JEE Main 202329 Jan, Evening shiftMathematicsMediumComplex Numbers and Quadratic Equations264× since 2002Let α=8−14i,A={z∈c:αz−α‾z‾z2−(z‾)2−112i=1}\alpha = 8 - 14i,A = \left\{ {z \in c:{{\alpha z - \overline \alpha \overline z } \over {{z^2} - {{\left( {\overline z } \right)}^2} - 112i}}=1} \right\}α=8−14i,A={z∈c:z2−(z)2−112iαz−αz=1} and B={z∈c:∣z+3i∣=4}B = \left\{ {z \in c:\left| {z + 3i} \right| = 4} \right\}B={z∈c:∣z+3i∣=4}. Then ∑z∈A∩B(Rez−Imz)\sum\limits_{z \in A \cap B} {({\mathop{\rm Re}\nolimits} z - {\mathop{\rm Im}\nolimits} z)}z∈A∩B∑(Rez−Imz) is equal to ____________.Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution
