} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{1} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{2} \right) = A^{4}
(z2−z3)(z¯3−z¯1)(z¯2−z¯3)⋅(z3−z2)(z¯3−z¯1)(z¯3−z¯2)=A4\frac{\left( z_{2} - z_{3} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{1} \right)}{\left( {\overset{¯}{z}}_{2} - {\overset{¯}{z}}_{3} \right)} \cdot \left( z_{3} - z_{2} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{1} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{2} \right) = A^{4}
(z2−z3)2(z¯1−z¯3)2=A4\left( z_{2} - z_{3} \right)^{2}\left( {\overset{¯}{z}}_{1} - {\overset{¯}{z}}_{3} \right)^{2} = A^{4}
z2−z3=A2z¯1−z¯3 so λ=1z_{2} - z_{3} = \frac{A^{2}}{{\overset{¯}{z}}_{1} - {\overset{¯}{z}}_{3}}\ so\ \lambda = 1

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

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